Question 12 Marks
If $\left(\frac{1+i}{1-i}\right)^{m} = 1$ then find the least positive integral value of m.
AnswerGiven $\left(\frac{1+i}{1-i}\right)^{m} = 1$
Now, $\left(\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right)^{m}=1$ [multiply divide numerator and denominator by 1+i]
$\Rightarrow$$\left[\frac{(1+i)^{2}}{1^{2}-i^{2}}\right]^{m}=1$
$\Rightarrow$$\left(\frac{1^{2}+\mathrm{i}^{2}+2 \mathrm{i}}{1+1}\right)^{\mathrm{m}}=1$
$\Rightarrow$$\left(\frac{1-1+2 i}{2}\right)^{m}=1$
$\Rightarrow$$\left(\frac{2 i}{2}\right)^{m}=1$
$\Rightarrow i^m = 1$
We can also write,$ i^m = i^{4k}$
On equating the powers,
Thus, m = 4k, Where k is some integer.
$\therefore$1 is the least positive integer.
Least positive integral value of m is $4 \times 1 = 4$
View full question & answer→Question 22 Marks
$\text { If }(a+i b)(c+i d)(e+i f)(g+i h)=A+i B \text { then show that }$
$\left(a^2+b^2\right)\left(c^2+d^2\right)\left(e^2+f^2\right)\left(g^2+h^2\right)=A^2+B^2$
AnswerHere $(a+i b)(c+i d)(e+i f)(g+i h)=A+i B$
Taking modulus on both sides
$|(a+i b)(c+i d)(e+i f)(g+i h)|=|A+i B|$
$\Rightarrow|a+i b||c+i d||e+i f||g+i h|=|A+i B|$
$\Rightarrow\left(\sqrt{a^2+b^2}\right)\left(\sqrt{c^2+d^2}\right)\left(\sqrt{e^2+f^2}\right)\left(g^2+h^2\right)=\sqrt{A^2+B^2}$
Squaring both sides
$\left(a^2+b^2\right)\left(c^2+d^2\right)\left(e^2+f^2\right)\left(g^2+h^2\right)=A^2+B^2$
View full question & answer→Question 32 Marks
Find the number of non-zero integral solutions of the equation ${\left| {1 - i} \right|^x} = {2^x}$.
AnswerHere ${\left| {1 - i} \right|^x} = {2^x}$
$ \Rightarrow {\left[ {\sqrt {{{(1)}^2} + {{( - 1)}^2}} } \right]^x} = {2^x}$$ \Rightarrow {(\sqrt 2 )^x} = {2^x}$
$ \Rightarrow {2^{\frac{x}{2}}} = {2^x} \Rightarrow \frac{x}{2} = x \Rightarrow \frac{x}{2} - x = 0$$ \Rightarrow \frac{{ - x}}{2} = 0$
$ \Rightarrow x = 0$
Thus the given equation has no non-zero integral solution.
View full question & answer→Question 42 Marks
Find the modulus of $\frac{{1 + i}}{{1 - i}} - \frac{{1 - i}}{{1 + i}}$.
Answer$\left| {\frac{{1 + i}}{{1 - i}} - \frac{{1 - i}}{{1 + i}}} \right| = \left| {\frac{{{{(1 + i)}^2} - {{(1 - i)}^2}}}{{(1 - i)(1 + i)}}} \right|$
$ = \left| {\frac{{1 + {i^2} + 2i - 1 - {i^2} + 2i}}{{1 - {i^2}}}} \right|$
$\left| {\frac{{4i}}{2}} \right| = |2i| = \sqrt 4 = 2$.
