Chapter 4: Complex Numbers and Quadratic Equations

Sol. (5i)
$
\begin{aligned}
\left(-\frac{3}{5} i\right) & =(5)\left(-\frac{3}{5}\right) i^2=-3(-1)=3\left[\because i^2=-1\right] \\
& =3+i 0
\end{aligned}
$

Sol.
$
\begin{aligned}
i^9+i^{19} & =i^8 \cdot i+i^{16} \cdot i^3=\left(i^4\right)^2 i+\left(i^4\right)^4 i^3=(1)^2 i+(1)^4 \cdot(-i) \\
& =i-i=0=0+0 i
\end{aligned}
$

Sol. $i^{-39}=\frac{1}{i^{39}}=\frac{1}{i^{36} \cdot i^3}=\frac{1}{\left(i^4\right)^9 \cdot i^3}$
$
=\quad \frac{1}{(1)^9 \cdot(-i)}=\frac{1}{-i} \quad\left[\because \quad i^4=1 \text { and } i^3=-i\right]
$

Rationalising the denominator
$
\begin{aligned}
& =\frac{1}{-i} \times \frac{i}{i}=\frac{i}{-i^2}=\frac{i}{1} \quad\left[\because i^2=-1\right] \\
& =i=0+i
\end{aligned}
$

Sol. $3(7+i 7)+i(7+i 7)=21+21 i+7 i+7 i^2$
$
\begin{array}{ll}
=21+28 i-7 & {\left[\because i^2=-1\right]} \\
=14+28 i &
\end{array}
$

Sol. $(1-i)-(-1+i 6)=1-i+1-6 i=2-7 i$.

Sol. $\left(\frac{1}{5}+i \frac{2}{5}\right)-\left(4+i \frac{5}{2}\right)=\frac{1}{5}+\frac{2}{5} i-4-\frac{5}{2} i$
$
\begin{aligned}
& =\left(\frac{1}{5}-4\right)+\left(\frac{2}{5}-\frac{5}{2}\right) i \\
& =\frac{1-20}{5}+\frac{4-25}{10} i \\
& =-\frac{19}{5}-\frac{21}{10} i
\end{aligned}
$

Sol. $\left[\left(\frac{1}{3}+i \frac{7}{3}\right)+\left(4+i \frac{1}{3}\right)\right]-\left(-\frac{4}{3}+i\right)$
$
\begin{aligned}
& \quad=\left(\frac{1}{3}+\frac{7}{3} i+4+\frac{1}{3} i\right)+\frac{4}{3}-i \\
& =\left[\left(\frac{1}{3}+4\right)+\left(\frac{7}{3}+\frac{1}{3}\right) i\right]+\frac{4}{3}-i \\
& =\left(\frac{13}{3}+\frac{8}{3} i\right)+\frac{4}{3}-i=\left(\frac{13}{3}+\frac{4}{3}\right)+\left(\frac{8}{3}-1\right) i \\
& =\frac{17}{3}+\frac{5}{3} i
\end{aligned}
$

Sol.
$
\begin{aligned}
(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=-4+0 i
\end{aligned}
$

Sol. $\left(\frac{1}{3}+3 i\right)^3=\left(\frac{1}{3}\right)^3+(3 i)^3+3 \cdot \frac{1}{3} \cdot 3 i\left(\frac{1}{3}+3 i\right)$
$
\left[\because \quad(a+b)^3=a^3+b^3+3 a b(a+b)\right]
$




$
\begin{array}{ll}
=\frac{1}{27}+27 i^3+3 i\left(\frac{1}{3}+3 i\right) & \\
=\frac{1}{27}-27 i+i+9 i^2 & \quad\left[\because \quad i^3=-i\right] \\
=\frac{1}{27}-26 i-9 & \quad\left[\because \quad i^2=-1\right] \\
=\left(\frac{1}{27}-9\right)-26 i=\frac{1-243}{27}-26 i & \\
=-\frac{242}{27}-26 i . &
\end{array}
$

