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Maths M.C.Q (1 Marks)

Complex Numbers · CBSE STD 11 Science (CBSE - English Medium). 147 questions.

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M.C.Q (1 Marks)

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147 questions · auto-graded multiple-choice test.

MCQ 11 Mark
Let $x=\alpha +\beta ,\,y=\alpha \omega +\beta {{\omega }^{2}},\,z=\alpha {{\omega }^{2}}+\beta \omega ,\,\omega $ is an imaginary cube root of unity. Product of xyz is  [Orissa JEE 2005]
  • A
     ${{\alpha }^{2}}+{{\beta }^{2}}$
  • B
     ${{\alpha }^{2}}-{{\beta }^{2}}$
  • C
     ${{\alpha }^{3}}+{{\beta }^{3}}$
  • ✓
     ${{\alpha }^{3}}-{{\beta }^{3}}$
Answer
Correct option: D.
 ${{\alpha }^{3}}-{{\beta }^{3}}$
D
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MCQ 21 Mark
If 1, $\omega ,\,{{\omega }^{2}}$ are the cube roots of unity then ${{\omega }^{2}}{{(1+\omega )}^{3}}-(1+{{\omega }^{2}})\omega =$ [Orissa JEE 2005]
  • A
     1
  • B
     -1
  • C
     i
  • ✓
     0
Answer
Correct option: D.
 0
D
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MCQ 31 Mark
If $\omega $ is a cube root of unity but not equal to 1 then minimum value of $|a+b\omega +c{{\omega }^{2}}|$ (where a, b, c are integers but not all equal) is [IIT Screening 2005]
  • A
     0
  • B
     $\frac{\sqrt{3}}{2}$
  • ✓
     1
  • D
     2
Answer
Correct option: C.
 1
C
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MCQ 41 Mark
If ${{\tan }^{-1}}(\alpha +i\beta )=x+iy,$ then x = [RPET 2002]
  • ✓
     $\frac{1}{2}{{\tan }^{-1}}\left( \frac{2\alpha }{1-{{\alpha }^{2}}-{{\beta }^{2}}} \right)$
  • B
     $\frac{1}{2}{{\tan }^{-1}}\left( \frac{2\alpha }{1+{{\alpha }^{2}}+{{\beta }^{2}}} \right)$
  • C
     ${{\tan }^{-1}}\left( \frac{2\alpha }{1-{{\alpha }^{2}}-{{\beta }^{2}}} \right)$
  • D
     None of these
Answer
Correct option: A.
 $\frac{1}{2}{{\tan }^{-1}}\left( \frac{2\alpha }{1-{{\alpha }^{2}}-{{\beta }^{2}}} \right)$
A
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MCQ 51 Mark
If $\tan (u+iv)=i$, then the value of v is [RPET 2001]
  • A
     0
  • ✓
     $\infty $
  • C
     1
  • D
     None of these
Answer
Correct option: B.
 $\infty $
B
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MCQ 61 Mark
$\cos (x+iy)$is equal to [RPET 2001]
  • A
     $\sin \,x\,\,\cosh \,y+i\,\cos \,x\,\,\sinh \,y$
  • B
     $\cos \,x\,\,\cosh \,y+i\,\sin \,x\,\,\sinh \,y$
  • ✓
     $\cos \,x\,\,\cosh \,y-i\,\sin \,x\,\,\sinh \,y$
  • D
     None of these
Answer
Correct option: C.
 $\cos \,x\,\,\cosh \,y-i\,\sin \,x\,\,\sinh \,y$
C
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MCQ 71 Mark
Which one is correct from the following [RPET 2001]
  • ✓
     $\sin (ix)=i\,\sinh \,x$
  • B
     $\cos (ix)=i\,\cosh \,x$
  • C
     $\sin (ix)=-i\,\sinh \,x$
  • D
     $\tan (ix)=-i\,\tanh \,x$
Answer
Correct option: A.
 $\sin (ix)=i\,\sinh \,x$
A
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MCQ 81 Mark
The imaginary part of $\cosh (\alpha +i\beta )$is [RPET 2000]
  • A
     $\cosh \,\alpha \,\,\cos \,\beta $
  • ✓
     $\sinh \,\alpha \,\,\sin \,\beta $
  • C
     $\cos \alpha \cosh \beta $
  • D
     $\cos \alpha \cos \beta $
Answer
Correct option: B.
 $\sinh \,\alpha \,\,\sin \,\beta $
B
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MCQ 91 Mark
$\cosh (\alpha +i\beta )-\cosh (\alpha -i\beta )$ is equal to [RPET 2000]
  • A
     $2\,\,\sinh \,\alpha \,\,\sinh \,\beta $
  • B
     $2\,\,\cosh \,\alpha \,\,\cosh \,\beta $
  • ✓
     $2i\,\,\sinh \,\alpha \,\,\sin \,\beta $
  • D
     $2\,\,\cosh \,\alpha \,\,\cos \,\beta $
Answer
Correct option: C.
 $2i\,\,\sinh \,\alpha \,\,\sin \,\beta $
C
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MCQ 111 Mark
If $\cos (u+iv)=\alpha +i\beta ,$ then ${{\alpha }^{2}}+{{\beta }^{2}}+1$ equals [RPET 1999]
  • A
     ${{\cos }^{2}}u+{{\sinh }^{2}}v$
  • B
     ${{\sin }^{2}}u+{{\cosh }^{2}}v$
  • ✓
     ${{\cos }^{2}}u+{{\cosh }^{2}}v$
  • D
     ${{\sin }^{2}}u+{{\sinh }^{2}}v$
Answer
Correct option: C.
 ${{\cos }^{2}}u+{{\cosh }^{2}}v$
C
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MCQ 121 Mark
$\sinh ix$ is [EAMCET 2002]
  • A
     $i\sin (ix)$
  • ✓
     $i\sin x$
  • C
     $-i\sin x$
  • D
     $\sin (ix)$
Answer
Correct option: B.
 $i\sin x$
B
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MCQ 131 Mark
The real part of  ${{\sin }^{-1}}({{e}^{i\theta }})$  is [RPET 1997]
  • ✓
     ${{\cos }^{-1}}(\sqrt{\sin \theta })$
  • B
     ${{\sinh }^{-1}}(\sqrt{\sin \theta })$
  • C
     ${{\sin }^{-1}}(\sqrt{\sin \theta })$
  • D
     ${{\sin }^{-1}}(\sqrt{\cos \theta })$
Answer
Correct option: A.
 ${{\cos }^{-1}}(\sqrt{\sin \theta })$
A
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MCQ 141 Mark
If $\omega $ is a complex cube root of unity, then the value of ${{\omega }^{99}}+{{\omega }^{100}}+{{\omega }^{101}}$ is [Pb. CET 2004]
  • A
     1
  • B
     - 1
  • C
     3
  • ✓
     0
Answer
Correct option: D.
 0
D
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MCQ 151 Mark
If $1,\,\omega ,\,{{\omega }^{2}}$ are the roots of unity, then ${{(1-2\omega +{{\omega }^{2}})}^{6}}$ is equal to [Pb. CET 2001]
  • ✓
     729
  • B
     246
  • C
     243
  • D
     81
Answer
Correct option: A.
 729
A
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MCQ 161 Mark
If $\omega =\frac{-1+\sqrt{3}i}{2}$then ${{(3+\omega +3{{\omega }^{2}})}^{4}}$= [Karnataka CET 2004; Pb. CET 2000]
  • A
     16
  • B
     -16
  • ✓
     16 $\omega $
  • D
     16${{\omega }^{2}}$
Answer
Correct option: C.
 16 $\omega $
C
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MCQ 171 Mark
If $1,\omega ,{{\omega }^{2}}$ are the cube roots of unity, then$\Delta =\left| \,\begin{matrix}    1\,\,\,\, & {{\omega }^{n}} & {{\omega }^{2n}}  \\    {{\omega }^{n}}\,\, & \,\,\,{{\omega }^{2n}}\,\, & 1  \\    {{\omega }^{2n}}\, & 1\,\, & {{\omega }^{n}}  \\ \end{matrix} \right|$= [AIEEE 2003]
  • ✓
     0
  • B
     1
  • C
     $\omega $
  • D
     ${{\omega }^{2}}$
Answer
Correct option: A.
 0
A
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MCQ 181 Mark
If $\omega $ is a complex cube root of unity, then$225+$${{(3\omega +8{{\omega }^{2}})}^{2}}$$+{{(3{{\omega }^{2}}+8\omega )}^{2}}=$ [EAMCET 2003]
  • A
     72
  • B
     192
  • C
     200
  • ✓
     248
Answer
Correct option: D.
 248
D
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MCQ 191 Mark
The value of (8)1/3 is [RPET 2003]
  • A
     $-1+i\sqrt{3}$
  • B
     $-1-i\sqrt{3}$
  • C
     2
  • ✓
     All of these
Answer
Correct option: D.
 All of these
D
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MCQ 201 Mark
. Which of the following is a fourth root of $\frac{1}{2}+\frac{i\sqrt{3}}{2}$ [Karnataka CET 2003]
  • A
     $cis\left( \frac{\pi }{2} \right)$
  • ✓
     $cis\left( \frac{\pi }{12} \right)$
  • C
     $cis\left( \frac{\pi }{6} \right)$
  • D
     $cis\left( \frac{\pi }{3} \right)$
Answer
Correct option: B.
 $cis\left( \frac{\pi }{12} \right)$
B
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MCQ 211 Mark
If $\omega $ is a non real cube root of unity, then $(a+b)$ $(a+b\omega )$ $(a+b{{\omega }^{2}})$ is [Kerala (Engg.) 2002]
  • ✓
     ${{a}^{3}}+{{b}^{3}}$
  • B
     ${{a}^{3}}-{{b}^{3}}$
  • C
     ${{a}^{2}}+{{b}^{2}}$
  • D
     ${{a}^{2}}-{{b}^{2}}$
Answer
Correct option: A.
 ${{a}^{3}}+{{b}^{3}}$
A
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MCQ 221 Mark
Find the value of ${{(1+2\omega +{{\omega }^{2}})}^{3n}}-{{(1+\omega +2{{\omega }^{2}})}^{3n}}=$ [UPSEAT 2002]
  • ✓
     0
  • B
     1
  • C
     $\omega $
  • D
     ${{\omega }^{2}}$
Answer
Correct option: A.
 0
A
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MCQ 231 Mark
If ${{\left( \frac{1+i\sqrt{3}}{1-i\sqrt{3}} \right)}^{n}}$ is an integer, then n is  [UPSEAT 2002]
  • A
     1
  • B
     2
  • ✓
     3
  • D
     4
Answer
Correct option: C.
 3
C
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MCQ 241 Mark
 If $\frac{1+\sqrt{3}\,i}{2}$ is a root of equation ${{x}^{4}}-{{x}^{3}}+x-1=0$  then its real roots are [EAMCET 2002]
  • A
     1, 1
  • B
     - 1, - 1
  • ✓
     1, - 1
  • D
     1, 2
Answer
Correct option: C.
 1, - 1
C
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MCQ 251 Mark
If $z+{{z}^{-1}}=1,\,\text{then }\,{{z}^{100}}+{{z}^{-100}}$ is equal to [UPSEAT 2001]
  • A
     i
  • B
     - i
  • C
     1
  • ✓
     - 1
Answer
Correct option: D.
 - 1
D
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MCQ 261 Mark
Let ${{\omega }_{n}}=\cos \left( \frac{2\pi }{n} \right)+i\,\sin \left( \frac{2\pi }{n} \right)\,,\,{{i}^{2}}=-1$, then $(x+y{{\omega }_{3}}+z{{\omega }_{3}}^{2})$ $(x+y{{\omega }_{3}}^{2}+z{{\omega }_{3}})$ is equal to [AMU 2001]
  • A
     0
  • B
     ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}$
  • ✓
     ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}-yz-zx-xy$$$
  • D
     ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}+yz+zx+xy$
Answer
Correct option: C.
 ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}-yz-zx-xy$$$
C
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MCQ 271 Mark
If $z=\frac{\sqrt{3}+i}{-2}$, then ${{z}^{69}}$ is equal to [RPET 2001]
  • A
     1
  • B
     - 1
  • ✓
     i
  • D
     - i
Answer
Correct option: C.
 i
C
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MCQ 281 Mark
If $1,\omega ,{{\omega }^{2}}$ are the cube roots of unity, then their product is [Karnataka CET 1999, 2001]
  • A
     0
  • B
     $\omega $
  • C
     - 1
  • ✓
     1
Answer
Correct option: D.
 1
D
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MCQ 291 Mark
The value of $(1-\omega +{{\omega }^{2}})\,{{(1-{{\omega }^{2}}+\omega )}^{6}}$, where $\omega ,{{\omega }^{2}}$ are cube roots of unity [DCE 2001]
  • A
     128$\omega $
  • B
     $-128{{\omega }^{2}}$
  • ✓
     $-128\omega $
  • D
     $128{{\omega }^{2}}$
Answer
Correct option: C.
 $-128\omega $
C
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MCQ 301 Mark
If cube root of 1 is $\omega $, then the value of ${{(3+\omega +3{{\omega }^{2}})}^{4}}$ is [MP PET 2001]
  • A
     0
  • B
     16
  • ✓
     $16\,\omega $
  • D
     $16\,{{\omega }^{2}}$
Answer
Correct option: C.
