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Maths 3 Marks Question

Complex Numbers and Quadratic Equations · Jharkhand Board (JAC) STD 11 Science (Jharkhand - English Medium). 22 questions.

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Question 13 Marks
Solve $21x^2- 28x + 10 = 0$
Answer
$21x^2 - 28x + 10 = 0$
Comparing the given quadratic equation with $ax^2 + bx + c = 0$ we have,
$a= 21, b = -28$ and $c = 10$
$\therefore x = \frac{{ - ( - 28) \pm \sqrt {{{( - 28)}^2} - 4 \times 21 \times 10} }}{{2 \times 21}}$
$ = \frac{{28 \pm \sqrt {784 - 840} }}{{42}}$
$ = \frac{{28 \pm \sqrt { - 56} }}{{42}} = \frac{{28 \pm 2\sqrt {14} i}}{{42}}$$ = \frac{{14 \pm \sqrt {14} i}}{{21}}$
Thus $x = \frac{2}{3} + \frac{{\sqrt {14} }}{{21}}i$ and $x = \frac{2}{3} - \frac{{\sqrt {14} }}{{21}}i$
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Question 23 Marks
Solve: $27x^2 - 10x+ 1 = 0$
Answer
Here $27x^2 - 10x + 1 = 0$
Comparing the given quadratic equation with $ax^2 + bx + c = 0$ we have,
$a = 27, b = - 10$ and $c = 1$
$x = \frac{{ - ( - 10) \pm \sqrt {{{( - 10)}^2} - 4 \times 27 \times 1} }}{{2 \times 27}}$
$ = \frac{{10 \pm \sqrt {100 - 108} }}{{54}} = \frac{{10 \pm 2\sqrt 2 i}}{{54}}$
$ = \frac{{5 \pm \sqrt 2 i}}{{27}}$
Thus $x = \frac{{5 + \sqrt 2 i}}{{27}}$ and $x = \frac{{5 - \sqrt 2 i}}{{27}}$
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Question 33 Marks
Solve: ${x^2} - 2x + \frac{3}{2} = 0$
Answer
Here ${x^2} - 2x + \frac{3}{2} = 0$
Comparing the given quadratic equation with $ax^2 + bx + c = 0$ we have,
a=1,b=-2, $c = \frac{3}{2}$
$\therefore x = \frac{{ - ( - 2) \pm \sqrt {{{( - 2)}^2} - 4 \times 1 \times \frac{3}{2}} }}{{2 \times 1}}$$ = \frac{{2 \pm \sqrt {4 - 6} }}{2} = \frac{{2 \pm \sqrt 2 }}{2}$
$ = \frac{{2 \pm \sqrt 2 i}}{2} = 1 \pm \frac{{\sqrt 2 }}{2}i$
Thus $x = 1 + \frac{{\sqrt 2 }}{2}i$ and $x = 1 - \frac{{\sqrt 2 }}{2}i$
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Question 43 Marks
Solve $3{x^2} - 4x + \frac{{20}}{3} = 0$
Answer
Here $3{x^2} - 4x + \frac{{20}}{3} = 0$
Comparing the given quadratic equation with $ax^2 + bx + c = 0,$ we have
$a = 3, b = -4$ and $c = \frac{{20}}{3}$
$\therefore x = \frac{{ - ( - 4) \pm \sqrt {{{( - 4)}^2} - 4 \times 3 \times \frac{{20}}{3}} }}{{2 \times 3}}$$ = \frac{{4 \pm \sqrt {16 - 80} }}{6}$
$ = \frac{{4 \pm \sqrt { - 64} }}{6} = \frac{{4 \pm 8\sqrt { - 1} }}{6} = \frac{{4 \pm 8i}}{6}$$ = \frac{{2 \pm 4i}}{3}$