View full question & answer→Question 52 Marks
Let $z_1 = 2 - i, z_2 = -2 + i$. Find $\operatorname{Re} \left( {\frac{{{z_1}{z_2}}}{{{{\overline z }_1}}}} \right)$
AnswerHere $z_1=2-i$ and $z_2=-2+i$
$\therefore \overline{z_1}=2+i$
$\text { Now } z_1 z_2=(2-i)(-2+i) $
$=-4+2 i+2 i-i^2=(-4+1)+4 i$
$=-3+4 i$
$\therefore \frac{{{z_1}{z_2}}}{{\overline {{z_1}} }} = \frac{{ - 3 + 4i}}{{2 + i}} \times \frac{{2 - i}}{{2 - i}}$$ = \frac{{ - 6 + 3i + 8i - 4{i^2}}}{{4 - {i^2}}}$
$ = \frac{{( - 6 + 4) + 11i}}{{4 + 1}} = \frac{{ - 2 + 11i}}{5} = \frac{{ - 2}}{5} + \frac{{11}}{5}i$
$\therefore \operatorname{Re} \left( {\frac{{{z_1}{z_2}}}{{\overline {{z_1}} }}} \right) = \frac{{ - 2}}{5}$
View full question & answer→Question 62 Marks
Solve $x^2 + x + \frac{1}{{\sqrt 2}} = 0$
AnswerHere, ${x^2} + x + \frac{1}{{\sqrt 2}} = 0$
Comparing the given quadratic equation with $ax^2 + bx + c = 0$ we have
$a = 1, b =1$ and $c = \frac{1}{{\sqrt 2}}$
$\therefore x = \frac{-1\pm\sqrt{1-2\sqrt2}}{2} = \frac{-1 \pm\sqrt{\left(2\sqrt{2 }-1\right)}i}{2}$
Thus, $x = \frac{-1 +\sqrt{\left(2\sqrt{2 }-1\right)}i}{2} \ x = \frac{-1 - \sqrt{\left(2\sqrt{2 }-1\right)}i}{2}$
View full question & answer→Question 72 Marks
Solve $\sqrt 3 {x^2} - \sqrt 2 x + 3\sqrt 3 = 0$
AnswerHere $\sqrt 3 {x^2} - \sqrt 2 x + 3\sqrt 3 = 0$
Comparing the given quadratic equation with $ax^2 + bx + c = 0$ we have
$a = \sqrt 3 ,b = - \sqrt 2 $ and $c = 3\sqrt 3 $
$\therefore x = \frac{{ - ( - \sqrt 2) \pm \sqrt {{{( - \sqrt 2 )}^2} - 4 \times \sqrt 3 \times 3\sqrt 3 } }}{{2 \times \sqrt 3 }}$$ = \frac{{\sqrt 2 \pm \sqrt { - 34} }}{{2\sqrt 3 }}$.
$ = \frac{{\sqrt 2 \pm \sqrt { 34} i}}{{2\sqrt 3 }}$
Thus $x = \frac{{\sqrt 2 + \sqrt {34} i}}{{2\sqrt 3 }}$ and $x = \frac{{\sqrt 2 - \sqrt {34} i}}{{2\sqrt 3 }}$.
View full question & answer→Question 82 Marks
Solve $\sqrt 2 \;{x^2} + x + \sqrt 2 = 0$
AnswerHere $\sqrt 2 \;{x^2} + x + \sqrt 2 = 0$
Comparing the given quadratic equation with $ax^2 + bx + c = 0$ we have
$a = \sqrt 2 $, b = 1 and $c = \sqrt 2 $
$\therefore x = \frac{{ - 1 \pm \sqrt {{{(1)}^2} - 4 \times \sqrt 2 \times \sqrt 2 } }}{{2 \times \sqrt 2 }}$$ = \frac{{ - 1 \pm \sqrt { - 7} }}{{2\sqrt 2 }} = \frac{{ - 1 \pm \sqrt 7 i}}{{2\sqrt 2 }}$
Thus $x = \frac{{ - 1 + \sqrt 7 i}}{{2\sqrt 2 }}$ and $x = \frac{{ - 1 - \sqrt 7 i}}{{2\sqrt 2 }}$
View full question & answer→Question 92 Marks
Solve $x^2 - x + 2 = 0$
AnswerHere $x^2 - x + 2 = 0$
Comparing the given quadratic equation with $ax^2 + bx + c = 0$ we have
a=1,b=-1and c = 2
$\therefore x = \frac{{ - (1) \pm \sqrt {{{(1)}^2} - 4 \times 1 \times 2} }}{{2 \times 1}}$$ = \frac{{1 \pm \sqrt { - 7} }}{2} = \frac{{1 \pm \sqrt 7 i}}{2}$
Thus $x = \frac{{1 + \sqrt 7 i}}{2}$ and $x = \frac{{1 - \sqrt 7 i}}{2}$
View full question & answer→Question 102 Marks
Solve $x^2 + 3x + 5 = 0$
AnswerHere $x^2 + 3x + 5 = 0$
Comparing the given quadratic equation with $ax^2 + bx + c = 0$ we have
a = 1 , b= 3 and c = 5