 $16\,\omega $
C
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MCQ 311 Mark
If $\pi /3$ is a complex root of the equation ${{z}^{3}}=1$, then $\omega +{{\omega }^{\left( \frac{1}{2}\,+\,\frac{3}{8}\,+\,\frac{9}{32}\,+\,\frac{27}{128}\,+... \right)}}$ is equal to [Roorkee 2000; AMU 2005]
  • ✓
    - 1
  • B
    0
  • C
    9
  • D
    i
Answer
Correct option: A.
- 1
A
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MCQ 321 Mark
$\frac{{{(-1+i\sqrt{3})}^{15}}}{{{(1-i)}^{20}}}+\frac{{{(-1-i\sqrt{3})}^{15}}}{{{(1+i)}^{20}}}$ is equal to [AMU 2000]
  • ✓
     - 64
  • B
     - 32
  • C
     - 16
  • D
     $\frac{1}{16}$
Answer
Correct option: A.
 - 64
A
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MCQ 331 Mark
If $\omega $ is an imaginary cube root of unity, ${{(1+\omega -{{\omega }^{2}})}^{7}}$equals [IIT 1998; MP PET 2000]
  • A
     $128\omega $
  • B
     $-128\omega $
  • C
     $128{{\omega }^{2}}$
  • ✓
     $-128{{\omega }^{2}}$
Answer
Correct option: D.
 $-128{{\omega }^{2}}$
D
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MCQ 341 Mark
${{\left( \frac{\sqrt{3}+i}{2} \right)}^{6}}+{{\left( \frac{i-\sqrt{3}}{2} \right)}^{6}}$is equal to [RPET 1997]
  • ✓
     $-2$
  • B
     0
  • C
     2
  • D
     1
Answer
Correct option: A.
 $-2$
A
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MCQ 351 Mark
If $\omega $ is an imaginary cube root of unity, then the value of  $\sin \,\left[ ({{\omega }^{10}}+{{\omega }^{23}})\,\pi -\frac{\pi }{4} \right]$ is        [IIT Screening 1994]
  • A
     $-\sqrt{3}/2$
  • B
     $-1/\sqrt{2}$
  • ✓
     $1/\sqrt{2}$
  • D
     $\sqrt{3}/2$
Answer
Correct option: C.
 $1/\sqrt{2}$
C
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MCQ 361 Mark
If  $\omega $ is the cube root of unity, then ${{(3+5\omega +3{{\omega }^{2}})}^{2}}$ + ${{(3+3\omega +5{{\omega }^{2}})}^{2}}$ = [MP PET 1999]
  • A
     4
  • B
     0
  • ✓
     - 4
  • D
     None of these
Answer
Correct option: C.
 - 4
C
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MCQ 371 Mark
If  $\alpha $ and $\beta $ are imaginary cube roots of unity, then the value of  ${{\alpha }^{4}}+{{\beta }^{28}}+\frac{1}{\alpha \beta }$,is [MP PET 1998]
  • A
     1
  • B
     $-1$
  • ✓
     0
  • D
     None of these
Answer
Correct option: C.
 0
C
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MCQ 381 Mark
${{\left( \frac{-1+i\sqrt{3}}{2} \right)}^{20}}+{{\left( \frac{-1-i\sqrt{3}}{2} \right)}^{20}}=$
  • A
     $20\sqrt{3}i$
  • B
     1
  • C
     $\frac{1}{{{2}^{19}}}$
  • ✓
     $-1$
Answer
Correct option: D.
 $-1$
D
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MCQ 391 Mark
If $\alpha $ is an imaginary cube root of unity, then for  $n\in N$,  the value of ${{\alpha }^{3n+1}}+{{\alpha }^{3n+3}}+{{\alpha }^{3n+5}}$ is [MP PET 1996; Pb. CET 2000]
  • A
     $-1$
  • ✓
     0
  • C
     1
  • D
     3
Answer
Correct option: B.
 0
B
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MCQ 401 Mark
If ${{z}_{1}},{{z}_{2}}{{z}_{3}},{{z}_{4}}$are the roots of the equation ${{z}^{4}}=1$, then the value of  $\sum\limits_{i=1}^{4}{z_{i}^{3}}$is  [Kurukshetra CEE 1996]
  • ✓
     0
  • B
     1
  • C
     $i$
  • D
     $1+i$
Answer
Correct option: A.
 0
A
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MCQ 411 Mark
The common roots of the equations ${{x}^{12}}-1=0$, ${{x}^{4}}+{{x}^{2}}+1=0$ are [EAMCET 1989]
  • A
     $\pm \omega $
  • B
     $\pm {{\omega }^{2}}$
  • ✓
     $\pm \omega ,\,\pm {{\omega }^{2}}$
  • D
     None of these
Answer
Correct option: C.
 $\pm \omega ,\,\pm {{\omega }^{2}}$
C
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MCQ 421 Mark
If $1,\omega ,{{\omega }^{2}}$ are three cube roots of unity, then ${{(a+b\omega +c{{\omega }^{2}})}^{3}}$ + ${{(a+b{{\omega }^{2}}+c\omega )}^{3}}$ is equal to, if $a+b+c=0$ [West Bengal JEE 1992]
  • ✓
     $27\,abc$
  • B
     0
  • C
     $3\,abc$
  • D
     None of these
Answer
Correct option: A.
 $27\,abc$
A
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MCQ 431 Mark
If ${{z}_{1}},{{z}_{2}},{{z}_{3}}......{{n}_{n}}$ are nth, roots of unity, then for $k=1,\,2,.....,n$
  • A
     $|{{z}_{k}}|=k|{{z}_{k+1}}|$
  • B
     $|{{z}_{k+1}}|=k|{{z}_{k}}|$
  • C
     $|{{z}_{k+1}}|\,=\,|{{z}_{k}}|+|{{z}_{k+1}}|$
  • ✓
     $|{{z}_{k}}|=|{{z}_{k+1}}|$
Answer
Correct option: D.
 $|{{z}_{k}}|=|{{z}_{k+1}}|$
D
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MCQ 441 Mark
If $\omega $ is an nth root of unity, other than unity, then the value of $1+\omega +{{\omega }^{2}}+...+{{\omega }^{n-1}}$ is     [Karnataka CET 1999]
  • ✓
     0
  • B
     1
  • C
     $-1$
  • D
     None of these
Answer
Correct option: A.
 0
A
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MCQ 451 Mark
If $n$ is a positive integer greater than unity and $z$ is a complex number satisfying the equation ${{z}^{n}}={{(z+1)}^{n}}$,  then
  • ✓
     $\operatorname{Re}(z)<0$
  • B
     $\operatorname{Re}(z)>0$
  • C
     $\operatorname{Re}(z)=0$
  • D
     None of these
Answer
Correct option: A.
 $\operatorname{Re}(z)<0$
A
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MCQ 461 Mark
Let $\Delta =\left| \,\begin{matrix}    1 & \omega  & 2{{\omega }^{2}}  \\    2 & 2{{\omega }^{2}} & 4{{\omega }^{3}}  \\    3 & 3{{\omega }^{3}} & 6{{\omega }^{4}}  \\ \end{matrix}\, \right|$ where $\omega $ is the cube root of unity, then
  • ✓
     $\Delta =0$
  • B
     $\Delta =1$
  • C
     $\Delta =2$
  • D
     $\Delta =3$
Answer
Correct option: A.
 $\Delta =0$
A
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MCQ 471 Mark
$(1-\omega +{{\omega }^{2}})(1-{{\omega }^{2}}+{{\omega }^{4}})(1-{{\omega }^{4}}+{{\omega }^{8}})...........$to $2n$ factors is [EAMCET 1988]
  • A
     ${{2}^{n}}$
  • ✓
     ${{2}^{2n}}$
  • C
     0
  • D
     1
Answer
Correct option: B.
 ${{2}^{2n}}$
B
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MCQ 481 Mark
If  $1,\omega ,{{\omega }^{2}}$ are the three cube roots of unity, then ${{(3+{{\omega }^{2}}+{{\omega }^{4}})}^{6}}=$ [MP PET 1995]
  • ✓
     64
  • B
     729
  • C
     2
  • D
     0
Answer
Correct option: A.
 64
A
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MCQ 491 Mark
The ${{n}^{th}}$roots of unity are in [Orissa JEE 2004]
  • A
     A.P.
  • ✓
     G.P.
  • C
     H.P.
  • D
     None of these
Answer
Correct option: B.
 G.P.
B
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MCQ 501 Mark
If $\omega (\ne 1)$ is a cube root of unity, then $\left| \begin{matrix}    1 & 1+i+{{\omega }^{2}} & {{\omega }^{2}}  \\    1-i & -1 & {{\omega }^{2}}-1  \\    -i & -i+\omega -1 & -1  \\ \end{matrix} \right|$ is equal to [IIT 1995]
  • ✓
     0
  • B
     1
  • C
     $\omega $
  • D
     $i$
Answer
Correct option: A.
 0
A
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MCQ 511 Mark
If  $\omega (\ne 1)$is a cube root of unity and ${{(1+\omega )}^{7}}=A+B\omega $,  then $A$ and $B$ are respectively, the numbers [IIT 1995]
  • A
     0, 1
  • B
     1, 0
  • ✓
     1, 1
  • D
     $-1,\ 1$
Answer
Correct option: C.
 1, 1
C
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MCQ 521 Mark
One of the cube roots of unity is [MP PET 1994, 2003]
  • ✓
     $\frac{-1+i\sqrt{3}}{2}$
  • B
     $\frac{1+i\sqrt{3}}{2}$
  • C
     $\frac{1-i\sqrt{3}}{2}$
  • D
     $\frac{\sqrt{3}-i}{2}$
Answer
Correct option: A.
 $\frac{-1+i\sqrt{3}}{2}$
A
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MCQ 531 Mark
If  $\omega $ is a complex cube root of unity, then for positive integral value of$n$, the product of $\omega .{{\omega }^{2}}.{{\omega }^{3}}........{{\omega }^{n}}$, will be [Roorkee 1991]
  • A
     $\frac{1-i\sqrt{3}}{2}$
  • B
     $-\frac{1-i\sqrt{3}}{2}$
  • C
     1
  • ✓
     (b) and (c) both
Answer
Correct option: D.
 (b) and (c) both
D
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MCQ 541 Mark
The roots of the equation ${{x}^{4}}-1=0$,  are [MP PET 1986]
  • A
     $1,\,1,i,-i$
  • ✓
     $1,\,-1,i,-i$
  • C
     $1,-1,\omega ,{{\omega }^{2}}$
  • D
     None of these
Answer
Correct option: B.
 $1,\,-1,i,-i$
B
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MCQ 551 Mark
If  $z=\frac{\sqrt{3}+i}{2}$, then the value of  ${{z}^{69}}$ is [RPET 2002]
  • ✓
     $-i$
  • B
     $i$
  • C
     1
  • D
     $-1$
Answer
Correct option: A.
 $-i$
A
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MCQ 561 Mark
If  $\alpha ,\beta ,\gamma $ are the cube roots of  $p(p<0)$, then for any $x,y$ and $z,\,\,\frac{x\alpha +y\beta +z\gamma }{x\beta +y\gamma +z\alpha }=$ [IIT 1989]
  • ✓
     $\frac{1}{2}(-1+i\sqrt{3})$
  • B
     $\frac{1}{2}(1+i\sqrt{3})$
  • C
     $\frac{1}{2}(1-i\sqrt{3})$
  • D
     None of these
Answer
Correct option: A.
 $\frac{1}{2}(-1+i\sqrt{3})$
A
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MCQ 571 Mark
${{\left( -\frac{1}{2}+\frac{\sqrt{3}}{2}i \right)}^{1000}}=$
  • A
     $\frac{1}{2}+\frac{\sqrt{3}}{2}i$
  • B
     $\frac{1}{2}-\frac{\sqrt{3}}{2}i$
  • ✓
     $-\frac{1}{2}+\frac{\sqrt{3}}{2}i$
  • D
     None of these
Answer
Correct option: C.
 $-\frac{1}{2}+\frac{\sqrt{3}}{2}i$
C
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MCQ 581 Mark
The cube roots of unity when represented on the Argand plane form the vertices of an       [IIT 1988; Pb. CET 2004]
  • ✓
     Equilateral triangle
  • B
     Isosceles triangle
  • C
     Right angled triangle
  • D
     None of these
Answer
Correct option: A.
 Equilateral triangle
A
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MCQ 591 Mark
The value of $\frac{a+b\omega +c{{\omega }^{2}}}{b+c\omega +a{{\omega }^{2}}}+\frac{a+b\omega +c{{\omega }^{2}}}{c+a\omega +b{{\omega }^{2}}}$ will be [BIT Ranchi 1989; Orissa JEE 2003]
  • A
     1
  • ✓
     - 1
  • C
     2
  • D
     - 2
Answer
Correct option: B.
 - 1
B
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MCQ 601 Mark
If $x=a+b,y=a\omega +b{{\omega }^{2}},z=a{{\omega }^{2}}+b\omega $,  then the value of ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}$ is equal to        [Roorkee 1977; IIT 1970]
  • A
     ${{a}^{3}}+{{b}^{3}}$
  • ✓
     $3({{a}^{3}}+{{b}^{3}})$
  • C
     $3({{a}^{2}}+{{b}^{2}})$
  • D
     None of these
Answer
Correct option: B.