Thus $x = \frac{{2 + 4i}}{3}$ and $x = \frac{{2 - 4i}}{3}$
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Question 53 Marks
Convert in the polar form: $\frac{{1 + 3i}}{{1 - 2i}}$
Answer
$\frac{{1 + 3i}}{{1 - 2i}} \times \frac{{1 + 2i}}{{1 + 2i}} = \frac{{1 + 2i + 3i + 6{i^2}}}{{1 - 4{i^2}}}$$ = \frac{{ - 5 + 5i}}{5} = - 1 + i$
Let $z = - 1 + i = r(\cos \theta + i\sin \theta )$
$ \Rightarrow r\cos \theta = - 1$ and $r\sin \theta = 1$
Squaring both sides of (i) and adding
${r^2}({\cos ^2}\theta + {\sin ^2}\theta ) = 1 + 1$$ \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 $\sqrt 2 \left( {\cos \frac{3\pi }{4} + i\sin \frac{{3\pi }}{4}} \right)$
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Question 63 Marks
Convert in the polar form: $\frac{{1 + 7i}}{{{{(2 - i)}^2}}}$
Answer
$\frac{{1 + 7i}}{{{{(2 - i)}^2}}}$$ = \frac{{1 + 7i}}{{4 + {i^2} - 4i}} = \frac{{1 + 7i}}{{3 - 4i}} \times \frac{{3 + 4i}}{{3 + 4i}}$
$ = \frac{{3 + 4i + 21i + 28{i^2}}}{{9 - 16{i^2}}}$
$ = \frac{{ - 25 + 25i}}{{25}} = - 1 + i$
Let $z = - 1 + i = r(\cos \theta + i\sin \theta )$
$ \Rightarrow r\;\cos \theta = - 1$ and $r\sin \theta = 1$ .... (i)
Squaring both sides of (i) and adding
${r^2}({\cos ^2}\theta + {\sin ^2}\theta ) = 1 + 1$$ \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 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 $\sqrt 2 \left( {\cos \frac{{3\pi }}{4} + i\sin \frac{{3\pi }}{4}} \right)$
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Question 73 Marks
If $x - iy = \sqrt {\frac{{a - ib}}{{c - id}}} $ prove that ${({x^2} + {y^2})^2} = \frac{{{a^2} + {b^2}}}{{{c^2} + {d^2}}}$
Answer
Here $x - iy = \sqrt {\frac{{a - ib}}{{c - id}}} $
Squaring both sides, we get
${(x - iy)^2} = \frac{{a - ib}}{{c - id}}$
$ \Rightarrow \left| {{{(x - iy)}^2}} \right| = \left| {\frac{{a - ib}}{{c - id}}} \right|$$ \Rightarrow \left| {(x - iy)} \right|\left| {x - iy} \right| = \left| {\frac{{a - ib}}{{c - id}}} \right|$
$ \Rightarrow \left( {\sqrt {{x^2} + {y^2}} } \right)\left( {\sqrt {{x^2} + {y^2}} } \right)$$ = \frac{{\sqrt {{a^2} + {b^2}} }}{{\sqrt {{c^2} + {d^2}} }} \Rightarrow ({x^2} + {y^2}) = \sqrt {\frac{{{a^2} + {b^2}}}{{{c^2} + {d^2}}}} $
Squaring both sides
${({x^2} + {y^2})^2} = \frac{{{a^2} + {b^2}}}{{{c^2} + {d^2}}}$
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Question 83 Marks
Reduce $\left( \frac { 1 } { 1 - 4 i } - \frac { 2 } { 1 + i } \right) \left( \frac { 3 - 4 i } { 5 + i } \right)$ to the standard form.