$\therefore x = \frac{{ - 3 \pm \sqrt {{{(3)}^2} - 4 \times 1 \times 5} }}{{2 \times 1}}$$ = \frac{{ - 3 \pm \sqrt { - 11} }}{2} = \frac{{ - 3 \pm \sqrt {11} i}}{2}$
Thus $x = \frac{{ - 3 + \sqrt {11} i}}{2}$ and $x = \frac{{ - 3 - \sqrt {11} i}}{2}$
View full question & answer→Question 112 Marks
Solve $- x^2 + x - 2 = 0$
AnswerGiven:
$- x^2 + x - 2 = 0$
Comparing the given quadratic equation with $ax^2 + bx + c = 0$ we have
a = -1,b = 1 and c = -2x = ${-b \pm \sqrt{b^2-4ac} \over 2a}$
$\therefore x = \frac{{ - (1) \pm \sqrt {{{(1)}^2} - 4 \times (-1) \times (-2)} }}{{2 \times (-1)}}$ = $ \frac{{-1 \pm \sqrt { 7} i}}{-2}$
Thus $x = \frac{{-1 \pm \sqrt 7 i}}{-2}$
View full question & answer→Question 122 Marks
Solve $x^2 + 3x + 9 = 0$
AnswerHere $x^2 + 3x + 9 = 0$
Comparing the given quadratic equation with $ax^2 + bx + c = 0$ we have
a = 1 , b= 3 and c = 9
$\therefore x = \frac{{ - 3 \pm \sqrt {{{(3)}^2} - 4 \times 9} }}{{2 \times 1}}$
$ = \frac{{ - 3 \pm \sqrt { - 27} }}{2} = \frac{{ - 3 \pm 3\sqrt 3 i}}{2}$
Thus $x = \frac{{ - 3 + 3\sqrt 3 i}}{2}$ and $x = \frac{{ - 3 - 3\sqrt 3 i}}{2}$
View full question & answer→Question 132 Marks
Solve: $2x^2 + x + 1 = 0$
AnswerHere $2x^2 + x + 1 = 0$
Comparing the given quadratic equation with $ax^2 + bx +c= 0$, we have
a = 2, b = 1 and c = 1
$\therefore x = \frac{{ - 1 \pm \sqrt {{{(1)}^2} - 4 \times 2 \times 1} }}{{2 \times 2}}$$ = \frac{{ - 1 \pm \sqrt { - 7} }}{4} = \frac{{ - 1 \pm \sqrt 7 i}}{4}$
Thus $x = \frac{{ - 1 + \sqrt 7 i}}{4}$ and $x = \frac{{ - 1 - \sqrt 7 i}}{4}$
View full question & answer→Question 142 Marks
Solve ${x^2} + \frac{x}{{\sqrt 2 }} + 1 = 0$
AnswerHere ${x^2} + \frac{x}{{\sqrt 2 }} + 1 = 0$
Comparing the given quadratic equation with $ax^2 + bx + c = 0$ we have
$a = 1,b = \frac{1}{{\sqrt 2 }}$ and c = 1
$\therefore x=\frac{\frac{-1}{\sqrt2}+\sqrt{\left(\frac1{\sqrt2}\right)^2-4\times1\times1}}{2\times1}$
$ = \frac{{\frac{{ - 1}}{{\sqrt 2 }} \pm \sqrt {\frac{1}{2} - 4} }}{2}$
$ = \frac{{\frac{{ - 1}}{{\sqrt 2 }} \pm \sqrt {\frac{{ - 7}}{2}} }}{2} = \frac{{ - 1 \pm \sqrt 7 i}}{{2\sqrt 2 }}$
Thus $x = \frac{{ - 1 + \sqrt 7 i}}{{2\sqrt 2 }}$ and $x = \frac{{ - 1 - \sqrt 7 i}}{{2\sqrt 2 }}$
View full question & answer→Question 152 Marks
Solve $x^2 + 3 = 0$
AnswerHere $x^2 + 3 = 0 \Rightarrow {x^2} = - 3 \Rightarrow x = \pm \sqrt { - 3} = \pm \sqrt 3 i$
View full question & answer→Question 162 Marks
Convert the complex number in the polar form: i
AnswerHere z = $i = r(\cos \theta + i\sin \theta )$
$ \Rightarrow r\cos \theta = 0$ and $r\;\sin \theta = 1$ . . . (i)
Squaring both sides of (i) and adding
$$${r^2}({\cos ^2}\theta + {\sin ^2}\theta ) \Rightarrow {r^2} = 1 \Rightarrow r = 1$
$\therefore \cos \theta = 0$ and $\sin \theta = 1$
Since $\sin \theta $ and $\cos \theta $ are both positive
$\therefore \theta $ lies in first quadrant
$\therefore \theta = \frac{\pi }{2}$
Hence polar form of z is $\left( {\cos \frac{\pi }{2} + i\sin \frac{\pi }{2}} \right)$
View full question & answer→Question 172 Marks