 $3({{a}^{3}}+{{b}^{3}})$
B
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MCQ 611 Mark
If $x=a+b,y=a\alpha +b\beta $ and $z=a\beta +b\alpha ,$ where $\alpha $and $\beta $ are complex cube roots of unity, then $xyz$ = [IIT 1978; Roorkee 1989; RPET 1997]
  • A
     ${{a}^{2}}+{{b}^{2}}$
  • ✓
     ${{a}^{3}}+{{b}^{3}}$
  • C
     ${{a}^{3}}{{b}^{3}}$
  • D
     ${{a}^{3}}-{{b}^{3}}$
Answer
Correct option: B.
 ${{a}^{3}}+{{b}^{3}}$
B
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MCQ 621 Mark
If  $\omega $ is a cube root of unity, then a root of the equation $\left| \begin{matrix}    x+1 & \omega  & {{\omega }^{2}}  \\    \omega  & x+{{\omega }^{2}} & 1  \\    {{\omega }^{2}} & 1 & x+\omega   \\ \end{matrix} \right|=0$ is   [MNR 1990; MP PET 1999]
  • A
     $x=1$
  • B
     $x=\omega $
  • C
     $x={{\omega }^{2}}$
  • ✓
     $x=0$
Answer
Correct option: D.
 $x=0$
D
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MCQ 631 Mark
The product of all the roots of ${{\left( \cos \frac{\pi }{3}+i\sin \frac{\pi }{3} \right)}^{3/4}}$  is  [MNR 1984; EAMCET 1985]
  • A
     $-1$
  • ✓
     1
  • C
     $\frac{3}{2}$
  • D
     $-\frac{1}{2}$
Answer
Correct option: B.
 1
B
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MCQ 641 Mark
If  $\omega $ is a complex cube root of unity, then $(1+\omega )(1+{{\omega }^{2}})$ $(1+{{\omega }^{4}})(1+{{\omega }^{8}})...$to $2n$ factors = [AMU 2000]
  • A
      0
  • ✓
     1
  • C
     $-1$
  • D
     None of these
Answer
Correct option: B.
 1
B
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MCQ 651 Mark
If $\omega $ is a complex cube root of unity, then $(x-y)(x\omega -y)$ $(x{{\omega }^{2}}-y)=$
  • A
     ${{x}^{2}}+{{y}^{2}}$
  • B
     ${{x}^{2}}-{{y}^{2}}$
  • ✓
     ${{x}^{3}}-{{y}^{3}}$
  • D
     ${{x}^{3}}+{{y}^{3}}$
Answer
Correct option: C.
 ${{x}^{3}}-{{y}^{3}}$
C
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MCQ 661 Mark
If  $x=a,y=b\omega ,z=c{{\omega }^{2}}$, where $\omega $ is a complex cube root of unity, then $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=$ [AMU 1983]
  • A
     3
  • B
     1
  • ✓
     0
  • D
     None of these
Answer
Correct option: C.
 0
C
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MCQ 671 Mark
If $\omega $ is a cube root of unity, then the value of ${{(1-\omega +{{\omega }^{2}})}^{5}}+{{(1+\omega -{{\omega }^{2}})}^{5}}=$ [IIT 1965; MP PET 1997; RPET 1997]
  • A
     16
  • ✓
     32
  • C
     48
  • D
     - 32
Answer
Correct option: B.
 32
B
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MCQ 681 Mark
If w is a complex cube root of unity, then $(1-\omega )(1-{{\omega }^{2}})$ $(1-{{\omega }^{4}})(1-{{\omega }^{8}})=$
  • A
     0
  • B
     1
  • C
     - 1
  • ✓
     9
Answer
Correct option: D.
 9
D
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MCQ 691 Mark
If $\alpha $and $\beta $ are imaginary cube roots of unity, then ${{\alpha }^{4}}+{{\beta }^{4}}$ + $\frac{1}{\alpha \beta }=$ [IIT 1977]
  • A
     3
  • ✓
     0
  • C
     1
  • D
     2
Answer
Correct option: B.
 0
B
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MCQ 701 Mark
If $\omega $ is a cube root of unity, then ${{(1+\omega )}^{3}}-{{(1+{{\omega }^{2}})}^{3}}=$
  • ✓
     0
  • B
     $\omega $
  • C
     ${{\omega }^{2}}$
  • D
     None of these
Answer
Correct option: A.
 0
A
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MCQ 711 Mark
Square of either of the two imaginary cube roots of unity will be
  • A
     Real root of unity
  • ✓
     Other imaginary cube root of unity
  • C
     Sum of two imaginary roots of unity
  • D
     None of these
Answer
Correct option: B.
 Other imaginary cube root of unity
B
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MCQ 721 Mark
If $n$ is a positive integer not a multiple of 3, then $1+{{\omega }^{n}}+{{\omega }^{2n}}$ = [MP PET 2004]
  • A
     3
  • B
     1
  • ✓
     0
  • D
     None of these
Answer
Correct option: C.
 0
C
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MCQ 731 Mark
${{(27)}^{1/3}}=$
  • A
     3
  • B
     $3,\,\,3i,\,3{{i}^{2}}$
  • ✓
     $3,\,3\omega ,\,3{{\omega }^{2}}$
  • D
     None of these
Answer
Correct option: C.
 $3,\,3\omega ,\,3{{\omega }^{2}}$
C
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MCQ 741 Mark
If $\omega $ is a cube root of unity, then $(1+\omega -{{\omega }^{2}})$ $(1-\omega +{{\omega }^{2}})$ =                 [MNR 1990; MP PET 1993, 2002]
  • A
     1
  • B
     0
  • C
     2
  • ✓
     4
Answer
Correct option: D.
 4
D
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MCQ 751 Mark
The two numbers such that each one is square of the other, are [MP PET 1987]
  • A
     $\omega ,\,{{\omega }^{3}}$
  • B
     $-i,\,\,i$
  • C
     $-1,\,1$
  • ✓
     $\omega ,\,\,{{\omega }^{2}}$
Answer
Correct option: D.
 $\omega ,\,\,{{\omega }^{2}}$
D
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MCQ 761 Mark
If $i{{z}^{4}}+1=0$, then $z$ can take the value [UPSEAT 2004]
  • A
     $\frac{1+i}{\sqrt{2}}$
  • ✓
     $\cos \frac{\pi }{8}+i\,\sin \frac{\pi }{8}$
  • C
     $\frac{1}{4i}$
  • D
     i
Answer
Correct option: B.
 $\cos \frac{\pi }{8}+i\,\sin \frac{\pi }{8}$
B
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MCQ 771 Mark
If $\frac{1}{x}+x=2\cos \theta ,$ then ${{x}^{n}}+\frac{1}{{{x}^{n}}}$ is equal to [UPSEAT 2001]
  • ✓
     $2\cos n\theta $
  • B
     $2\sin n\theta $
  • C
     $\cos n\,\theta $
  • D
     $\sin \,n\theta $
Answer
Correct option: A.
 $2\cos n\theta $
A
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MCQ 781 Mark
If n is a positive integer, then ${{(1+i)}^{n}}+{{(1-i)}^{n}}$ is equal to [Orissa JEE 2003]
  • A
     ${{(\sqrt{2})}^{n-2}}\cos \left( \frac{n\pi }{4} \right)$
  • B
     ${{(\sqrt{2})}^{n-2}}\sin \left( \frac{n\pi }{4} \right)$
  • ✓
     ${{(\sqrt{2})}^{n+2}}\cos \left( \frac{n\pi }{4} \right)$
  • D
     ${{(\sqrt{2})}^{n+2}}\sin \left( \frac{n\pi }{4} \right)$
Answer
Correct option: C.
 ${{(\sqrt{2})}^{n+2}}\cos \left( \frac{n\pi }{4} \right)$
C
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MCQ 791 Mark
${{\left( \frac{1+\sin \theta +i\,\cos \theta }{1+\sin \theta -i\,\cos \theta } \right)}^{n}}$= [Kerala (Engg.) 2002]
  • ✓
     $\cos \left( \frac{n\pi }{2}-n\theta  \right)+i\,\sin \left( \frac{n\pi }{2}-n\theta  \right)$
  • B
     $\cos \left( \frac{n\pi }{2}+n\theta  \right)+i\,\sin \left( \frac{n\pi }{2}+n\theta  \right)$
  • C
     $\sin \left( \frac{n\pi }{2}-n\theta  \right)+i\,\cos \left( \frac{n\pi }{2}-n\theta  \right)$
  • D
     $\cos \,n\left( \frac{\pi }{2}+2\theta  \right)+i\,\sin \,n\left( \frac{\pi }{2}+2\theta  \right)$
Answer
Correct option: A.
 $\cos \left( \frac{n\pi }{2}-n\theta  \right)+i\,\sin \left( \frac{n\pi }{2}-n\theta  \right)$
A
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MCQ 801 Mark
Given $z={{(1+i\sqrt{3})}^{100}},$ then $\frac{\operatorname{Re}(z)}{\operatorname{Im}(z)}$ equals [AMU 2002]
  • A
     2100
  • B
     250
  • ✓
     $\frac{1}{\sqrt{3}}$
  • D
     $\sqrt{3}$
Answer
Correct option: C.
 $\frac{1}{\sqrt{3}}$
C
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MCQ 811 Mark
The value of i1/3 is [UPSEAT 2002]
  • ✓
     $\frac{\sqrt{3}\,+i}{2}$
  • B
     $\frac{\sqrt{3}\,-i}{2}$
  • C
     $\frac{1+i\sqrt{3}}{2}$
  • D
     $\frac{1-i\sqrt{3}}{2}$
Answer
Correct option: A.
 $\frac{\sqrt{3}\,+i}{2}$
A
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MCQ 821 Mark
$\frac{{{(\cos \alpha +i\,\sin \alpha )}^{4}}}{{{(\sin \beta +i\,\cos \beta )}^{5}}}=$ [RPET 2002]
  • A
     $\cos (4\alpha +5\beta )+i\,\sin (4\alpha +5\beta )$
  • B
     $\cos (4\alpha +5\beta )-i\,\sin (4\alpha +5\beta )$
  • ✓
     $\sin (4\alpha +5\beta )-i\cos (4\alpha +5\beta )$
  • D
     None of these
Answer
Correct option: C.
 $\sin (4\alpha +5\beta )-i\cos (4\alpha +5\beta )$
C
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MCQ 831 Mark
 If ${{x}_{n}}=\cos \,\left( \frac{\pi }{{{4}^{n}}} \right)+i\,\sin \,\left( \frac{\pi }{{{4}^{n}}} \right)\,,$ then ${{x}_{1}}.\,{{x}_{2}}.\,{{x}_{3}}....\infty =$ [EAMCET 2002]
  • ✓
     $\frac{1+i\sqrt{3}}{2}$
  • B
     $\frac{-1+i\sqrt{3}}{2}$
  • C
     $\frac{1-i\sqrt{3}}{2}$
  • D
     $\frac{-1-i\sqrt{3}}{2}$
Answer
Correct option: A.
 $\frac{1+i\sqrt{3}}{2}$
A
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MCQ 841 Mark
${{\left[ \frac{1+\cos (\pi /8)+i\,\sin (\pi /8)}{1+\cos (\pi /8)-i\,\sin (\pi /8)} \right]}^{8}}$ is equal to [RPET 2001]
  • ✓
     - 1
  • B
     0
  • C
     1
  • D
     2
Answer
Correct option: A.
 - 1
A
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MCQ 851 Mark
The value of $\frac{(\cos \alpha +i\,\sin \alpha )\,(\cos \beta +i\,\sin \beta )}{(\cos \gamma +i\,\sin \gamma )\,(\cos \,\delta +i\,\sin \delta )}$ is [RPET 2001]
  • A
     $\cos (\alpha +\beta -\gamma -\delta )-i\,\sin (\alpha +\beta -\gamma -\delta )$
  • ✓
     $\cos (\alpha +\beta -\gamma -\delta )+i\,\sin (\alpha +\beta -\gamma -\delta )$
  • C
     $\sin (\alpha +\beta -\gamma -\delta )-i\,\cos (\alpha +\beta -\gamma -\delta )$
  • D
     $\sin (\alpha +\beta -\gamma -\delta )+i\,\cos (\alpha +\beta -\gamma -\delta )$
Answer
Correct option: B.
 $\cos (\alpha +\beta -\gamma -\delta )+i\,\sin (\alpha +\beta -\gamma -\delta )$
B
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MCQ 861 Mark
${{(\sin \theta +i\,\cos \theta )}^{n}}\,$is equal to [RPET 2001]
  • A
     $\cos n\theta +i\,\sin n\theta $
  • B
     $\sin n\theta +i\,\cos n\theta $
  • ✓
     $\cos n\left( \frac{\pi }{2}-\theta  \right)+i\,\sin n\left( \frac{\pi }{2}-\theta  \right)$
  • D
     None of these
Answer
Correct option: C.
 $\cos n\left( \frac{\pi }{2}-\theta  \right)+i\,\sin n\left( \frac{\pi }{2}-\theta  \right)$
C
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MCQ 871 Mark
We express $\frac{{{(\cos 2\theta -i\sin 2\theta )}^{4}}{{(\cos 4\theta +i\sin 4\theta )}^{-5}}}{{{(\cos 3\theta +i\sin 3\theta )}^{-2}}{{(\cos 3\theta -i\sin 3\theta )}^{-9}}}$  in the form of $x+iy$, we get [Karnataka CET 2001]
  • ✓
     $\cos 49\theta -i\,\sin 49\theta $
  • B
     $\cos 23\theta -i\,\sin 23\theta $
  • C
     $\cos 49\theta +i\,\sin 49\theta $
  • D
     $\cos 21\theta +i\,\sin 21\theta $
Answer
Correct option: A.