Answer
We have, $\left( \frac { 1 } { 1 - 4 i } - \frac { 2 } { 1 + i } \right) \left( \frac { 3 - 4 i } { 5 + i } \right)$
$= \left[ \frac { 1 + i - 2 ( 1 - 4 i ) } { ( 1 - 4 i ) ( 1 + i ) } \right] \left( \frac { 3 - 4 i } { 5 + i } \right)$
$= \left( \frac { 1 + i - 2 + 8 i } { 1 + i - 4 i - 4 i ^ { 2 } } \right) \left( \frac { 3 - 4 i } { 5 + i } \right)$
$= \left( \frac { - 1 + 9 i } { 1 - 3 i + 4 } \right) \left( \frac { 3 - 4 i } { 5 + i } \right)  [ \because i^2 = - 1]$
$=\left( \frac { - 1 + 9 i } { 5 - 3 i } \right) \left( \frac { 3 - 4 i } { 5 + i } \right)$
$= \frac { - 3 + 4 i + 27 i - 36 i ^ { 2 } } { 25 + 5 i - 15 i - 3 i ^ { 2 } }$
$= \frac { - 3 + 31 i + 36 } { 25 - 10 i + 3 } = \frac { 33 + 31 i } { 28 - 10 i }$
$= \frac { ( 33 + 31 i ) } { ( 28 - 10 i ) } \times \frac { ( 28 + 10 i ) } { ( 28 + 10 i ) }$
$[$multiplying numerator and denominator by $28 + 10i]$
$= \frac { 924 + 868 i + 330 i + 310 i ^ { 2 } } { 784 - 100 i ^ { 2 } }$
$= \frac { 924 + 1198 i - 310 } { 784 + 100 }$
$= \frac { 614 + 1198 i } { 884 } = \frac { 307 } { 442 } + \frac { 599 } { 442 }$i
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Question 93 Marks
For any two complex numbers $z_1$ and $z_2,$ prove that $Re (z_1 z_2) = Re (Z_1) Re (z_2) - Im (z_1) Im (z_2).$
Answer
Let $z_1 = x_1 + iy_1 $ and $z_2 = x_2 + iy_2$_
Then, $z_1 z_2 = (x_1 x_2 - y_1y_2) + i (x_1y_2 + y_1 x_2)$
$\therefore Re (z_1 z_2) = x_1 x_2 - y_1 y_2$
$= Re (z_1) Re (z_2) - Im (z_1) Im (Z_2)$
Hence proved.
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Question 103 Marks
If $\alpha $ and $\beta $ are different complex numbers with $\left| \beta \right| = 1$ then find $\left| {\frac{{\beta - \alpha }}{{1 - \overline \alpha \beta }}} \right|$
Answer
Now ${\left| {\frac{{\beta - \alpha }}{{1 - \overline \alpha \beta }}} \right|^2} = \left[ {\frac{{\beta - \alpha }}{{1 - \overline \alpha \beta }}} \right]\left[ {\frac{{\overline {\beta - \alpha } }}{{1 - \overline \alpha \beta }}} \right]$$\left[ {\because {{\left| z \right|}^2} = z\overline z } \right]$
$ = \left[ {\frac{{\beta - \alpha }}{{1 - \overline \alpha \beta }}} \right]\left[ {\frac{{\overline \beta - \overline \alpha }}{{1 - \alpha \overline \beta }}} \right]$
$ = \frac{{\beta \overline \beta - \beta \overline \alpha - \alpha \overline \beta + \alpha \overline \alpha }}{{1 - \overline \alpha \beta - \alpha \overline \beta + \alpha \overline \alpha \beta \overline \beta }}$$ = \frac{{{{\left| \beta \right|}^2} - \alpha \overline \beta - \alpha \overline \beta + {{\left| \alpha \right|}^2}}}{{1 - \overline \alpha \beta - \alpha \overline \beta + {{\left| \alpha \right|}^2}{{\left| \beta \right|}^2}}}$

$ = \frac{{1 - \overline \alpha \beta - \alpha \overline \beta + {{\left| \alpha \right|}^2}}}{{1 - \overline \alpha \beta - \overline \alpha \beta + {{\left| \alpha \right|}^2}}} = 1$
$\therefore \left| {\frac{{\beta - \alpha }}{{1 - \alpha \beta }}} \right| = 1$

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Question 113 Marks
If $(x + iy)^3 = u + iv,$ then show that $\frac{u}{x} + \frac{v}{y} = 4({x^2} - {y^2})$
Answer
$(x + iy)^3 =u + iv$
$ \Rightarrow {x^3} + {i^3}{y^3} + 3{x^2}yi + 3x{y^2}{i^2} = u + iv$
$ \Rightarrow ({x^3} - 3x{y^2}) + (3{x^2}y - {y^3})i = u + iv$
Comparing both sides
$u = x (x^2 - 3y^2)$ and $v = y(3x^2 - y^2)$
Now $\frac{u}{x} + \frac{v}{y} = \frac{{x({x^2} - 3{y^2})}}{x} + \frac{{y(3{x^2} - {y^2})}}{y}$
=$ x^2 - 3y^2 + 3x^2 - y^2 = 4x^2 - 4y^2 = 4(x^2 - y^2)$
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Question 123 Marks
Find the real numbers x and y if (x - iy) (3+ 5i) is the conjugate of -6 - 24i.