Convert the complex number in the polar form: $\sqrt 3 + i$
AnswerHere $z = \sqrt 3 + i = r(\cos \theta + i\sin \theta )$
$ \Rightarrow r\cos \theta = \sqrt 3 $ and $r\;\sin \theta = 1$
Squaring both sides of (i) and adding
${r^2}({\cos ^2}\theta + {\sin ^2}\theta ) = 3 + 1$$ \Rightarrow {r^2} = 4 \Rightarrow r = 2$
$\therefore 2\cos \theta = \sqrt 3 $ and $2\sin \theta = 1$
$\therefore \cos \theta = \frac{{\sqrt 3 }}{2}$ and $\sin \theta = \frac{1}{2}$
Since $\sin \theta $ and $\cos \theta $ are both positive
$\therefore \theta $ lies in first quadrant
$\therefore \theta = \frac{\pi }{6}$
Hence polar form of z is $2\left( {\cos \frac{\pi }{6} + i\;\sin \frac{\pi }{6}} \right)$
View full question & answer→Question 182 Marks
Convert the complex number in the polar form: $-3$
AnswerHere $z = -3 = r(\cos \theta + i\sin \theta )$
$ \Rightarrow r\;\cos \theta = - 3$ and $r\;\sin \theta = 0$
Squaring both sides of (i) and adding
${r^2}({\cos ^2}\theta + {\sin ^2}\theta ) = 9 + 0$$ \Rightarrow {r^2} = 9 \Rightarrow r = 3$
$\therefore 3\cos \theta = - 3$ and $3\sin \theta = 0$
$ \Rightarrow \cos \theta = - 1$ and $\sin \theta = 0$
Since $\sin \theta $ is positive and $\cos \theta $ is negative
$\therefore \theta $ lies in second quadrant
$\therefore\theta=(\pi-0)\;=\;\mathrm\pi$
Hence polar form of z is $3(\cos \pi + i\sin \pi )$
View full question & answer→Question 192 Marks
Convert the complex number in the polar form: $ -1 - i $
AnswerHere $ z = - 1 - i = r(\cos \theta + i\sin \theta )$
$ \Rightarrow r\;\cos \theta = - 1$ and $r\sin \theta = - 1$
Squaring both sides of (i) and adding
${r^2}({\cos ^2}\theta + {\sin ^2}\theta ) = 1 + 1$$ \Rightarrow {r^2} = 2 \Rightarrow r = \sqrt 2 $
$\therefore \sqrt 2 \cos \theta = - 1$ and $\sqrt 2 \sin \theta = - 1$
$ \Rightarrow \cos \theta = \frac{{ - 1}}{{\sqrt 2 }}$ and ${\mathrm{sinθ}=\frac{-1}{\sqrt2}}$
Since $\sin \theta $ and $\cos \theta $ are both negative
$\therefore \theta $ lies in third quadrant
$\therefore \theta = \left( { - \pi + \frac{\pi }{4}} \right) = \frac{{ - 3\pi }}{4}$
Hence polar form of z is $\sqrt 2 \left[ {\cos \left( {\frac{{ - 3\pi }}{4}} \right) + i\sin \left( {\frac{{ - 3\pi }}{4}} \right)} \right]$
View full question & answer→Question 202 Marks
Convert the complex number in the polar form: -1 + i
AnswerHere z = -1 + i $ = r(\cos \theta + i\sin \theta )$
$ \Rightarrow r\;\cos \theta = - 1$ and $r\;\sin \theta = 1$
Squaring both sides of (i) and adding
${r^2}({\cos ^2}\theta + {\sin ^2}\theta ) = 1 + 1$$ \Rightarrow {r^2} = 2 \Rightarrow r = \sqrt 2 $
$\therefore \sqrt 2 \cos \theta = - 1$ and $\sqrt 2 \sin \theta = 1$
$ \Rightarrow \cos \theta = - \frac{1}{{\sqrt 2 }}$ and $\sin \theta = \frac{1}{{\sqrt 2 }}$
Since $\sin \theta $ is positive and $\cos \theta $ is negative
$\therefore \theta $ lies in second quadrant
$\therefore \theta = \left( {\pi - \frac{\pi }{4}} \right) = \frac{{3\pi }}{4}$
Hence polar form of z is $\sqrt 2 \left( {\cos \frac{{3\pi }}{4} + i\sin \frac{{3\pi }}{4}} \right)$.