 $\cos 49\theta -i\,\sin 49\theta $
A
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MCQ 881 Mark
The value of ${{\left[ \frac{1-\cos \frac{\pi }{10}+i\sin \frac{\pi }{10}}{1-\cos \frac{\pi }{10}-i\sin \frac{\pi }{10}} \right]}^{10}}=$ [Karnataka CET 2001]
  • A
     0
  • ✓
     - 1
  • C
     1
  • D
     2
Answer
Correct option: B.
 - 1
B
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MCQ 891 Mark
${{(-\sqrt{3}+i)}^{53}}$ where ${{i}^{2}}=-1$ is equal to [AMU 2000]
  • A
     ${{2}^{53}}(\sqrt{3}+2i)$
  • B
       ${{2}^{52}}(\sqrt{3}-i)$
  • ✓
     ${{2}^{53}}\,\left( \frac{\sqrt{3}}{2}+\frac{1}{2}i \right)$
  • D
     ${{2}^{53}}(\sqrt{3}-i)$
Answer
Correct option: C.
 ${{2}^{53}}\,\left( \frac{\sqrt{3}}{2}+\frac{1}{2}i \right)$
C
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MCQ 901 Mark
If $\cos \alpha +\cos \beta +\cos \gamma =0=$$\sin \alpha +\sin \beta +\sin \gamma $ then $\cos 2\alpha +\cos 2\beta +\cos 2\gamma $ equals [RPET 2000]
  • A
     $2\cos (\alpha +\beta +\gamma )$
  • B
     $\cos 2(\alpha +\beta +\gamma )$
  • ✓
     0
  • D
     1
Answer
Correct option: C.
 0
C
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MCQ 911 Mark
If $\sin \alpha +\sin \beta +\sin \gamma =0=$$\cos \alpha +\cos \beta +\cos \gamma ,$ then the value of  ${{\sin }^{2}}\alpha +{{\sin }^{2}}\beta +{{\sin }^{2}}\gamma $ is    [RPET 1999]
  • A
     2/3
  • ✓
     3/2
  • C
     1/2
  • D
     1
Answer
Correct option: B.
 3/2
B
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MCQ 921 Mark
${{\left( \frac{\cos \theta +i\sin \theta }{\sin \theta +i\cos \theta } \right)}^{4}}$equals [RPET 1996]
  • A
     $\sin 8\theta -i\cos 8\theta $
  • B
     $\cos 8\theta -i\sin 8\theta $
  • C
     $\sin 8\theta +i\cos 8\theta $
  • ✓
     $\cos 8\theta +i\sin 8\theta $
Answer
Correct option: D.
 $\cos 8\theta +i\sin 8\theta $
D
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MCQ 931 Mark
The value of expression $\left( \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} \right)$ $\,\left( \cos \frac{\pi }{{{2}^{2}}}+i\sin \frac{\pi }{{{2}^{2}}} \right)$........to $\infty $ is    [Kurukshetra CEE 1998]
  • ✓
     $-1$
  • B
      $1$
  • C
     0
  • D
     2
Answer
Correct option: A.
 $-1$
A
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MCQ 941 Mark
If  ${{\left( \frac{1+\cos \theta +i\sin \theta }{i+\sin \theta +i\cos \theta } \right)}^{4}}=\cos n\theta +i\sin n\theta $, then $n$ is equal to [EAMCET 1986]
  • A
     1
  • B
     2
  • C
     3
  • ✓
     4
Answer
Correct option: D.
 4
D
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MCQ 951 Mark
${{\left( \frac{1+\cos \varphi +i\sin \varphi }{1+\cos \varphi -i\sin \varphi } \right)}^{n}}=$
  • A
     $\cos n\varphi -i\sin n\varphi $
  • ✓
     $\cos n\varphi +i\sin n\varphi $
  • C
     $\sin n\varphi +i\cos n\varphi $
  • D
     $\sin n\varphi -i\cos n\varphi $
Answer
Correct option: B.
 $\cos n\varphi +i\sin n\varphi $
B
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MCQ 961 Mark
If $(\cos \theta +i\sin \theta )(\cos 2\theta +i\sin 2\theta )........$ $(\cos n\theta +i\sin n\theta )=1$, then the value of $\theta $ is[Karnataka CET 1992; Kurukshetra CEE 2002]
  • A
     $4m\pi $
  • B
     $\frac{2m\pi }{n(n+1)}$
  • ✓
     $\frac{4m\pi }{n(n+1)}$
  • D
     $\frac{m\pi }{n(n+1)}$
Answer
Correct option: C.
 $\frac{4m\pi }{n(n+1)}$
C
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MCQ 971 Mark
If $a=\sqrt{2i}$ then which of the following is correct [Roorkee 1989]
  • ✓
     $a=1+i$
  • B
     $a=1-i$
  • C
     $a=-(\sqrt{2})i$
  • D
     None of these
Answer
Correct option: A.
 $a=1+i$
A
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MCQ 981 Mark
The following in the form of $A+iB$ ${{(\cos 2\theta +i\sin 2\theta )}^{-5}}$ ${{(\cos 3\theta -i\sin 3\theta )}^{6}}$${{(\sin \theta -i\cos \theta )}^{3}}$ in the form of $A+iB$ is [MNR 1991]
  • A
     $(\cos 25\theta +i\sin 25\theta )$
  • B
     $i(\cos 25\theta +i\sin 25\theta )$
  • C
     $i\,(\cos 25\theta -i\sin 25\theta )$
  • ✓
     $(\cos 25\theta -i\sin 25\theta )$
Answer
Correct option: D.
 $(\cos 25\theta -i\sin 25\theta )$
D
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MCQ 991 Mark
The value of $\frac{4(\cos {{75}^{o}}+i\sin {{75}^{o}})}{0.4(\cos {{30}^{o}}+i\sin {{30}^{o}})}$ is
  • A
     $\frac{\sqrt{2}}{10}(1+i)$
  • B
     $\frac{\sqrt{2}}{10}(1-i)$
  • C
     $\frac{10}{\sqrt{2}}(1-i)$
  • ✓
     $\frac{10}{\sqrt{2}}(1+i)$
Answer
Correct option: D.
 $\frac{10}{\sqrt{2}}(1+i)$
D
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MCQ 1001 Mark
The roots of ${{(2-2i)}^{1/3}}$ are
  • ✓
      $\sqrt{2}\left( \cos \frac{\pi }{12}-i\sin \frac{\pi }{12} \right),\sqrt{2}\left( -\sin \frac{\pi }{12}+i\cos \frac{\pi }{12} \right),-1-i$
  • B
      $\sqrt{2}\left( \cos \frac{\pi }{12}+i\sin \frac{\pi }{12} \right),\sqrt{2}\left( -\sin \frac{\pi }{12}-i\cos \frac{\pi }{12} \right)\,,\,1+i$
  • C
     $1+\sqrt{2}i,-1-i,-2-2i$
  • D
     None of the above
Answer
Correct option: A.
  $\sqrt{2}\left( \cos \frac{\pi }{12}-i\sin \frac{\pi }{12} \right),\sqrt{2}\left( -\sin \frac{\pi }{12}+i\cos \frac{\pi }{12} \right),-1-i$
A
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MCQ 1011 Mark
If  $z={{\left( \frac{\sqrt{3}}{2}+\frac{i}{2} \right)}^{5}}+{{\left( \frac{\sqrt{3}}{2}-\frac{i}{2} \right)}^{5}}$, then [MP PET 1997]
  • A
     $\operatorname{Re}(z)=0$
  • ✓
     $\operatorname{Im}(z)=0$
  • C
     $\operatorname{Re}(z)>0,\operatorname{Im}(z)>0$
  • D
     $\operatorname{Re}(z)>0,\operatorname{Im}(z)<0$
Answer
Correct option: B.
 $\operatorname{Im}(z)=0$
B
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MCQ 1021 Mark
$\frac{{{(\cos \theta +i\sin \theta )}^{4}}}{{{(\sin \theta +i\cos \theta )}^{5}}}$ is equal to   [MNR 1985; UPSEAT 2000]
  • A
     $\cos \theta -i\sin \theta $
  • B
     $\cos 9\theta -i\sin 9\theta $
  • C
     $\sin \theta -i\cos \theta $
  • ✓
     $\sin 9\theta -i\cos 9\theta $
Answer
Correct option: D.
 $\sin 9\theta -i\cos 9\theta $
D
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MCQ 1031 Mark
If  ${{x}_{r}}=\cos \left( \frac{\pi }{{{2}^{r}}} \right)+i\sin \left( \frac{\pi }{{{2}^{r}}} \right)$, then${{x}_{1}}.{{x}_{2}}......\infty $is [RPET 1990, 2000; BIT Mesra 1996; Karnataka CET 2000]
  • A
     $-3$
  • B
     $-2$
  • ✓
     $-1$
  • D
       0
Answer
Correct option: C.
 $-1$
C
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MCQ 1041 Mark
$\sqrt{i}=$
  • A
     $\frac{1\pm i}{\sqrt{2}}$
  • B
     $\pm \frac{1-i}{\sqrt{2}}$
  • ✓
     $\pm \frac{1+i}{\sqrt{2}}$
  • D
     None of these
Answer
Correct option: C.
 $\pm \frac{1+i}{\sqrt{2}}$
C
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MCQ 1051 Mark
The polar form of $(\text{i}^{25})^3$ is:
  • A
    $\cos\frac{\pi}{2}+\text{i}\sin\frac{\pi}{2}$
  • B
    $\cos\pi+\text{i}\sin\pi$
  • C
    $\cos\pi-\text{i}\sin\pi$
  • ✓
    $\cos\frac{\pi}{2}-\text{i}\sin\frac{\pi}{2}$
Answer
Correct option: D.
$\cos\frac{\pi}{2}-\text{i}\sin\frac{\pi}{2}$
$(\text{i}^{25})^3=(\text{i})^{75}$
$=(\text{i})^{4\times18+3}$
$=(\text{i})^3$
$=-\text{i} \ (\because\text{i}^4=1)$
Let $\text{z}=0-\text{i}$
Since, the point (0,−1) lies on the negative direction of imaginary axis.Therefore, $\text{arg(z)}=\frac{-\pi}{2}$
Modulus, $\text{r}=|\text{z}|=|1|=1$
$\therefore$ Polar form $=\text{r}(\cos\theta+\text{i}\sin\theta)$
$=\cos\Big(\frac{-\pi}{2}\Big)+\text{i}\sin\Big(\frac{-\pi}{2}\Big)$
$=\cos\frac{\pi}{2}-\text{i}\sin\frac{\pi}{2}$
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MCQ 1061 Mark
The argument of $\frac{1-\text{i}\sqrt{3}}{1+\text{i}\sqrt{3}}$ is:
  • A
    $60^\circ$
  • B
    $120^\circ$
  • C
    $210^\circ$
  • ✓
    $240^\circ$
Answer
Correct option: D.
$240^\circ$
$\frac{1-\text{i}\sqrt{3}}{1+\text{i}\sqrt{3}}$
Rationalising the denominator,
$\frac{1-\text{i}\sqrt{3}}{1+\text{i}\sqrt{3}}\times\frac{1-\text{i}\sqrt{3}}{1-\text{i}\sqrt{3}}$
$=\frac{1+3\text{i}^2-2\sqrt{3}\text{i}}{1-3\text{i}^2}$
$=\frac{-2-2\sqrt{3}\text{i}}{4} \ (\because\text{i}^2=-1)$
$=\frac{-1}{2}-\text{i}\frac{\sqrt{3}}{2}$
Then, $\Rightarrow\tan\alpha=\Big|\frac{\text{Im(z)}}{\text{Re(z)}}\Big|$
$=\sqrt{3}$
$\Rightarrow\alpha=60^\circ$
Since the points $(\frac{-1}{2},-\frac{-\sqrt{3}}{2})$ lie in the third quadrant, the argument is given by:
$\theta=180^\circ+60^\circ$
$=240^\circ$
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MCQ 1071 Mark
If $\text{a}=\cos\theta+\text{i}\sin\theta,$ then $\frac{1+\text{a}}{1-\text{a}}=$
  • A
    $\cot\frac{\theta}{2}$
  • B
    $\cot\theta$
  • ✓
    $\text{i}\cot\frac{\theta}{2}$
  • D
    $\text{i}\tan\frac{\theta}{2}$
Answer
Correct option: C.