Answer
Here $\overline { - 6 - 24i} = - 6 + 24i$
Now (x - iy) (3 + 5i) = -6 + 24i
$ \Rightarrow 3x + 5xi - 3yi - 5y{i^2} = 6 + 24i$
$ \Rightarrow (3x + 5y) + (5x - 3y)i = - 6 + 24i$
Comparing both sides, we have
3x + 5y = -6. . . . (i)
and 5x - 3y = 24 .... (ii)
Multiplying (i) by 3 and (ii) by 5 and then adding
$\begin{gathered} \underline {\begin{array}{*{20}{c}} {9x + 15y = - 18} \\ {25x - 15y = 120} \end{array}} \hfill \\ 34x = 102 \hfill \\ \end{gathered} $$ \Rightarrow x = 3$
Putting x = 3 in (i)
3(3)+5y=-6
Thus y=-3
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Question 133 Marks
Find the modulus and argument of the complex number $\frac{{1 + 2i}}{{1 - 3i}}$.
Answer
Let $z = \frac{{1 + 2i}}{{1 - 3i}} \times \frac{{1 + 3i}}{{1 + 3i}} = \frac{{1 + 3i + 2i + 6{i^2}}}{{1 - 9{i^2}}}$
$ = \frac{{ - 5 + 5i}}{{10}} = \frac{{ - 1 + i}}{2}$
Now $z = \frac{{ - 1}}{2} + \frac{i}{2} = r(\cos \theta + i\sin \theta )$
$\therefore r\;\cos \theta = \frac{{ - 1}}{2}$ and $r\sin \theta = \frac{1}{2}$
Squaring both sides of (i) and adding
${r^2}({\cos ^2}\theta + {\sin ^2}\theta ) = \frac{1}{4} + \frac{1}{4}$$ \Rightarrow {r^2} = \frac{1}{2} \Rightarrow r = \frac{1}{{\sqrt 2 }}$
$\therefore \frac{1}{{\sqrt 2 }}\cos \theta = \frac{{ - 1}}{2}$ and $\frac{1}{{\sqrt 2 }}\sin \theta = \frac{1}{2}$
$ \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
$\therefore \theta = \left( {\pi - \frac{\pi }{4}} \right) = \frac{{3\pi }}{4}$
$\therefore $ Modulus of $z = \frac{1}{{\sqrt 2 }}$ and argument of $z = \frac{{3\pi }}{4}$.
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Question 143 Marks
If a + ib $ = \frac{{{{(x + i)}^2}}}{{2{x^2} + 1}}$, prove that ${a^2} + {b^2} = \frac{{{{({x^2} + 1)}^2}}}{{{{(2{x^2} + 1)}^2}}}$.