View full question & answer→Question 212 Marks
Convert the complex number in the polar form: 1 - i
AnswerHere $z = 1 - i = r(\cos \theta + i\sin \theta )$
$\ \Rightarrow r\cos \theta = 1$ and $r\sin \theta = - 1$ .... (i)
Squaring both sides of (i) and adding
${r^2}({\cos ^2}\theta + {\sin ^2}\theta ) = 1 + 1$$ \Rightarrow {r^2} = 2 \Rightarrow r = \sqrt 2 $
$\therefore \sqrt 2 \cos \theta = 1$ and $\sqrt 2 \sin \theta = - 1$
$ \Rightarrow \cos \theta = \frac{1}{{\sqrt 2 }}$ and $\sin \theta = \frac{{ - 1}}{{\sqrt 2 }}$
Since $\sin \theta $ is negative and $\cos \theta $ is positive
$\therefore \theta $ lies in fourth quadrant
$\therefore \theta = \frac{{ - \pi }}{4}$
Hence polar form of z is $\sqrt 2 \left[ {\cos \left( {\frac{{ - \pi }}{4}} \right) + i\sin \left( {\frac{{ - \pi }}{4}} \right)} \right]$
View full question & answer→Question 222 Marks
Find the modulus and the arguments of the complex number:$z = - \sqrt 3 + i$
AnswerHere $z = - \sqrt 3 + i$$ = r(\cos \theta + i\sin \theta )$
$ \Rightarrow r\cos \theta = - \sqrt 3 $ and $r\sin \theta = 1$
Squaring both sides of (i) and adding
${r^2}({\cos ^2}\theta + {\sin ^2}\theta ) = 3 + 1$$ \Rightarrow {r^2} = 4 \Rightarrow r = 2$
$\therefore 2\cos \theta = \frac{{ - \sqrt 3 }}{2}$ and $2\sin \theta = 1$
$ \Rightarrow \cos \theta = \frac{{ - \sqrt 3 }}{2}$ and $\sin \theta = \frac{1}{2}$
Since $\sin \theta $ is positive and $\cos \theta $ is negative
$\therefore \theta $ lies in second quadrant
$\therefore \theta = \left( {\pi - \frac{\pi }{6}} \right) = \frac{{5\pi }}{6}$
$\therefore |z| = 2$ and arg $(z) = \frac{{5\pi }}{6}$
View full question & answer→Question 232 Marks
Find the modulus and the argument of the complex number: $z = - 1 - i\sqrt 3 $
AnswerHere z $ = - 1 - i\sqrt 3 = r(\cos \theta + i\sin \theta )$
$ \Rightarrow r\cos \theta = - 1$ and $r\sin \theta = - \sqrt 3 $
Squaring both sides of (i) and adding
${r^2}({\cos ^2}\theta + {\sin ^2}\theta ) = 1 + 3$$ \Rightarrow {r^2} = 4 \Rightarrow r = 2$
$\therefore 2\cos \theta = - 1$ and $2\sin \theta = - \sqrt 3 $
$ \Rightarrow \cos \theta = \frac{{ - 1}}{2}$ and $\sin \theta = \frac{{ - \sqrt 3 }}{2}$
Since both $\sin \theta $ and $\cos \theta $ are negative.
$\therefore \theta $ lies in third quadrant.
$\therefore \theta = \left( { - \pi + \frac{\pi }{3}} \right) = \frac{{ - 2\pi }}{3}$
$\therefore |z| = 2$ and arg (z) $ = \frac{{ - 2\pi }}{3}$
View full question & answer→Question 242 Marks
Express the complex number ${\left( {\frac{1}{3} + 3i} \right)^3}$ in the form of a + ib.
Answer$\left(\frac13+3i\right)^3\;=\left(\frac13\right)^3+{(3i)^3}+3\times\left(\frac13\right)(3i)\left(\frac13+3i\right)$
$=\frac1{27}+27i^3+i+3i\left(\frac13+3i\right)$
$=\frac1{27}+27(-i)+i+9i^2$$\left[ {\because {i^3} = - i\;and\;{i^2} = - 1} \right]$
$ =\frac1{27}-27i+i-9$
$=\left(\frac1{27}-9\right)-26i\;$
$=\frac{-242}{27}-26i$
View full question & answer→Question 252 Marks
Express the following in the form of a + ib.