$\text{i}\cot\frac{\theta}{2}$
$\text{a}=\cos\theta+\text{i}\sin\theta$ (given)
$\Rightarrow\frac{1+\text{a}}{1-\text{a}}=\frac{1+\cos\theta+\text{i}\sin\theta}{1-\cos\theta-\text{i}\sin\theta}$
$\Rightarrow\frac{1+\text{a}}{1-\text{a}}=\frac{1+\cos\theta+\text{i}\sin\theta}{1-\cos\theta-\text{i}\sin\theta}\times\frac{1-\cos\theta+\text{i}\sin\theta}{1-\cos\theta+\text{i}\sin\theta}$
$\Rightarrow\frac{1+\text{a}}{1-\text{a}}=\frac{(1+\text{i}\sin\theta)^2-\cos^2\theta}{(1-\cos\theta)^2-(\text{i}\sin\theta)^2}$
$\Rightarrow\frac{1+\text{a}}{1-\text{a}}=\frac{1-\text{i}\sin^2\theta+2\text{i}\sin\theta-\cos^2\theta}{1+\cos^2\theta-2\cos\theta+\sin^2\theta}$
$\Rightarrow\frac{1+\text{a}}{1-\text{a}}=\frac{1-(\sin^2\theta+\cos^2\theta)+2\text{i}\sin\theta}{1+(\sin^2\theta+\cos^2\theta)-2\cos\theta}$
$\Rightarrow\frac{1+\text{a}}{1-\text{a}}=\frac{2\text{i}\sin\theta}{2(1-\cos\theta)}$
$\Rightarrow\frac{1+\text{a}}{1-\text{a}}=\frac{2\text{i}\sin\frac{\theta}{2}-\cos\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}}$
$\Rightarrow\frac{1+\text{a}}{1-\text{a}}=\frac{\text{i}\cos\frac{\theta}{2}}{\sin\frac{\theta}{2}}$
$\Rightarrow\frac{1+\text{a}}{1-\text{a}}=\text{i}\cot\frac{\theta}{2}$
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MCQ 1081 Mark
The complex number $z$ which satisfies the condition $\Big|\frac{\text{i}+\text{z}}{\text{i}-\text{z}}\Big|=1$ lies on:
  • A
    Circle $x^2+ y^2 = 1$
  • ✓
    The $x-$axis
  • C
    The $y-$axis
  • D
    The line $x + y = 1$
Answer
Correct option: B.
The $x-$axis
$\Big|\frac{\text{i}+\text{z}}{\text{i}-\text{z}}\Big|=1$
$\Rightarrow \Big|\frac{\text{i}+\text{z}}{\text{i}-\text{z}}\Big|^2=1^2$
$\Rightarrow\Big(\frac{\text{i}+\text{z}}{\text{i}-\text{z}}\Big)\overline{\Big(\frac{\text{i}+\text{z}}{\text{i}-\text{z}}\Big)}=1$
$\Rightarrow\Big(\frac{\text{i}+\text{z}}{\text{i}-\text{z}}\Big)\Big(\frac{-\text{i}+\overline{\text{z}}}{-\text{i}-\overline{\text{z}}}\Big)=1$
$\Rightarrow\Big(\frac{-\text{i}^2-\text{zi}+\overline{\text{z}}\text{i}+\text{z}\overline{\text{z}}}{-\text{i}^2+\text{z}\text{i}-\overline{\text{z}}\text{i}+\text{z}\overline{\text{z}}}\Big)=1$
$\Rightarrow-\text{i}^2-\text{zi}+\overline{\text{z}}\text{i}+\text{z}\overline{\text{z}}=-\text{i}^2+\text{zi}-\overline{\text{z}}\text{i}+\text{z}\overline{\text{z}}$
$\Rightarrow-\text{zi}+\overline{\text{z}}\text{i}=\text{zi}-\overline{\text{z}}\text{i}$
$\Rightarrow\overline{\text{z}}\text{i}+\overline{\text{z}}\text{i}=\text{zi}-\text{zi}$
$\Rightarrow2\overline{\text{z}}\text{i}=2\text{zi}$
$\Rightarrow\overline{\text{z}}=\text{z}$
$ \Rightarrow\text{z}$ is purely real.
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MCQ 1091 Mark
If z is a complex number, then:
  • A
    $|\text{z}|^2>|\text{z}|^2$
  • ✓
    $|\text{z}|^2=|\text{z}|^2$
  • C
    $|\text{z}|^2<|\text{z}|^2$
  • D
    $|\text{z}|^2\geq|\text{z}|^2$
Answer
Correct option: B.
$|\text{z}|^2=|\text{z}|^2$
It is obvious that, for any complex number z,
$|\text{z}|^2=|\text{z}|^2$
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MCQ 1101 Mark
If $\text{z}=1-\cos\theta+\text{i}\sin\theta,$ then $|\text{z}|=$
  • A
    $2\sin\frac{\theta}{2}$
  • B
    $2\cos\frac{\theta}{2}$
  • ✓
    $2\Big|\sin\frac{\theta}{2}\Big|$
  • D
    $2\Big|\cos\frac{\theta}{2}\Big|$
Answer
Correct option: C.
$2\Big|\sin\frac{\theta}{2}\Big|$
$\therefore\text{z}=1-\cos\theta+\text{i}\sin\theta$
$\Rightarrow|\text{z}|=\sqrt{(1-\cos\theta)^2+\sin^2\theta}$
$\Rightarrow|\text{z}|=\sqrt{1+\cos^2\theta-2\cos\theta+\sin^2\theta}$
$\Rightarrow|\text{z}|=\sqrt{1+1-2\cos\theta}$
$\Rightarrow|\text{z}|=\sqrt{2(1-2\cos\theta)}$
$\Rightarrow|\text{z}|=\sqrt{4\sin^2\frac{\theta}{2}}$
$\Rightarrow|\text{z}|=2\Big|\sin\frac{\theta}{2}\Big|$
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MCQ 1111 Mark
If $\text{x}+\text{iy}=(1+\text{i})(1+2\text{i})(1+3\text{i}),$ then $\text{x}^2+\text{y}^2=$
  • A
    0
  • B
    1
  • ✓
    100
  • D
    none of these
Answer
Correct option: C.
100
$\therefore\text{x}+\text{iy}=(1+\text{i})(1+2\text{i})(1+3\text{i})$
Taking modulus on both the sides:
$|\text{x}+\text{iy}|=|(1+\text{i})(1+2\text{i})(1+3\text{i})|$
$\Rightarrow|\text{x}+\text{iy}|=|(1+\text{i})|\times|(1+2\text{i})|\times|(1+3\text{i})|$
$\Rightarrow\sqrt{\text{x}^2+\text{y}^2}=\sqrt{1^2+1^2}\sqrt{1^2+2^2}\sqrt{1^2+3^2}$
$\Rightarrow\sqrt{\text{x}^2+\text{y}^2}=\sqrt{2}\sqrt{5}\sqrt{10}$
$\Rightarrow\sqrt{\text{x}^2+\text{y}^2}=\sqrt{100}$
Squaring both the sides,
$\Rightarrow\text{x}^2+\text{y}^2=100$
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MCQ 1121 Mark
If $\text{i}^2=-1,$ then the sum $\text{i}+\text{i}^2+\text{i}^3+...$ upto 1000 terms is equal to:
  • A
    1
  • B
    -1
  • C
    i
  • ✓
    0
Answer
Correct option: D.
0
$\text{i}+\text{i}^2+\text{i}^3+\text{i}^4...\text{i}^{1000}$
$\text{i}+\text{i}^2+\text{i}^3+\text{i}^4 \ [\because\text{i}^2=-1,\text{i}^3=-\text{i} \ \text{and} \ \text{i}^4=1]$
$=\text{i}-1-\text{i}+1$
$=0$
Similarly, the sum of the next four terms of the series will be equal to 0. This is because the powers of i follow a cyclicity of 4. Hence, the sum of all terms, till 1000, will be zero.
$\text{i}+\text{i}^2+\text{i}^3+\text{i}^4...\text{i}^{1000}=0$
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MCQ 1131 Mark
$\big(\sqrt{-2}\big)\big(\sqrt{-3}\big)$ is equal to:
  • A
    $\sqrt{6}$
  • ✓
    $-\sqrt{6}$
  • C
    $\text{i}\sqrt{6}$
  • D
    none of these.
Answer
Correct option: B.
$-\sqrt{6}$
$\sqrt{-2}\times\sqrt{-3}$
$=\sqrt{2}\times\sqrt{3}\times\sqrt{-1}\times\sqrt{-1}$
$=\sqrt{6}\times\text{i}\times\text{i}$
$=\sqrt{6}\times\text{i}^2$
$=-\sqrt{6} \ [\because\text{i}^2=-1]$
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MCQ 1141 Mark
If $\frac{3+2\text{i}\sin\theta}{1-2\text{i}\sin\theta}$ is a real number and $0 < \theta < 2\pi,$ then $\theta=$
  • ✓
    $\pi$
  • B
    $\frac{\pi}{2}$
  • C
    $\frac{\pi}{3}$
  • D
    $\frac{\pi}{6}$
Answer
Correct option: A.
$\pi$
Given:
$\frac{3+2\text{i}\sin\theta}{1-2\text{i}\sin\theta}$ is a real number
On rationalising, we get,
$\frac{3+2\text{i}\sin\theta}{1-2\text{i}\sin\theta}\times\frac{{1+2\text{i}\sin\theta}}{{1+2\text{i}\sin\theta}}$
$=\frac{(3+2\text{i}\sin\theta)({1+2\text{i}\sin\theta})}{(1)^2-(2\text{i}\sin\theta)^2}$
$=\frac{3+2\text{i}\sin\theta+{6\text{i}\sin\theta}+4\text{i}^2\sin^2\theta}{1+4\sin^2\theta}$
$=\frac{3-4\text{i}\sin^2\theta+{8\text{i}\sin\theta}}{1+4\sin^2\theta} \ [\because\text{i}^2=-1]$
$=\frac{3-4\text{i}\sin^2\theta}{1+4\sin^2\theta}+\text{i}\frac{8\sin\theta}{1+4\sin^2\theta}$
For the above term to be real, the imaginary part has to be zero.
$\therefore\frac{8\sin\theta}{1+4\sin^2\theta}=0$
$\Rightarrow8\sin\theta=0$
For this to be zero,
$\sin\theta=0$
$\Rightarrow\theta=0,\pi,2\pi,3\pi...$
But $0<\theta<2\pi$
Hence, $\theta=\pi$
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MCQ 1151 Mark
The value of $(1+\text{i})(1+\text{i}^2)(1+\text{i}^3)(1+\text{i}^4)$ is
  • A
    2
  • ✓
    0
  • C
    1
  • D
    i
Answer
Correct option: B.
0
$(1+\text{i})(1+\text{i}^2)(1+\text{i}^3)(1+\text{i}^4)$
$=(1+\text{i})(1-1)(1-\text{i})(1+1) \ \big(\because\text{i}^2=-1, \beta=-\text{i and} \ \text{i}^4=1\big)$
$=(1+\text{i})(0)(1-\text{i})(2)$
$=0$
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MCQ 1161 Mark
If $\text{z}=\frac{-2}{1+\sqrt{3}},$ then the value of arg (z) is:
  • A
    $\pi$
  • B
    $\frac{\pi}{3}$
  • ✓
    $\frac{2\pi}{3}$
  • D
    $\frac{\pi}{4}$
Answer
Correct option: C.
$\frac{2\pi}{3}$
$\text{z}=\frac{-2}{1+\sqrt{3}}$
Rationalising z, we get,
$\text{z}=\frac{-2}{1+\text{i}\sqrt{3}}\times\frac{1-\text{i}\sqrt{3}}{1-\text{i}\sqrt{3}}$
$\Rightarrow\text{z}=\frac{-2+\text{i}2\sqrt{3}}{1+3}$
$\Rightarrow\text{z}=\frac{-1+\text{i}\sqrt{3}}{2}$
$\Rightarrow\text{z}=\frac{-1}{2}+\frac{\text{i}\sqrt{3}}{2}$
$\Rightarrow\tan\alpha=\Big|\frac{\text{Im(z)}}{\text{Re(z)}}\Big|$
$=\sqrt{3}$
$\Rightarrow\alpha=\frac{\pi}{3}$
Since, z lies in the second quadrant.
Therefore, $\text{arg(z)}=\pi-\frac{\pi}{3}$
$=\frac{2\pi}{3}$
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MCQ 1171 Mark
If $\sqrt{\text{a}+\text{ib}}=\text{x}+\text{iy},$ then possible value of $\sqrt{\text{a}-\text{ib}}$ is:
  • A
    $\text{x}^2+\text{y}^2$
  • B
    $\sqrt{\text{x}^2+\text{y}^2}$
  • C
    $\text{x}+\text{iy}$
  • ✓
    $\text{x}-\text{iy}$
Answer
Correct option: D.
$\text{x}-\text{iy}$
$\sqrt{\text{a}+\text{ib}}=\text{x}+\text{iy}$
Squaring on both the sides, we get,
$\text{a}+\text{ib}=\text{x}^2+(\text{iy})^2+2\text{ixy}$
$\Rightarrow\text{a}+\text{ib}=(\text{x}^2-\text{y}^2)+2\text{ixy}$
$\therefore\text{a}=(\text{x}^2-\text{y}^2)$
and $\text{b}=2\text{xy}$
$\therefore\text{a}-\text{ib}=(\text{x}^2-\text{y}^2)-2\text{ixy}$
$\Rightarrow\text{a}-\text{ib}=\text{x}^2+\text{i}^2\text{y}^2-2\text{ixy} \ [\because\text{i}^2=-1]$
Taking square root on both the sides, we get:
$\sqrt{\text{a}-\text{ib}}=\text{x}-\text{iy}$
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MCQ 1181 Mark
If the complex number $\text{z}=\text{x}+\text{iy}$ satisfies the condition $|\text{z}+1|=1,$ then z lies on:
  • A
    x−axis
  • ✓
    circle with centre (-1, 0) and radius 1
  • C
    y−axis
  • D
    none of these
Answer
Correct option: B.
circle with centre (-1, 0) and radius 1
$|\text{z}+1|=1$
$\Rightarrow|\text{z+1|}^2=1^2$
$\Rightarrow(\text{z}+1)\overline{(\text{z}+1)}=1$
$\Rightarrow(\text{z}+1)(\overline{z}+1)=1$
$\Rightarrow\text{z}\overline{\text{z}}+\text{z}+\overline{\text{z}}+1=1$
$\Rightarrow\text{z}\overline{\text{z}}+\text{z}+\overline{\text{z}}=0$
Since, $\text{z}=\text{x}+\text{iy}$
$\therefore\text{z}\overline{\text{z}}+\text{z}+\overline{\text{z}}=0$
$\Rightarrow(\text{x}+\text{iy})(\text{x}-\text{iy})+\text{x}+\text{iy}+\text{x}- \text{iy}=0$
$\Rightarrow\text{x}^2+\text{y}^2+\text{2x}=0$
$\Rightarrow(\text{x}+1)^2+(\text{y}-0)^2=1^2$
which is the equation of a circle with the center $(-1,\ 0)$ and radius 1.