Sol. $\left(-2-\frac{1}{3} i\right)^3=\left[-\left(2+\frac{1}{3} i\right)\right]^3=-\left(2+\frac{1}{3} i\right)^3$
$
\begin{aligned}
& =-\left[2^3+\left(\frac{1}{3} i\right)^3+3.2 . \frac{1}{3} i\left(2+\frac{1}{3} i\right)\right] \\
& =\left[\because(a+b)^3=a^3+b^3+3 a b(a+b)\right] \\
& =-\left[8+\frac{1}{27} i^3+2 i\left(2+\frac{1}{3} i\right)\right] \\
& =-\left[8+\frac{1}{27} i^3+2 i\left(2+\frac{1}{3} i\right)\right] \\
& =-\left[8-\frac{1}{27} i+4 i+\frac{2}{3} i^2\right] \quad \quad\left[\because \quad i^3=-i\right] \\
& =-\left(8-\frac{1}{27} i+4 i-\frac{2}{3}\right) \quad\left[\because \quad i^2=-1\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
\end{aligned}
$





Find the multiplicative inverse of each of the complex numbers given in the Exercises 11 to 13.

Sol. The multiplicative inverse of $z=4-3 i$ is
$
\begin{aligned}
z^{-1}=\frac{1}{z} & =\frac{1}{4-3 i}=\frac{1}{4-3 i} \times \frac{4+3 i}{4+3 i}=\frac{4+3 i}{4^2-(3 i)^2} \\
& =\frac{4+3 i}{16-9 i^2}=\frac{4+3 i}{16+9}=\frac{4+3 i}{25} \\
& =\frac{4}{25}+\frac{3}{25} i
\end{aligned}
$

Alternative Approach
The multiplicative inverse of $z=4-3 i$ is
$
z^{-1}=\frac{\bar{z}}{|z|^2}=\frac{4+3 i}{(4)^2+(-3)^2}=\frac{4+3 i}{16+9}=\frac{4+3 i}{25}=\frac{4}{25}+\frac{3}{25} i
$
$\left[\because\right.$ If $z=x+i y$, then $\bar{z}=x-i y$ and $|z|=\sqrt{x^2+y^2}$ and hence $\left.|z|^2=x^2+y^2\right]$

Sol. The multiplicative inverse of $z=\sqrt{5}+3 i$ is
$
\begin{aligned}
z^{-1}=\frac{1}{z} & =\frac{1}{\sqrt{5}+3 i}=\frac{1}{\sqrt{5}+3 i} \times \frac{\sqrt{5}-3 i}{\sqrt{5}-3 i} \\
& =\frac{\sqrt{5}-3 i}{(\sqrt{5})^2-(3 i)^2}=\frac{\sqrt{5}-3 i}{5-9 i^2}=\frac{\sqrt{5}-3 i}{5+9} \\
& =\frac{\sqrt{5}-3 i}{14}=\frac{\sqrt{5}}{14}-\frac{3}{14} i
\end{aligned}
$

Alternative Approach
The multiplicative inverse of $z=\sqrt{5}+3 i$ is
$
\begin{aligned}
z^{-1} & =\frac{\bar{z}}{|z|^2}=\frac{\sqrt{5}-3 i}{(\sqrt{5})^2+(3)^2}=\frac{\sqrt{5}-3 i}{5+9=14} \\
& =\frac{\sqrt{5}}{14}-\frac{3 i}{14}
\end{aligned}
$

Sol. The multiplicative inverse of $z=-i$ is
$
\begin{aligned}
z^{-1}=\frac{1}{z} & =\frac{1}{-i}=\frac{1}{-i} \times \frac{i}{i}=\frac{i}{-i^2}=\frac{i}{-(-1)}=i \\
& =0+i
\end{aligned}
$

Alternative Approach
The multiplicative inverse of $z=-i=0-i$ is
$
z^{-1}=\frac{\overline{\boldsymbol{z}}}{|\boldsymbol{z}|^2}=\frac{0+i}{(0)^2+(-1)^2}=\frac{0+i}{1}=0+i
$