Answer
Here a + ib $ = \frac{{{{(x + i)}^2}}}{{2{x^2} + 1}} = \frac{{{x^2} + {i^2} + 2ix}}{{2{x^2} + 1}} = \frac{{{x^2} - 1}}{{2{x^2} + 1}} + i$$\frac{{2x}}{{2{x^2} + 1}}$
Comparing both sides, we have
$a = \frac{{{x^2} - 1}}{{2{x^2} + 1}}$ amd $b = \frac{{2x}}{{2{x^2} + 1}}$
$\therefore {a^2} + {b^2} = \left( {\frac{{{x^2} - 1}}{{2{x^2} + 1}}} \right)^2 + {\left( {\frac{{2x}}{{2{x^2} + 1}}} \right)^2}$
$ = \frac{{{{({x^2} - 1)}^2}}}{{{{(2{x^2} + 1)}^2}}} + \frac{{{{(2x)}^2}}}{{{{(2{x^2} + 1)}^2}}}$
$ = \frac{{{{({x^2} - 1)}^2} + {{(2x)}^2}}}{{{{(2{x^2} + 1)}^2}}}$$ = \frac{{{x^4} + 1 - 2{x^2} + 4{x^2}}}{{{{(2{x^2} + 1)}^2}}}$
$ = \frac{{{x^4} + 1 + 2{x^2}}}{{{{(2{x^2} + 1)}^2}}} = \frac{{{{({x^2} + 1)}^2}}}{{{{(2{x^2} + 1)}^2}}}$
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Question 153 Marks
If $z_1 = 2 - i, z_2 = 1 + i,$ find ${\left|\frac{z_1+z_2+1}{z_1-z_2+1}\right|}$
Answer
Here $z_1 = 2 - i$ and $z_2 = 1 + i$
$\therefore \left| {\frac{{{z_1} + {z_2} + 1}}{{{z_1} - {z_2} + 1}}} \right| = \left| {\frac{{2 - i + 1 + i + 1}}{{2 - i - 1 - i + 1}}} \right|$$ = \left| {\frac{4}{{2 - 2i}}} \right| = \frac{{\left| 4 \right|}}{{\left| {2 - 2i} \right|}}$
$ = \frac{4}{{\sqrt {{{(2)}^2} + {{( - 2)}^2}} }} = \frac{4}{{\sqrt {4 + 4} }} = \frac{4}{{\sqrt 8 }}$$ = \frac{4}{{2\sqrt 2 }} = \sqrt 2 $
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Question 163 Marks
Evaluate ${\left[ {{i^{18}} + {{\left( {\frac{1}{i}} \right)}^{25}}} \right]^3}$
Answer
$=\left[(-1)^9+\frac{1}{i^{24} \cdot i}\right]^3=\left[-1+\frac{1}{\left(i^2\right)^{12} \cdot i}\right]^3$
$=\left[-1+\frac{1}{(-1)^{12} \cdot i}\right]^3=\left[-1+\frac{1}{1 \times i}\right]^3$
$=\left[-1+\frac{1}{i}\right]^3=[-1-i]^3$
$=-(1+i)^3=-\left[1+i^3+3 \times 1 \times i(1+i)\right]$
$=-[1-i+3 i(1+i)]=-\left[1-i+3 i+3 i^2\right]$
$=-[1-i+3 i-3]=-[-2+2 i]$
$=2-2 i$
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Question 173 Marks
Express the $i^{–35}$ in the form $a + ib$
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
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Question 183 Marks
Express $\frac { 5 + \sqrt { 2 } i } { 1 - \sqrt { 2 } i }$ in the form of a + ib.
Answer
Let 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
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Question 193 Marks
Write the complex number $ z = \frac { i - 1 } { \cos \frac { \pi } { 3 } + i \sin \frac { \pi } { 3 } }$ in the polar form.
Answer
We have, $z = \frac { i - 1 } { \cos \frac { \pi } { 3 } + i \sin \frac { \pi } { 3 } }$
Let, $-1 + i = r (\cos \theta + i \sin \theta)$
$\Rightarrow r \cos \theta = - 1 ...(i)$
and $r \sin \theta = 1 ...(ii)$
On squaring and adding Eqs. $(i)$ and $(ii)$, we get
$r^2 (\cos^2 \theta + \sin^2 \theta) = 1 + 1$
$\Rightarrow r^2 = 2$
$\therefore r = \sqrt { 2 }[$ taking positive square root$]$
On putting the value of r in Eqs. $(i)$ and $(ii),$ we get
$\cos \theta = \frac { - 1 } { \sqrt { 2 } }$ and $\sin \theta = \frac { 1 } { \sqrt { 2 } }$
Since, $\sin \theta$ is positive and \cos\theta is negative.
So, $\theta$ lies in II quadrant.