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MCQ 1191 Mark
If $\text{f(z)}=\frac{7-\text{z}}{1-\text{z}^2},$ where $\text{z}=1+2\text{i},$ then $|\text{f(z)}|$ is:
  • ✓
    $\frac{|\text{z}|}{2}$
  • B
    $|\text{z}|$
  • C
    $2|\text{z}|$
  • D
    None of these
Answer
Correct option: A.
$\frac{|\text{z}|}{2}$
$\text{f(z)}=\frac{7-\text{z}}{1-\text{z}^2}$
$=\frac{7-(1+2\text{i})}{1-(1+2\text{i})^2}$
$=\frac{7-1+2\text{i}}{1-(1^2+2^2\text{i}^2+4\text{i})}$
$=\frac{6-2\text{i}}{1-1+4-4\text{i}}$
$=\frac{6-2\text{i}}{4-4\text{i}}$
$=\frac{6-2\text{i}}{4-4\text{i}}\times\frac{4+4\text{i}}{4+4\text{i}}$
$=\frac{24+24\text{i}-8\text{i}-8\text{i}^2}{4^2-4^2\text{i}^2}$
$=\frac{24+16\text{i}+8}{16+16}$
$=\frac{32+16\text{i}}{32}$
$=1+\frac{1}{2}\text{i}$
since $\text{z}=1+2\text{i},$
$\because|\text{z}|=\sqrt{(1)^2+(2)^2}$
$=\sqrt{1+4}$
$=\sqrt{5}$
$\therefore|\text{f}\text{(z)|}=\sqrt{(1)^1+(\frac{1}{2})^2}$
$=\sqrt{1+\frac{1}{4}}$
$=\frac{\sqrt5}{2}$
$=\frac{|\text{z}|}{2}$
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MCQ 1201 Mark
If $(1+\text{i})(1+2\text{i})(1+3\text{i})...(1+\text{ni})=\text{a}+\text{ib},$ then 2.5.10.17...$(1+\text{n}^2)=$
  • A
    $\text{a}-\text{ib}$
  • B
    $\text{a}^2-\text{b}^2$
  • ✓
    $\text{a}^2+\text{b}^2$
  • D
    none of these
Answer
Correct option: C.
$\text{a}^2+\text{b}^2$
$(1+\text{i})(1+2\text{i})(1+3\text{i})...(1+\text{ni})=\text{a}+\text{ib}$
Taking modulus on both the sides, we get,
$|(1+\text{i})(1+2\text{i})(1+3\text{i})...(1+\text{ni})|=\text{a}+\text{ib}$
$|(1+\text{i})(1+2\text{i})(1+3\text{i})...(1+\text{ni})|$ can be wriiten as $|(1+\text{i})||(1+2\text{i})||(1+3\text{i})|...|(1+\text{ni})|$
$\therefore\sqrt{1^2+1^2}\times\sqrt{1^2+2^2}\times\sqrt{1^2+3^2}...\times\sqrt{1^2+\text{n}^2}=\sqrt{\text{a}^2+\text{b}^2}$
$\therefore\sqrt{2}\times\sqrt{5}\times\sqrt{10}...\times\sqrt{1^2+\text{n}^2}=\sqrt{\text{a}^2+\text{b}^2}$
Squaring on both the sides, we get:
$2\times5\times10...\times(1+\text{n}^2)=\sqrt{\text{a}^2+\text{b}^2}$
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MCQ 1211 Mark
If $\frac{1-\text{ix}}{1+\text{ix}}=\text{a}+\text{ib},$ then $\text{a}^2+\text{b}^2=$
  • ✓
    1
  • B
    -1
  • C
    0
  • D
    none of these
Answer
Correct option: A.
1
$\frac{1-\text{ix}}{1+\text{ix}}=\text{a}+\text{ib}$
Taking modulus on both the sides, we get:
$\Big|\frac{1-\text{ix}}{1+\text{ix}}\Big|=\big|\text{a}+\text{ib}\big|$
$\Rightarrow\frac{\sqrt{1^2+\text{x}^2}}{\sqrt{1^2+\text{x}^2}}=\sqrt{\text{a}^2+\text{b}^2}$
$\Rightarrow\sqrt{\text{a}^2+\text{b}^2}=1$
Squaring both the sides, we get:
$\text{a}^2+\text{b}^2=1$
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MCQ 1221 Mark
A real value of x satisfies the equation $\frac{3-4\text{ix}}{3+4\text{ix}}=\text{a}-\text{ib}\Big(\text{a},\text{b}\in\text{R}\Big),$ if $\text{a}^2+\text{b}^2=$
  • ✓
    1
  • B
    -1
  • C
    2
  • D
    -2
Answer
Correct option: A.
1
$\text{a}-\text{ib}=\frac{3-4\text{ix}}{3+4\text{ix}}$
$=\frac{3-4\text{ix}}{3+4\text{ix}}\times\frac{3-4\text{ix}}{3-4\text{ix}}$
$=\frac{9+16\text{x}^2\text{i}^2-24\text{xi}}{9-16\text{x}^2\text{i}^2}$
$=\frac{(9-16\text{x}^2)-\text{i}(24\text{x})}{9+16\text{x}^2}$
$\Rightarrow\ |\text{a}-\text{ib}|^2=\Bigg|\frac{(9-16\text{x}^2)-\text{i}(24\text{x})}{9+16\text{x}^2}\Bigg|^2$
$\Rightarrow\text{a}^2+\text{b}^2=\frac{(9-16\text{x}^2)^2+(24\text{x})^2}{(9+16\text{x}^2)^2}$
$=\frac{81+256\text{x}^4-288\text{x}^2+576\text{x}^2}{(9+16\text{x}^2)^2}$
$=\frac{81+256\text{x}^4+288\text{x}^2}{(9+16\text{x}^2)^2}$
$=\frac{(9+16\text{x}^2)^2}{(9+16\text{x}^2)^2}$
$=1$
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MCQ 1231 Mark
If $\text{z}=\frac{1}{(1-\text{i})(2+3\text{i})},$ then $|\text{z}|=$
  • A
    1
  • ✓
    $\frac{1}{\sqrt{26}}$
  • C
    $\frac{5}{\sqrt{26}}$
  • D
    none of these.
Answer
Correct option: B.
$\frac{1}{\sqrt{26}}$
Let $\text{z}=\frac{1}{(1-\text{i})(2+3\text{i})}$
$\Rightarrow\text{z}=\frac{1}{2+\text{i}-3\text{i}^2}$
$\Rightarrow\text{z}=\frac{1}{2+\text{i}+3}$
$\Rightarrow\text{z}=\frac{1}{5+\text{i}}\times\frac{5-\text{i}}{5-\text{i}}$
$\Rightarrow\text{z}=\frac{5-\text{i}}{25-\text{i}^2}$
$\Rightarrow\text{z}=\frac{5-\text{i}}{25+1}$
$\Rightarrow\text{z}=\frac{5-\text{i}}{26}$
$\Rightarrow\text{z}=\frac{5}{26}-\frac{\text{i}}{26}$
$\Rightarrow|\text{z}|=\sqrt{\frac{25}{676}+\frac{1}{676}}$
$\Rightarrow\text{z}=\frac{1}{\sqrt{26}}$
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MCQ 1241 Mark
If $\theta$ is the amplitude of $\frac{\text{a}+\text{ib}}{\text{a}-\text{ib}},$ then $\tan\theta=$
  • A
    $\frac{2\text{a}}{\text{a}^2+\text{b}^2}$
  • ✓
    $\frac{2\text{ab}}{\text{a}^2-\text{b}^2}$
  • C
    $\frac{\text{a}^2-\text{b}^2}{\text{a}^2+\text{b}^2}$
  • D
    none of these
Answer
Correct option: B.
$\frac{2\text{ab}}{\text{a}^2-\text{b}^2}$
$\text{z}=\frac{\text{a}+\text{ib}}{\text{a}-\text{ib}}\times\frac{\text{a}+\text{ib}}{\text{a}+\text{ib}}$
$\Rightarrow\text{z}=\frac{\text{a}^2+\text{i}^2\text{b}^2+2\text{abi}}{\text{a}^2-\text{i}^2\text{b}^2}$
$\Rightarrow\text{z}=\frac{\text{a}^2-\text{b}^2+2\text{abi}}{\text{a}^2+\text{b}^2}$
$\Rightarrow\text{z}=\frac{\text{a}^2-\text{b}^2}{\text{a}^2+\text{b}^2}+\text{i}\frac{2\text{a}\text{b}}{\text{a}^2+\text{b}^2}$
$\Rightarrow\text{Re(z)}=\frac{\text{a}^2-\text{b}^2}{\text{a}^2+\text{b}^2},\text{Im(z)}\frac{2\text{a}\text{b}}{\text{a}^2+\text{b}^2}$
$\Rightarrow\tan\alpha=\Big|\frac{\text{Im(z)}}{\text{Re(z)}}\Big|$
$=\frac{2\text{ab}}{\text{a}^2-\text{b}^2}$
$\alpha=\tan^{-1}\Big(\frac{2\text{ab}}{\text{a}^2-\text{b}^2}\Big)$
Since, z lies in the first quadrant . Therefore,
$\text{arg(z)}=\alpha=\tan^{-1}\Big(\frac{2\text{ab}}{\text{a}^2-\text{b}^2}\Big)$
$\tan\theta=\frac{2\text{ab}}{\text{a}^2-\text{b}^2}$
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MCQ 1251 Mark
If $\text{z}=\cos\frac{\pi}{4}+\text{i}\sin\frac{\pi}{6},$ then
  • A
    $|\text{z}|=1,\text{arg(z)}=\frac{\pi}{4}$
  • B
    $|\text{z}|=1,\text{arg(z)}=\frac{\pi}{6}$
  • C
    $|\text{z}|=\frac{\sqrt{3}}{2},\text{arg(z)}=\frac{5\pi}{24}$
  • ✓
    $|\text{z}|=\frac{\sqrt{3}}{2},\text{arg(z)}=\tan^{-1}\frac{1}{\sqrt{2}}$
Answer
Correct option: D.
$|\text{z}|=\frac{\sqrt{3}}{2},\text{arg(z)}=\tan^{-1}\frac{1}{\sqrt{2}}$
$\text{z}=\cos\frac{\pi}{4}+\text{i}\sin\frac{\pi}{6}$
$\Rightarrow\text{z}=\frac{1}{\sqrt{2}}+\frac{1}{2}\text{i}$
$\Rightarrow|\text{z}|=\sqrt{\Big(\frac{1}{\sqrt{2}}\Big)^2+\frac{1}{4}}$
$\Rightarrow|\text{z}|=\sqrt{\frac{1}{2}+\frac{1}{4}}$
$\Rightarrow|\text{z}|=\sqrt{\frac{3}{4}}$
$\Rightarrow|\text{z}|=\frac{\sqrt{3}}{2}$
$\Rightarrow\tan\alpha=\Big|\frac{\text{Im(z)}}{\text{Re(z)}}\Big|$
$=\frac{1}{\sqrt{2}}$
$\Rightarrow\alpha=\tan^{-1}\Big(\frac{1}{\sqrt{2}}\Big)$
Since, the point z lies in the first quadrant.
Therefore, $|\text{z}|=\frac{\sqrt{3}}{2},\text{arg(z)}=\tan^{-1}\frac{1}{\sqrt{2}}$
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MCQ 1261 Mark
The value of $\frac{(\text{i}^5+\text{i}^6+\text{i}^7+\text{i}^8+\text{i}^9)}{(1+\text{i})}$ is:
  • ✓
    $\frac{1}{2}(1+\text{i})$
  • B
    $\frac{1}{2}(1-\text{i})$
  • C
    $1$
  • D
    $\frac{1}{2}$
Answer
Correct option: A.
$\frac{1}{2}(1+\text{i})$
$\frac{(\text{i}^5+\text{i}^6+\text{i}^7+\text{i}^8+\text{i}^9)}{(1+\text{i})}$
$=\frac{\text{i}-1-\text{i}+1+\text{i}}{1+\text{i}} \ [$ As, $\text{i}^5=\text{i},\text{i}^6=-1,\text{i}^7=-\text{i},\text{i}^8=1,\text{i}^9=\text{i}]$
$=\frac{\text{i}}{\text{i}+1}$
$=\frac{\text{i}}{\text{i}+1}\times\frac{\text{i}-1}{\text{i}-1}$
$=\frac{\text{i}(\text{i}-1)}{\text{i}^2-1}$
$=\frac{\text{i}^2-\text{i}}{-2}$
$=\frac{1}{2}(1+\text{i})$
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MCQ 1271 Mark
The value of $\frac{\text{i}^{592}+\text{i}^{590}+\text{i}^{588}+\text{i}^{586}+\text{i}^{584}}{\text{i}^{582}+\text{i}^{580}+\text{i}^{578}+\text{i}^{576}+\text{i}^{574}}-1$ is
  • A
    -1
  • ✓
    -2
  • C
    -3
  • D
    -4
Answer
Correct option: B.