$\frac { ( 3 + \sqrt { 5 } i ) ( 3 - \sqrt { 5 } i ) } { ( \sqrt { 3 } + \sqrt { 2 } i ) - ( \sqrt { 3 } - \sqrt { 2 }i ) }$
AnswerWe have, $\frac { ( 3 + \sqrt { 5 }i ) ( 3 - \sqrt { 5 }i ) } { ( \sqrt { 3 } + \sqrt { 2 } i ) - ( \sqrt { 3 } - \sqrt { 2 }i ) }$
= $\frac { 9 - 3 \sqrt { 5 }i+3 \sqrt {5 }i - \sqrt { 5 }i \sqrt { 5 }i } { \sqrt { 3 } + \sqrt { 2 } i - \sqrt { 3 } + \sqrt { 2 } i }$
= $\frac { 9 + 5 } { 2 \sqrt { 2 } i } = \frac { 14 } { 2 \sqrt { 2 } i } = \frac { 7 } { \sqrt { 2 } i } \times \frac { \sqrt { 2 } i } { \sqrt { 2 } i }$
= $\frac { 7 \sqrt { 2 } i } { 2 i ^ { 2 } } = \frac { 7 \sqrt { 2 } i } { - 2 }$ = 0 - i $\frac { 7 \sqrt { 2 } } { 2 }$ = a + ib [say]
where, a = 0 and b = $\frac { - 7 \sqrt { 2 } } { 2 }$
View full question & answer→Question 262 Marks
Find the multiplicative inverse of the complex numbers $ = \sqrt 5 + 3i$
AnswerM.I. of $ = \sqrt 5 + 3i$
$ = \frac{1}{{\sqrt 5 + 3i}} = \frac{1}{{\sqrt 5 + 3i}} \times \frac{{\sqrt 5 - 3i}}{{\sqrt 5 - 3i}}$
$ = \frac{{\sqrt 5 - 3i}}{{{{(\sqrt 5 )}^2} - {{(3i)}^2}}}$
$ = \frac{{\sqrt 5 - 3i}}{{5 - 9{i^2}}} = \frac{{\sqrt 5 - 3i}}{{5 + 9}} = \frac{1}{{14}}(\sqrt 5 - 3i)$
View full question & answer→Question 272 Marks
Find the multiplicative inverse of the complex numbers 4 - 3i
AnswerM.I. of (4 - 3i)$ = \frac{1}{{4 - 3i}} = \frac{1}{{4 - 3i}} \times \frac{{4 + 3i}}{{4 + 3i}} = \frac{{4 + 3i}}{{{{(4)}^2} - {{(3i)}^2}}}$
$ = \frac{{4 + 3i}}{{16 - 9{i^2}}} = \frac{{4 + 3i}}{{16 + 9}} = \frac{1}{{25}}(4 + 3i)$
View full question & answer→Question 282 Marks
Express the complex number ${\left( { - 2 - \frac{1}{3}i} \right)^3}$ in the form of a + ib.
Answer${\left( { - 2 - \frac{1}{3}i} \right)^3}$$ = - {\left( {2 + \frac{1}{3}i} \right)^3}$
$=-\left[{(2)}^3+\left(\frac13i\right)^3+3\times{(2)}^2\times\frac13i\right.$$\left. { + 3 \times 2 \times {{\left( {\frac{1}{3}i} \right)}^2}} \right]$
$=-\left[8+\frac1{27}i^3+4i+\frac23i^2\right]$$ = - \left[ {8 - \frac{1}{{27}}i + 4i - \frac{2}{3}} \right]\left[ {\begin{array}{*{20}{c}} {\because {i^3} = - i} \\ {{i^2} = - 1} \end{array}} \right]$
$ = \left[ {\left( {8 - \frac{2}{3}} \right) + \left( {4 - \frac{1}{{27}}} \right)i} \right.$
$ = - \left[ {\frac{{22}}{3} + \frac{{107}}{{27}}i} \right] = \frac{{ - 22}}{3} - \frac{{107}}{{27}}i$
View full question & answer→Question 292 Marks
Solve $x^2 + 2 = 0$
AnswerWe have $x^2 + 2 = 0$
or $x^2 = -2$
taking squareroot both sides
i.e., x $=\pm \sqrt{-2} = \pm \sqrt2 i$
View full question & answer→Question 302 Marks
Convert the complex number $\frac {-16}{1+\sqrt3i}$ into polar form.
