(Each question 1 marks)

Question 11 Mark

Let $z_1 = 2 - i, z_2 = -2 + i$. Find $\operatorname{Im} \left( {\frac{1}{{{z_1}{z_1}}}} \right)$

Answer

Here $z_1=2-i$ and $z_2=-2+i$
$\therefore \overline {{z_1}} = 2 + i$$\frac{1}{{{z_1}\overline {{z_1}} }} = \frac{1}{{(2 - i)(2 + i)}} = \frac{1}{{4 - {i^2}}} = \frac{1}{5}$ = $\frac 1 5$ + 0i
$\therefore \operatorname{Im} \left( {\frac{1}{{{z_1}\overline {{z_1}} }}} \right) = 0$

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Question 21 Mark

Express the complex numbers $(1 - i)^4$ in the form of a + ib.

Answer

$(1-i)^4=\left[(1-i)^2\right]^2$
$=\left(1+i^2-2 i\right)^2$
$=(1-1-2 i)^2=(-2 i)^2$
$=4 i^2=4 \times-1=-4$

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Question 31 Mark

Express the complex number$\left[ {\left( {\frac{1}{3} + \frac{7}{3}i} \right) + \left( {4 + \frac{1}{3}i} \right)} \right] - \left[ {\frac{{ - 4}}{3} + i} \right]$ in the form of a + ib.

Answer

$\left[ {\left( {\frac{1}{3} + \frac{7}{3}i} \right) + \left( {4 + \frac{1}{3}i} \right)} \right] - \left[ {\frac{{ - 4}}{3} + i} \right]$
$ =\left(\frac13+4+\frac43\right)+\left(\frac73i+\frac13i\;-i\right)$
$\;=\left(\frac13+\frac{4\times3}3+\frac43\right)+\left(\frac73i\;+\frac i3-\frac{3i}3\right)$
$=\left(\frac{1+12+4}3\right)+\left(\frac{7i+i\;-3i}3\right)$
$ = \frac{{17}}{3} + \frac{5}{3}i$

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Question 41 Mark

Express the complex number $\left( {\frac{1}{5} + \frac{2}{5}i} \right) - \left( {4 + \frac{5}{2}i} \right)$ in the form of a + ib.

Answer

$\left( {\frac{1}{5} + \frac{2}{5}i} \right) - \left( {4 + \frac{5}{2}i} \right)$
$ = \frac{1}{5} + \frac{2}{5}i - 4 - \frac{5}{2}i$
$ = \left( {\frac{1}{5} - 4} \right) + \left( {\frac{2}{5} - \frac{5}{2}} \right)i$
$ = \frac{{ - 19}}{5} - \frac{{21}}{{10}}i$

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Question 51 Mark

Express the complex number (1 + i) - (- 1 + i6) in form of a + ib.

Answer

(1 + i) - (-1 + i6)
1 + i + 1 - 6i = 2 - 5i

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Question 61 Mark

Express the complex numbers 3(7 + i7) + i(7 + i7) in the form of a + ib.

Answer

$3(7+i 7)+i(7+i 7)$
$=21+21 i+7 i+7 i^2=21+28 i-7$
$=14+28 i$

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Question 71 Mark

Express the complex numbers $i^{-39}$ in the form of a + ib.

Answer

$i^{-39}$ = $i^{4 \times -9-3}$
$= (i^4)^{-9} \times i^{-3}$
$=(1)^{-9} \times i^{-3}$
$=\frac {1}{i^3}$
$=\frac {1}{-i} \times \frac {i}{i} $
$= \frac {-i}{i^2} = \frac {-i}{-1} $ = i

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Question 81 Mark

Express the complex numbers $i^9 + i^{19}$ in the form a + ib.

Answer

$i^9 + i^{19} = {({i^2})^4} \cdot i + {({i^2})^9} \cdot i$
$ = {( - 1)^4} \cdot i + {( - 1)^9} \cdot i$
= i - i = 0

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Question 91 Mark

Find the multiplicative inverse of the complex numbers -i

Answer

M.I. of $ - i = \frac{1}{{ - i}} = \frac{i}{{ - {i^2}}} = \frac{i}{{ - (i)}} = i$

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Question 101 Mark

Express the complex numbers $(5i)\left( { - \frac{3}{5}i} \right)$ in the form a + ib.

Answer

$(5i)\left( { - \frac{3}{5}i} \right) = - 3{i^2} = - 3 \times - 1$$(\because {i^2} = - 1)$

= 3 = 3+0i

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Question 111 Mark

Find the modulus and argument of complex number $\frac { 1 } { 1 + i }$.

Answer

We have, $\frac { 1 } { 1 + i } = \frac { 1 } { 1 + i } \times \frac { 1 - i } { 1 - i } = \frac { 1 - i } { 1 ^ { 2 } - i ^ { 2 } } = \frac { 1 - i } { 1 + 1 }$
= $\frac { 1 - i } { 2 } = \frac { 1 } { 2 } - \frac { i } { 2 }$
Let r cos$\theta$ = $\frac { 1 } { 2 }$ ...(i)
and r sin$\theta$ = - $\frac { 1 } { 2 }$ ...(ii)
On squaring and adding Eqs. (i) and (ii), we get
$r^2 \cos ^2 \theta+r^2 \sin ^2 \theta$ = $\left( \frac { 1 } { 2 } \right) ^ { 2 } + \left( - \frac { 1 } { 2 } \right) ^ { 2 }$
$\Rightarrow r^2$ =$\frac { 1 } { 4 } + \frac { 1 } { 4 } = \frac { 1 } { 2 } \Rightarrow$ r = $\frac { 1 } { \sqrt { 2 } }$ [$\because$ r is positive]
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, cos$\theta$ is positive and sin$\theta$ is negative.
So, $\theta$ lies in IV quadrant.
$\therefore$ $\theta$ = - $\frac { \pi } { 4 }$
Hence, modulus of $\frac { 1 } { 1 + i }$ is $\frac { 1 } { \sqrt { 2 } }$ and argument is $\frac { - \pi } { 4 }$.

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Question 121 Mark

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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