-2
$\frac{\text{i}^{592}+\text{i}^{590}+\text{i}^{588}+\text{i}^{586}+\text{i}^{584}}{\text{i}^{582}+\text{i}^{580}+\text{i}^{578}+\text{i}^{576}+\text{i}^{574}}-1$
$=\frac{\text{i}^{4\times148}+\text{i}^{4\times147+2}+\text{i}^{4\times147}+\text{i}^{4\times146+2}+\text{i}^{4\times146}}{\text{i}^{4\times145+2}+\text{i}^{4\times145}+\text{i}^{4\times144+2}+\text{i}^{4\times144}+\text{i}^{4\times143+2}}-1$ $[\because\text{i}^4=1$ and $\text{i}^2=-1]$
$=\frac{1+\text{i}^2+1+\text{i}^2+1}{\text{i}^2+1+\text{i}^2+1+\text{i}^2}-1$
$=\frac{1}{-1}-1$
$=-2$
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MCQ 1281 Mark
The least positive integer n such that $\Big(\frac{2\text{i}}{1+\text{i}}\Big)^\text{n}$ is a positive integer, is:
  • A
    16
  • ✓
    8
  • C
    4
  • D
    2
Answer
Correct option: B.
8
Let $\text{z}=\Big(\frac{2\text{i}}{1+\text{i}}\Big)$
$\Rightarrow\text{z}=\frac{2\text{i}}{1+\text{i}}\times\frac{1-\text{i}}{1-\text{i}}$
$\Rightarrow\text{z}=\frac{2\text{i}(1-\text{i})}{1-\text{i}^2}$
$\Rightarrow\text{z}=\frac{2\text{i}(1-\text{i})}{1+1} \ [\because\text{i}^2=-1]$
$\Rightarrow\text{z}=\frac{2\text{i}(1-\text{i})}{2}$
$\Rightarrow\text{z}=\text{i}-\text{i}^2$
$\Rightarrow\text{z}=\text{i}+1$
Now, $\text{z}^\text{n}=(1+\text{i})^\text{n}$
For $\text{n}=2,$
$\text{z}^2=(1+\text{i})^2$
$=1+\text{i}^2+2\text{i}$
$=1-1+2\text{i}$
$=2\text{i} \ ...(1)$
Since this is not a positive integer,
For $\text{n}=4,$
$\text{z}^4=(1+\text{i})^4$
$=\big[(1+\text{i})^2\big]^2$
$=(2\text{i})^2$ [Using (1)]
$=(4\text{i})^2$
$=-4 \ ...(2)$
This is a negative integer.
For $\text{n}=8,$
$\text{z}^8=(1+\text{i})^8$
$=\big[(1+\text{i})^4\big]^2$
$=(-4)^2$ [Using (2)]
$=16$
This is a positive integer.
Thus, $\text{z}=\Big(\frac{2\text{i}}{1+\text{i}}\Big)^\text{n}$ is positive for $\text{n}=8.$
Therefore, 8 is the least positive integer such that $\Big(\frac{2\text{i}}{1+\text{i}}\Big)^\text{n}$ is a positive integer.
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MCQ 1291 Mark
The amplitude of $\frac{1}{\text{i}}$ is equal to:
  • A
    0
  • B
    $\frac{\pi}{2}$
  • ✓
    $-\frac{\pi}{2}$
  • D
    $\pi$
Answer
Correct option: C.
$-\frac{\pi}{2}$
Let $\text{z}=\frac{1}{\text{i}}$
$\text{z}=\frac{1}{\text{i}}\times\frac{\text{i}}{\text{i}}$
$\text{z}=\frac{\text{i}}{\text{i}^2}$
$\text{z}=-\text{i}$
Since, z (0, -1) lies on the negative imaginary axis . Therefore, $\text{arg(z)}=\frac{-\pi}{2}$
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MCQ 1301 Mark
If $\text{z}=\text{a}+\text{ib}$ lies in third quadrant, then $\frac{\bar{\text{z}}}{\text{z}}$ also lies in third quadrant if:
  • A
    $\text{a}>\text{b}>0$
  • B
    $\text{a}<\text{b}<0$
  • ✓
    $\text{b}<\text{a}<0$
  • D
    $\text{b}>\text{a}>0$
Answer
Correct option: C.
$\text{b}<\text{a}<0$
Since, $\text{z}=\text{a}+\text{ib}$ lies in third quadrant.
$\Rightarrow\text{a}<0$ and $\text{b}<0 \ ...(1)$
Now,
$\frac{\bar{\text{z}}}{\text{z}}=\frac{\overline{\text{a}+\text{ib}}}{\text{a}+\text{ib}}$
$=\frac{\text{a}-\text{ib}}{\text{a}+\text{ib}}$
$=\frac{\text{a}-\text{ib}}{\text{a}+\text{ib}}\times\frac{\text{a}-\text{ib}}{\text{a}-\text{ib}}$
$=\frac{\text{a}^2+\text{i}^2\text{b}^2-2\text{abi}}{\text{a}^2-\text{i}^2\text{b}^2}$
$=\frac{\text{a}^2-\text{b}^2-2\text{abi}}{\text{a}^2+\text{b}^2}$
Since, $\frac{\bar{\text{z}}}{\text{z}}$ also lies in third quadrant.
$\Rightarrow\text{a}^2-\text{b}^2<0$
$\Rightarrow(\text{a}-\text{b})(\text{a}+\text{b})<0$
$\Rightarrow\text{a}-\text{b}>0$ and $\text{a}+\text{b}<0$
$\Rightarrow\text{a}>\text{b} \ ...(2)$
From (1) and (2),
$\text{b}<\text{a}<0$
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MCQ 1311 Mark
If $\text{z}=\Big(\frac{1+2\text{i}}{1-(1-\text{i})^2}\Big),$ then arg(z) equal:
  • ✓
    0
  • B
    $\frac{\pi}{2}$
  • C
    $\pi$
  • D
    none of these.
Answer
Correct option: A.
0
$\text{z}=\frac{1+2\text{i}}{1-(1-\text{i})^2}$
$\Rightarrow\text{z}=\frac{1+2\text{i}}{1-(1+\text{i}^2-2\text{i})}$
$\Rightarrow\text{z}=\frac{1+2\text{i}}{1-(1-1-2\text{i})}$
$\Rightarrow\text{z}=\frac{1+2\text{i}}{1+2\text{i}}$
$\Rightarrow\text{z}=1$
Since point (1,0) lies on the positive direction of real axis, we have:
arg(z) = 0
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MCQ 1321 Mark
If $\text{z}=\frac{1}{1-\cos\theta-\text{i}\sin\theta},$ then $\text{Re(z)}=$
  • A
    0
  • ✓
    $\frac{1}{2}$
  • C
    $\cot\frac{\theta}{2}$
  • D
    $\frac{1}{2}\cot\frac{\theta}{2}$
Answer
Correct option: B.
$\frac{1}{2}$
$\text{z}=\frac{1}{1-\cos\theta-\text{i}\sin\theta}$
$\text{z}=\frac{1}{1-\cos\theta-\text{i}\sin\theta}\times\frac{1-\cos\theta+\text{i}\sin\theta}{1-\cos\theta+\text{i}\sin\theta}$
$\Rightarrow\text{z}=\frac{1-\cos\theta+\text{i}\sin\theta}{(1-\cos\theta)^2-(\text{i}\sin\theta)^2}$
$\Rightarrow\text{z}=\frac{1-\cos\theta+\text{i}\sin\theta}{1+\cos^2\theta-2\cos\theta+\text{i}\sin^2\theta}$
$\Rightarrow\text{z}=\frac{1-\cos\theta+\text{i}\sin\theta}{1+1-2\cos\theta}$
$\Rightarrow\text{z}=\frac{1-\cos\theta+\text{i}\sin\theta}{2(1-\cos\theta)}$
$\Rightarrow\text{Re(z)}=\frac{(1-\cos\theta)}{2(1-\cos\theta)}=\frac{1}{2}$
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MCQ 1331 Mark
The amplitude of $\frac{1+\text{i}\sqrt{3}}{\sqrt{3}+\text{i}}$ is:
  • A
    $\frac{\pi}{3}$
  • B
    $-\frac{\pi}{3}$
  • ✓
    $\frac{\pi}{6}$
  • D
    $-\frac{\pi}{6}$
Answer
Correct option: C.
$\frac{\pi}{6}$
Let $\text{z}=\frac{1+\text{i}\sqrt{3}}{\sqrt{3}+\text{i}}$
$\Rightarrow\text{z}=\frac{1+\text{i}\sqrt{3}}{\sqrt{3}+\text{i}}\times\frac{\sqrt{3}-\text{i}}{\sqrt{3}-\text{i}}$
$\Rightarrow\text{z}=\frac{\sqrt{3}+2\text{i}-\sqrt{3}\text{i}^2}{3-\text{i}^2}$
$\Rightarrow\text{z}=\frac{\sqrt{3}+\sqrt{3}+2\text{i}}{4}$
$\Rightarrow\text{z}=\frac{2\sqrt{3}+2\text{i}}{4}$
$\Rightarrow\text{z}=\frac{\sqrt{3}}{2}+\frac{1}{2}\text{i}$
$\Rightarrow\tan\alpha=\Big|\frac{\text{Im(z)}}{\text{Re(z)}}\Big|$
$=\frac{1}{\sqrt{3}}$
$\Rightarrow\alpha=\frac{\pi}{6}$
Since, z lies in the first quadrant. Therefore, $\text{arg(z)}=\tan^{-1}\Big(\frac{1}{\sqrt{3}}\Big)=\frac{\pi}{6}$
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MCQ 1341 Mark
If $\text{a}=1+\text{i},$ then $a^2$ equals:
  • A
    $1-\text{i}$
  • ✓
    $2\text{i}$
  • C
    $(1+\text{i})(1-\text{i})$
  • D
    $\text{i}-1.$
Answer
Correct option: B.
$2\text{i}$
$\text{a}=1+\text{i}$
On squaring both the sides, we get,
$\text{a}^2=(1+\text{i})^2$
$\Rightarrow\text{a}^2=1+\text{i}^2+2\text{i}$
$\Rightarrow\text{a}^2=1-1+2\text{i} \ (\because\text{i}^2=-1)$
$\Rightarrow\text{a}^2=2\text{i}$
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MCQ 1351 Mark
The principal value of the amplitude of (1 + i) is:
  • ✓
    $\frac{\pi}{4}$
  • B
    $\frac{\pi}{12}$
  • C
    $\frac{3\pi}{4}$
  • D
    $\pi$
Answer
Correct option: A.
$\frac{\pi}{4}$
Let $\text{z}=(1+\text{i})$
$\Rightarrow\tan\alpha=\Big|\frac{\text{Im(z)}}{\text{Re(z)}}\Big|$
$=1$
$\Rightarrow\alpha=\frac{\pi}{4}$
Since, z lies in the first quadrant.
Therefore, $\text{arg(z)}=\frac{\pi}{4}$
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MCQ 1361 Mark
If $(\text{x}+\text{iy})^{\frac{1}{3}}=\text{a}+\text{ib,}$ then $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=$
  • A
    0
  • B
    1
  • C
    -1
  • ✓
    none of these
Answer
Correct option: D.
none of these
$(\text{x}+\text{iy)}^{\frac{1}{3}}=\text{a}+\text{ib}$
Cubing on both the sides, we get:
$\text{x}+\text{iy}=(\text{a}+\text{ib})^3$
$\Rightarrow\text{x}+\text{iy}=\text{a}^3+(\text{ib})^3+3\text{a}^2\text{bi}+3\text{a}(\text{bi})^2$
$\Rightarrow\text{x}+\text{iy}=\text{a}^3+\text{i}^3\text{b}^3+3\text{a}^2\text{bi}+3\text{i}^2\text{ab}^2$
$\Rightarrow\text{x}+\text{iy}=\text{a}^3-\text{i}\text{b}^3+3\text{a}^2\text{bi}-3\text{ab}^2 \ (\because\text{i}^2=-1,\text{i}^3=-\text{i})$
$\Rightarrow\text{x}+\text{iy}=\text{a}^3-3\text{a}\text{b}^2+\text{i}(-\text{b}^3+3\text{a}^2\text{b})$
$\therefore\text{x}=\text{a}^3-3\text{a}\text{b}^2$ and $\text{y}=3\text{a}^2\text{b}-\text{b}^3$
or, $\frac{\text{x}}{\text{a}}=\text{a}^2-3\text{b}^2$ and $\frac{\text{y}}{\text{b}}=3\text{a}^2-\text{b}^2$
$\Rightarrow\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=\text{a}^2-3\text{b}^2+3\text{a}^2-\text{b}^2$
$\Rightarrow\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=4\text{a}^2-4\text{b}^2$
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MCQ 1371 Mark
If $\frac{1+7\text{i}}{(2-\text{i})^2},$ then:
  • A
    $|\text{z}|=2$
  • B
    $|\text{z}|=\frac{1}{2}$
  • C
    $\text{amp(z)}=\frac{\pi}{4}$
  • ✓
    $\text{amp(z)}=\frac{3\pi}{4}$
Answer
Correct option: D.