AnswerGiven complex number $\frac {-16}{1+\sqrt3i}$ convert the complex number in x +iy form
$= \frac {-16}{1+\sqrt3i} \times \frac {1-\sqrt3i}{1-\sqrt3i}$
$= \frac {-16(1-\sqrt3i)}{1-(\sqrt3i)^2} = \frac {-16(1-\sqrt3i)}{1+3}$
$= -4 (1 - \sqrt3i) = -4 + 4\sqrt3i$
Let -4 = $r\; cos \;\theta, 4\sqrt3 = r\; sin \;\theta$
By squaring and adding, we get
$16 + 48 = r^2(cos^2 \theta + sin^2 \theta)$
which gives $r^2 = 64, i.e., r = 8$
Hence $cos \theta = - \frac 12, sin \theta = \frac {\sqrt3}2$
$\theta = \pi - \frac {\pi}3 = \frac {2\pi}{3}$
Thus, the required polar form is r(cos$\theta$ + i sin$\theta$) = $8\left(cos \frac{2\pi}{3} + i sin \frac{2\pi}{3} \right)$
View full question & answer→Question 312 Marks
Represent the complex number z = 1 + $i \sqrt { 3 }$ in the polar form.
AnswerWe have, $z=1+i \sqrt{3}$
Let $1+i \sqrt{3}=r(\cos \theta+i \sin \theta)$,
On equating real and imaginary parts both sides, we get $r \cos \theta=1$ and $r \sin \theta=\sqrt{3} \ldots$...(ii)
On squaring and adding Eqs. (i) and (ii), we get
$r^2\left(\cos ^2 \theta+\sin ^2 \theta\right)=1+3$
$\Rightarrow r^2=4$
$\Rightarrow r=2$
$\therefore \cos \theta=\frac{1}{2}$ and $\sin \theta=\frac{\sqrt{3}}{2}$
Since, both $\cos \theta$ and $\sin \theta$ are positive.
So, $\theta$ lies in first quadrant.
$\therefore \theta=\frac{\pi}{3}$
On putting $r=2$ and $\theta=\frac{\pi}{3}$ in Eq. (i), we get polar form of $z=2\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)$
View full question & answer→Question 322 Marks
Express the $i^{-35}$ in the form $a+i b$
Answer$i^{-35}$ $= \frac 1{i^{35}}$ $\frac 1{(i^2)^{17} i}$
$= \frac 1{-i} \times \frac i{i}$ $\frac i{-i^2} $ = i
View full question & answer→Question 332 Marks
Express $\frac { 5 + \sqrt { 2 } i } { 1 - \sqrt { 2 } i }$ in the form of a + ib.
AnswerLet z = $\frac { 5 + \sqrt { 2 } i } { 1 - \sqrt { 2 } i } = \frac { 5 + \sqrt { 2 } i } { 1 - \sqrt { 2 } i } \times \frac { 1 + \sqrt { 2 } i } { 1 + \sqrt { 2 } i }$
[multiplying numerator and denominator by 1 + $\sqrt { 2 } i$]
= $\frac { 5 + 5 \sqrt { 2 } i + \sqrt { 2 } i - 2 } { 1 - ( \sqrt { 2 } i ) ^ { 2 } }$
= $\frac { 3 + 6 \sqrt { 2 } i } { 1 + 2 }$
= $\frac { 3 ( 1 + 2 \sqrt { 2 } i ) } { 3 }$
= 1 + $2 \sqrt { 2 }$ i
View full question & answer→Question 342 Marks
Find the multiplicative inverse of 2 - 3i.
AnswerLet $z=2-3 \mathrm{i}$
Then, $\bar{z}=2+3 \mathrm{i}$
and $|z|^2=2^2+(-3)^2=4+9=13$
Therefore, the multiplicative inverse of $2-3 \mathrm{i}$ is given by $z^{-1}=\frac{\bar{z}}{|z|^2}=\frac{2+3 i}{13}=\frac{2}{13}+\frac{3}{13} i$
View full question & answer→Question 352 Marks
Express $(-\sqrt3 + \sqrt{-2})(2\sqrt3 - i)$ in the form of a + ib.
AnswerLet z= $(-\sqrt3 + \sqrt{-2})(2\sqrt3 - i)$ = $(-\sqrt3 + \sqrt{2i^2})(2\sqrt3 - i)$ = $(-\sqrt3 + \sqrt{2}i)(2\sqrt3 - i)$
z $= -6 + \sqrt3 i + 2\sqrt6 i - \sqrt2 i^2$
z $= (-6 + \sqrt2)+\sqrt3(1 + 2\sqrt2)i$
View full question & answer→Question 362 Marks
Express $(5-3 i)^3$ in the form $a+i b$.