$\therefore \theta = \left( \pi - \frac { \pi } { 4 } \right) = \frac { 3 \pi } { 4 }$
$\Rightarrow i - 1 = \sqrt { 2 } \left( \cos \frac { 3 \pi } { 4 } + i \sin \frac { 3 \pi } { 4 } \right)$
$\therefore z = \frac { i - 1 } { \left( \cos \frac { \pi } { 3 } + i \sin \frac { \pi } { 3 } \right) } = \frac { \sqrt { 2 } \left( \cos \frac { 3 \pi } { 4 } + i \sin \frac { 3 \pi } { 4 } \right) } { \left( \cos \frac { \pi } { 3 } + i \sin \frac { \pi } { 3 } \right) }$
$= \frac { \sqrt { 2 } \left( \cos \frac { 3 \pi } { 4 } + i \sin \frac { 3 \pi } { 4 } \right) } { \left( \cos \frac { \pi } { 3 } + i \sin \frac { \pi } { 3 } \right) } \times \frac { \left( \cos \frac { \pi } { 3 } - i \sin \frac { \pi } { 3 } \right) } { \left( \cos \frac { \pi } { 3 } - i \sin \frac { \pi } { 3 } \right) } [$multiplying numerator and denominator by $\left( \cos \frac { \pi } { 3 } - i \sin \frac { \pi } { 3 } \right)]$
$ = \frac{{\sqrt 2\left[ { \left( {\cos \frac{{3\pi }}{4} \cdot \cos \frac{\pi }{3} + \sin \frac{{3\pi }}{4} \cdot \sin \frac{\pi }{3}} \right)} \right. + i\left( {\sin \frac{{3\pi }}{4} \cdot \cos \frac{\pi }{3} - \cos \frac{{3\pi }}{4} \cdot \sin \frac{\pi }{3}} \right)]}}{{\left( {{{\cos }^2}\frac{\pi }{3} + {{\sin }^2}\frac{\pi }{3}} \right)}}$
$= \frac { \sqrt { 2 } \left[ \cos \left( \frac { 3 \pi } { 4 } - \frac { \pi } { 3 } \right) + i \sin \left( \frac { 3 \pi } { 4 } - \frac { \pi } { 3 } \right) \right] } { 1 }$
$= \sqrt { 2 } \left[ \cos \frac { 5 \pi } { 12 } + i \sin \frac { 5 \pi } { 12 } \right]$
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Question 203 Marks
Find real $\theta$ such that $\frac { 3 + 2 i \sin \theta } { 1 - 2 i \sin \theta }$ is purely real.
Answer
We have, $\frac { 3 + 2 i \sin \theta } { 1 - 2 i \sin \theta } = \frac { 3 + 2 i \sin \theta } { 1 - 2 i \sin \theta } \times \frac { 1 + 2 i \sin \theta } { 1 + 2 i \sin \theta }$
$[$multiplying numerator and denominator by $1 + 2i \sin \theta]$
$= \frac { 3 + 6 i \sin \theta + 2 i \sin \theta - 4 \sin ^ { 2 } \theta } { 1 - 4 ( i ) ^ { 2 } \sin ^ { 2 } \theta }  [ \because(a - b) (a + b) = a^2 - b^2]$
$= \frac { 3 - 4 \sin ^ { 2 } \theta + 8 i \ \sin \theta } { 1 + 4 \sin ^ { 2 } \theta }$
$= \frac { 3 - 4 \sin ^ { 2 } \theta } { 1 + 4 \sin ^ { 2 } \theta } + \frac { 8 i \sin \theta } { 1 + 4 \sin ^ { 2 } \theta }$
We are given the complex number to be real.
$\therefore \frac { 8 \sin \theta } { 1 + 4 \sin ^ { 2 } \theta } = 0$
i.e., $\sin \theta = 0$
Thus, $\theta = n \pi , n \in Z$
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Question 213 Marks
If $x + iy = \frac{a+i b}{a-i b}$, prove that $x^2 + y^2 = 1$
Answer
We 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 $
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Question 223 Marks
Find the modulus and argument of the complex number $\frac{1+i}{1-i}$
Answer
We have $\frac{1+i}{1-i}=\frac{1+i}{1-i} \times \frac{1+i}{1+i}=\frac{1-1+2 i}{1+1} = i = 0 + i$
Now, let us put $0 = r \cos \theta , 1 = r \sin  \theta$
Squaring and adding, $r^2 = 1$ i.e., $r = 1$ so that
$\cos \theta = 0, \sin \theta = 1$
Therefore, $\theta = \frac {\pi}{2}$
Hence, the modulus of $\frac {1+i}{1-i}$ is $1$ and the argument is $\frac {\pi}2.$
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