$\text{amp(z)}=\frac{3\pi}{4}$
$\text{z}=\frac{1+7\text{i}}{(2-\text{i})^2}$
$\Rightarrow\text{z}=\frac{1+7\text{i}}{4-1-4\text{i}} \ [\because\text{i}^2=-1]$
$\Rightarrow\text{z}=\frac{1+7\text{i}}{3-4\text{i}}$
$\Rightarrow\text{z}=\frac{1+7\text{i}}{3-4\text{i}}\times\frac{3+4\text{i}}{3+4\text{i}}$
$\Rightarrow\text{z}=\frac{3+4\text{i}+21\text{i}+28\text{i}^2}{9-16\text{i}^2}$
$\Rightarrow\text{z}=\frac{3-28+25\text{i}}{9+16}$
$\Rightarrow\text{z}=\frac{-25+25\text{i}}{25}$
$\Rightarrow\text{z}=-1+\text{i}$
$\Rightarrow\tan\alpha=\Big|\frac{\text{Im(z)}}{\text{Re(z)}}\Big|$
$=1$
$\Rightarrow\alpha=\frac{\pi}{4}$
Since, z lies in the second quadrant.
Therefore, $\text{amp(z)}=\pi-\alpha$
$=\pi-\frac{\pi}{4}$
$=\frac{3\pi}{4}$
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MCQ 1381 Mark
If $\text{z}=\frac{1}{(2+3\text{i})^2},$ then $|\text{z}|=$
  • ✓
    $\frac{1}{13}$
  • B
    $\frac{1}{5}$
  • C
    $\frac{1}{12}$
  • D
    none of these.
Answer
Correct option: A.
$\frac{1}{13}$
Let $\text{z}=\frac{1}{(2+3\text{i})^2}$
$\Rightarrow\text{z}=\frac{1}{4+9\text{i}^2+12\text{i}}$
$\Rightarrow\text{z}=\frac{1}{4-9+12\text{i}}$
$\Rightarrow\text{z}=\frac{1}{-5+12\text{i}}$
$\Rightarrow\text{z}=\frac{1}{-5+12\text{i}}\times\frac{-5-12\text{i}}{-5-12\text{i}}$
$\Rightarrow\text{z}=\frac{-5-12\text{i}}{25+144}$
$\Rightarrow\text{z}=\frac{-5}{169}-\frac{12\text{i}}{169}$
$\Rightarrow|\text{z}|=\sqrt{\frac{25}{169^2}+\frac{144}{169^2}}$
$\Rightarrow|\text{z}|=\frac{1}{\sqrt{169}}$
$\Rightarrow|\text{z}|=\frac{1}{13}$
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MCQ 1391 Mark
If z is a non-zero complex number, then $\Big|\frac{|\bar{\text{z}}|^2}{\text{z}\bar{\text{z}}}\Big|$ is equal to:
  • ✓
    $\Big|\frac{\bar{\text{z}}}{\text{z}}\Big|$
  • B
    $\big|{\text{z}}\big|$
  • C
    $\big|{\bar{\text{z}}}\big|$
  • D
    none of these
Answer
Correct option: A.
$\Big|\frac{\bar{\text{z}}}{\text{z}}\Big|$
$\Big|\frac{|\bar{\text{z}}|^2}{\text{z}\bar{\text{z}}}\Big|=\Big|\frac{|\bar{\text{z}}|^2}{|\text{z}|^2}\Big| \ \Big(\because\text{z}\bar{\text{z}}=|\text{z}|^2\Big)$
Let $\text{z}=\text{a}+\text{ib}$
$\Rightarrow|\text{z}|=\sqrt{\text{a}^2+\text{b}^2}$
Let $\bar{\text{z}}=\text{a}-\text{ib}$
$\Rightarrow|\bar{\text{z}}|=\sqrt{\text{a}^2+\text{b}^2}$
$\therefore\Big|\frac{|\bar{\text{z}}|^2}{\text{z}\bar{\text{z}}}\Big|=\Big|\frac{|\bar{\text{z}}|^2}{|\text{z}|^2}\Big|$
$=\Big|\frac{\bar{\text{z}}}{\text{z}}\Big|$
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MCQ 1401 Mark
Which of the following is correct for any two complex numbers $z_1$ and $z_2?$
  • ✓
    $|\text{z}_1\text{z}_2|=|\text{z}_1||\text{z}_2|$
  • B
    $\text{arg}(\text{z}_1\text{z}_2)=\text{arg}(\text{z}_1)\text{arg}(\text{z}_2)$
  • C
    $|\text{z}_1+\text{z}_2|=|\text{z}_1|+|\text{z}_2|$
  • D
    $|\text{z}_1+\text{z}_2|\geq|\text{z}_1|+|\text{z}_2|$
Answer
Correct option: A.
$|\text{z}_1\text{z}_2|=|\text{z}_1||\text{z}_2|$
since we know that
$|\text{z}_1\text{z}_2|=|\text{z}_1||\text{z}_2|$
$\text{arg}(\text{z}_1\text{z}_2)=\text{arg}(\text{z}_1)+\text{arg}(\text{z}_2)$ and
$|\text{z}_1+\text{z}_2|\le|\text{z}_1|+|\text{z}_2|$
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MCQ 1411 Mark
If $\frac{(\text{a}^2+1)^2}{2\text{a}-\text{i}}=\text{x}+\text{iy},$ then $\text{x}^2+\text{y}^2$ is equal to:
  • ✓
    $\frac{(\text{a}^2+1)^4}{4\text{a}^2+1}$
  • B
    $\frac{(\text{a}+1)^2}{4\text{a}^2+1}$
  • C
    $\frac{(\text{a}^2-1)^2}{(4\text{a}^2-1)^2}$
  • D
    None of these
Answer
Correct option: A.
$\frac{(\text{a}^2+1)^4}{4\text{a}^2+1}$
$\text{x}+\text{iy}=\frac{(\text{a}^2+1)^2}{2\text{a}-\text{i}}$
Taking modulus on both the sides, we get:
$\sqrt{\text{x}^2+\text{y}^2}=\frac{(\text{a}^2+1)^2}{\sqrt{4\text{a}^2+1}}$
Squaring both sides, we get,
$\text{x}^2+\text{y}^2=\frac{(\text{a}^2+1)^4}{{4\text{a}^2+1}}$
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MCQ 1421 Mark
The argument of $\frac{1-\text{i}}{1+\text{i}}$ is:
  • ✓
    $-\frac{\pi}{2}$
  • B
    $\frac{\pi}{2}$
  • C
    $\frac{3\pi}{2}$
  • D
    $\frac{5\pi}{2}$
Answer
Correct option: A.
$-\frac{\pi}{2}$
Let $\text{z}=\frac{1-\text{i}}{1+\text{i}}$
$\Rightarrow\text{z}=\frac{1-\text{i}}{1+\text{i}}\times\frac{1-\text{i}}{1-\text{i}}$
$\Rightarrow\text{z}=\frac{1+\text{i}^2-2\text{i}}{1-\text{i}^2}$
$\Rightarrow\text{z}=\frac{1-1-2\text{i}}{1+1}$
$\Rightarrow\text{z}=\frac{-2\text{i}}{2}$
$\Rightarrow\text{z}=\text{i}$
Since, z lies on negative direction of imaginary axis . Therefore, $\text{arg(z)}=\frac{-\pi}{2}$
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MCQ 1431 Mark
The value of $(1+\text{i})^4+(1-\text{i})^4$ is:
  • A
    8
  • B
    4
  • ✓
    -8
  • D
    -4
Answer
Correct option: C.
-8
Using $\text{a}^4+\text{b}^4=(\text{a}^2+\text{b}^2)^2-2\text{a}^2\text{b}^2$
$(1+\text{i})^4+(1-\text{i})^4$
$=\Big((1+\text{i})^2+(1-\text{i})^2\Big)^2-2(1+\text{i})^2(1-\text{i})^2$
$=(1+\text{i}^2+2\text{i}+1+\text{i}^2-2\text{i})^2-2(1+\text{i}^2+2\text{i})(1+\text{i}^2-2\text{i})$
$=(1-1+2\text{i}+1-1-2\text{i})^2-2(1-1+2\text{i})(1-1-2\text{i})$
$=(0)-2(2\text{i})(-2\text{i}) \ (\because\text{i}^2=-1)$
$=8\text{i}^2$
$=-8$
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MCQ 1441 Mark
If $\text{x}+\text{iy}=\frac{3+5\text{i}}{7-6\text{i}},$ then $\text{y}=$
  • A
    $\frac{9}{85}$
  • B
    $\frac{-9}{85}$
  • ✓
    $\frac{53}{85}$
  • D
    none of these
Answer
Correct option: C.
$\frac{53}{85}$
$\text{x}+\text{iy}=\frac{3+5\text{i}}{7-6\text{i}}$
$\Rightarrow\text{x}+\text{iy}=\frac{3+5\text{i}}{7-6\text{i}}\times\frac{7+6\text{i}}{7+6\text{i}}$
$\Rightarrow\text{x}+\text{iy}=\frac{21+53\text{i}+30\text{i}^2}{49-36\text{i}^2}$
$\Rightarrow\text{x}+\text{iy}=\frac{21-30+53\text{i}}{49+36}$
$\Rightarrow\text{x}+\text{iy}=\frac{-9}{85}+\text{i}\frac{53}{85}$
On comparing both the sides:
$\text{y}=\frac{53}{85}$
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MCQ 1451 Mark
$\frac{1+2\text{i}+3\text{i}^2}{1-2\text{i}+3\text{i}^2}$ equals:
  • A
    i
  • B
    -1
  • ✓
    -i
  • D
    4
Answer
Correct option: C.
-i
Let $\text{z}=\frac{1+2\text{i}+3\text{i}^2}{1-2\text{i}+3\text{i}^2}$
$\Rightarrow\text{z}=\frac{1+2\text{i}-3}{1-2\text{i}-3}$
$\Rightarrow\text{z}=\frac{-2+2\text{i}}{-2-2\text{i}}\times\frac{-2+2\text{i}}{-2+2\text{i}}$
$\Rightarrow\text{z}=\frac{(-2+2\text{i})^2}{(-2)^2-(2\text{i})^2}$
$\Rightarrow\text{z}=\frac{4+4\text{i}^2-8\text{i}}{4+4}$
$\Rightarrow\text{z}=\frac{4-4-8\text{i}}{8}$
$\Rightarrow\text{z}=\frac{-8\text{i}}{8}$
$\Rightarrow\text{z}=-\text{i}$
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MCQ 1461 Mark
If $\text{z}=\Big(\frac{1+\text{i}}{1-\text{i}}\Big),$ then $z^4$ equals:
  • ✓
    $1$
  • B
    $-1$
  • C
    $0$
  • D
    none of these.
Answer
Correct option: A.
$1$
Let $\text{z}=\frac{1+\text{i}}{1-\text{i}}$
Rationalising the denominator:
$\text{z}=\frac{1+\text{i}}{1-\text{i}}\times\frac{1+\text{i}}{1+\text{i}}$
$\Rightarrow\text{z}=\frac{1+\text{i}^2+2\text{i}}{1-\text{i}^2}$
$\Rightarrow\text{z}=\frac{1-1+2\text{i}}{1+1}$
$\Rightarrow\text{z}=\frac{2\text{i}}{2}$
$\Rightarrow\text{z}=\text{i}$
$\Rightarrow\text{z}^4=\text{i}^4$
Since $\text{i}^2=-1,$ we have:
$\Rightarrow\text{z}^4=\text{i}^2\times\text{i}^2$
$\Rightarrow\text{z}^4=1$
 
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MCQ 1471 Mark
If $(1+\text{i})(1+2\text{i})(1+3\text{i})...(1+\text{ni})=\text{a}+\text{ib},$ then $2\times5\times10\times...\times(1+\text{n}^2)$ is equal to:
  • A
    $\sqrt{\text{a}^2+\text{b}^2}$
  • B
    $\sqrt{\text{a}^2-\text{b}^2}$
  • ✓
    $\text{a}^2+\text{b}^2$
  • D
    $\text{a}^2-\text{b}^2$
Answer
Correct option: C.
$\text{a}^2+\text{b}^2$
$(1+\text{i})(1+2\text{i})(1+3\text{i})...(1+\text{ni})=\text{a}+\text{ib}$
Taking modulus on both the sides, we get:
$|(1+\text{i})(1+2\text{i})(1+3\text{i})...(1+\text{ni})|=|\text{a}+\text{ib}|$
$|(1+\text{i})(1+2\text{i})(1+3\text{i})...(1+\text{ni})|$ can be written as $|(1+\text{i})||(1+2\text{i})||(1+3\text{i})|...|(1+\text{ni})|$
$\sqrt{1^2+1^2}\times\sqrt{1^2+2^2}\times\sqrt{1^2+3^2}\times...\times\sqrt{1+\text{n}^2}=\sqrt{\text{a}^2+\text{b}^2}$
$\Rightarrow\sqrt{2}\times\sqrt{5}\times\sqrt{10}\times...\times\sqrt{1+\text{n}^2}=\sqrt{\text{a}^2+\text{b}^2}$
Squaring on both the sides, we get:
$2\times5\times10\times...\times(1+\text{n}^2)=\text{a}^2+\text{b}^2$
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