AnswerWe have, $(5-3 i)^3=5^3-3 \times 5^2 \times(3 i)+3 \times 5(3 i)^2-(3 i)^3\left[(a-b)^3=a^3-3 a^2 b+3 b^2 a-b^3\right]$ $=125-225 i-135+27 i=-10-198 i$
View full question & answer→Question 372 Marks
Express the (-i)(2i)$\left (- \frac 18 i\right)^3$ in the form of a + bi:
Answer(-i)(2i)$\left (- \frac 18 i\right)^3$ = $2i^2 \times \frac 1{8\times 8\times 8} \times i^3$ = $2 \times \frac 1{8\times 8\times 8} \times i^5$
$= \frac 1{256} (i^2)^2$i
$i = \frac {1}{256} i$
View full question & answer→Question 382 Marks
Express the $(-5i)\left(\frac 18 i\right)$ in the form of a+bi.
Answer$(-5i)\left(\frac 18 i\right) = \frac {-5}{8} i^2$
$i^2 = -1$
$= \frac {-5}{8} (-1) = \frac 58 = \frac 58 + i0$
View full question & answer→Question 392 Marks
If x + iy = $\frac{a+i b}{a-i b}$, prove that $x^2 + y^2 = 1$
AnswerWe have $x+i y=\frac{(a+i b)}{(a-i b)}$
=> $|x+i y|=|\frac{(a+i b)}{(a-i b)}|$Squaring Both the sides,
=> $|x + iy|^2 =\frac{|(a+i b)|^2}{|(a-i b)|^2}$
=> $x^2 + y^2 =\frac {a^2 + b^2} {a^2 + b^2}$ = 1
View full question & answer→Question 402 Marks
Find the conjugate of $\frac{(3-2 i)(2+3 i)}{(1+2 i)(2-i)}$
AnswerWe have $\frac{(3-2 i)(2+3 i)}{(1+2 i)(2-i)}$
$=\frac{6+9 i-4 i+6}{2-i+4 i+2}=\frac{12+5 i}{4+3 i} \times \frac{4-3 i}{4-3 i}$
$=\frac{48-36 i+20 i+15}{16+9}=\frac{63-16 i}{25}=\frac{63}{25}-\frac{16}{25} i$
Therefore, conjugate of $\frac{(3-2 i)(2+3 i)}{(1+2 i)(2-i)}$is $\frac{63}{25}+\frac{16}{25} i$ [If z = x + iy then then conjugate is x - iy]
View full question & answer→Question 412 Marks
Solve $\sqrt{5} x^{2}+x+\sqrt{5}=0$
AnswerHere, a= $\sqrt 5$ = c and b = 1 the discriminant of the equation is
$1^{2}-4 \times \sqrt{5} \times \sqrt{5}=1-20=-19$
Therefore, the solutions are
= $\frac{-1 \pm \sqrt{-19}}{2 \sqrt{5}}=\frac{-1 \pm \sqrt{19} i}{2 \sqrt{5}}$
View full question & answer→Question 422 Marks
Solve $x^2 + x + 1= 0$
AnswerHere a = 1, b = 1 and c = 1$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$
$b^{2}-4 a c=1^{2}-4 \times 1 \times 1=1-4=-3$
Therefore, the solutions are given by $x=\frac{-1 \pm \sqrt{-3}}{2 \times 1}=\frac{-1 \pm \sqrt{3} i}{2}$
View full question & answer→Question 432 Marks
If 4x + i(3x - y) = 3 + i (-6), where x and y are real numbers, then find the values of x and y.
AnswerWe have,
4x + i (3x - y) = 3 - 6i
$\Rightarrow$ 4x + i (3x - y) = 3 + i (- 6)
On equating real and imaginary parts from both sides, we get
4x = 3 $\Rightarrow$ x = $\frac { 3 } { 4 }$ and 3x - y = - 6
$\Rightarrow$ 3 $\left( \frac { 3 } { 4 } \right)$ - y = - 6 [$\because$ x = $\frac { 3 } { 4 }$]
$\Rightarrow$ $\frac { 9 } { 4 }$ - y = - 6
$\Rightarrow$ y = $\frac { 9 } { 4 } + 6 = \frac { 33 } { 4 }$
$\therefore$ x = $\frac { 3 } { 4 }$ and y = $\frac { 33 } { 4 }$
View full question & answer→