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COMPLEX NUMBERS AND QUADRATIC EQUATIONS · Gujarat Board (GSEB) STD 11 Science (GSEB - English Medium). 18 questions.

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18 questions · self-marked practice — reveal the answer and mark yourself.

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}{\sqrt2}}$ 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 $\operatorname{Re}\left(z_1 z_2\right)=\operatorname{Re}\left(z_1\right) \operatorname{Re}\left(z_2\right)-\operatorname{Im}\left(z_1\right) \operatorname{Im}\left(z_2\right)$.
Answer
Let $z_1=x_1+i y_1$ and $z_2=x_2+i y_2$
Then, $z_1 z_2=\left(x_1 x_2-y_1 y_2\right)+i\left(x_1 y_2+y_1 x_2\right)$
$\therefore \operatorname{Re}\left(\mathrm{z}_1 \mathrm{z}_2\right)=\mathrm{x}_1 \mathrm{x}_2-\mathrm{y}_1 \mathrm{y}_2$
$=\operatorname{Re}\left(z_1\right) \operatorname{Re}\left(z_2\right)-\operatorname{Im}\left(z_1\right) \operatorname{Im}\left(z_2\right)$
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
$(\mathrm{x}+\mathrm{iy})^3=\mathrm{u}+\mathrm{iv}$
$\Rightarrow x^3+i^3 y^3+3 x^2 y i+3 x y^2 i^2=u+i v$
$\Rightarrow\left(x^3-3 x y^2\right)+\left(3 x^2 y-y^3\right) i=u+i v$
Comparing both sides
$\mathrm{u}=\mathrm{x}\left(\mathrm{x}^2-3 \mathrm{y}^2\right) \text { and } \mathrm{v}=\mathrm{y}\left(3 \mathrm{x}^2-\mathrm{y}^2\right)$
$\text { Now } \frac{u}{x}+\frac{v}{y}=\frac{x\left(x^2-3 y^2\right)}{x}+\frac{y\left(3 x^2-y^2\right)}{y}$
$=\mathrm{x}^2-3 \mathrm{y}^2+3 \mathrm{x}^2-\mathrm{y}^2=4 \mathrm{x}^2-4 \mathrm{y}^2=4\left(\mathrm{x}^2-\mathrm{y}^2\right)$
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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
$=\;\left[\left(-1\right)^9\;+\frac1{i^{24}.i}\right]^3\;=\;\left[-1\;+\frac1{\left(i^2\right)^{12}.i}\right]^3$
$=\;\left[-1\;+\frac1{\left(-1\right)^{12}.i}\right]^3\;=\;\left[-1\;+\frac1{1\times i}\right]^3$
$=\left[-1+\frac1i\right]^3\;=\;\left[-1-i\right]^3$
$=-\left(1\;+i\right)^3\;=\;-\left[1+i^3+3\times1\times i(1+i)\right]$
$=-\left[1-i+3i(1+i)\right]\;=\;-\left[1-i+3i+3i^2\right]$
$=-\left[1-i+3i-3\right]\;=\;-\left[-2+2i\right]$
$=2\;-\;2i$
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Question 173 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 183 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$\theta$]
= $\frac { 3 + 6 i \sin \theta + 2 i \sin \theta - 4 \sin ^ { 2 } \theta } { 1 - 4 ( i ) ^ { 2 } \sin ^ { 2 } \theta }$ $\left[\because(a-b)(a+b)=a^2-b^2\right]$
= $\frac { 3 - 4 \sin ^ { 2 } \theta + 8 i \sin \theta } { 1 + 4 \sin ^ { 2 } \theta }$
= $\frac { 3 - 4 \sin ^ { 2 } \theta } { 1 + 4 \sin ^ { 2 } \theta } + \frac { 8 i \sin \theta } { 1 + 4 \sin ^ { 2 } \theta }$
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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