Sol. $\frac{(3+i \sqrt{5})(3-i \sqrt{5})}{(\sqrt{3}+\sqrt{2} i)-(\sqrt{3}-i \sqrt{2})}$
Applying the identity $(a+b)(a-b)=a^2-b^2$,
$
\begin{aligned}
& =\frac{3^2-(i \sqrt{5})^2}{\sqrt{3}+\sqrt{2} i-\sqrt{3}+\sqrt{2} i}=\frac{9-5 i^2}{2 \sqrt{2} i} \\
& \quad=\frac{9+5}{2 \sqrt{2} i}=\frac{7}{\sqrt{2} i} \times \frac{-\sqrt{2} i}{-\sqrt{2} i} \quad \text { (Rationalising the denominator) } \\
& \quad=\frac{-7 \sqrt{2} i}{-2 i^2}=\frac{-7 \sqrt{2} i}{2} \quad\left[\because i^2=-1\right] \\
& \quad=0-\frac{7 \sqrt{2}}{2} i
\end{aligned}
$

Sol.
$
\begin{aligned}
& i^{18}= i^{16} i^2=\left(i^4\right)^4 i^2=(1)^4(-1)=-1 \quad\left[\because i^4=1\right. \\
&\text { and } \left.i^2=-1\right] \\
&\left(\frac{1}{i}\right)^{25}= \frac{1}{i^{25}}=\frac{1}{i^{24} \cdot i}=\frac{1}{\left(i^4\right)^6 \cdot i} \\
&= \frac{1}{i}=\frac{1}{i} \times \frac{i}{i} \quad\left[\because \quad\left(i^4\right)^6=1^6=1\right] \\
&= \frac{i}{i^2}=\frac{i}{-1}=-i=0-i \quad\left[\because \quad i^2=-1\right] \\
& \therefore \quad\left[\begin{array}{rl}
\left.i^{18}+\left(\frac{1}{i}\right)^{25}\right]^3 & =(-1-i)^3=[-(1+i)]^3 \\
{\left[\because \quad i^{18}\right.} & \left.=i^{16} \cdot i^2=\left(i^4\right)^4 i^2=(1)^4(-1)=-1 \text { and } \frac{1}{i}=-i\right] \\
= & -(1+i)^3 \\
= & -\left[1^3+i^3+3 \cdot 1 \cdot i(1+i)\right] \\
& =-\left(1-i+(a+b)^3=a^3+b^3+3 a b(a+b)\right. \\
= & -(1+2 i-3) \\
= & -(-2+2 i)=2-2 i .
\end{array}\right.
\end{aligned}
$

Sol. Suppose $z_1=x_1+i y_1$ and $z_2=x_2+i y_2$.
So we have $\operatorname{Re}(z_1)=x_1, \operatorname{Im}(z_1)=y_1$ and $\operatorname{Re}(z_2)=x_2, \operatorname{Im}(z_2)=y_2$.
Computing the product: $z_1 z_2=\left(x_1+i y_1\right)\left(x_2+i y_2\right)=x_1 x_2+i x_1 y_2+i y_1 x_2 +i^2 y_1 y_2$
$
=\left(x_1 x_2-y_1 y_2\right)+i\left(x_1 y_2+y_1 x_2\right) .
$
$
\begin{aligned}
\therefore \quad \operatorname{Re}\left(z_1 z_2\right) & =x_1 x_2-y_1 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)
\end{aligned}
$

Sol. $\left(\frac{1}{1-4 i}-\frac{2}{1+i}\right)\left(\frac{3-4 i}{5+i}\right)$
$
\begin{aligned}
& =\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) \\
& =\frac{(-1+9 i)(3-4 i)}{(5-3 i)(5+i)}=\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} \times \frac{28+10 i}{28+10 i}
\end{aligned}
$
(Rationalising the denominator)
$
\begin{aligned}
& =\frac{924+868 i+330 i+310 i^2}{(28)^2-(10 i)^2} \\
& =\frac{924+1198 i-310}{784-100 i^2} \\
& =\frac{614+1198 i}{784+100}=\frac{2(307+599 i)}{884} \\
& =\frac{307+599 i}{442}=\frac{307}{442}+\frac{599 i}{442}
\end{aligned}
$

Sol. We are given that $x-i y=\sqrt{\frac{a-i b}{c-i d}}$
Squaring both sides gives us
$
(x-i y)^2=\frac{a-i b}{c-i d}
$

Taking the modulus of both sides,
$
\begin{aligned}
& \Rightarrow \quad\left|(x-i y)^2\right|=\left|\frac{a-i b}{c-i d}\right| \\
& \Rightarrow \quad|x-i y|^2=\frac{|a-i b|}{|c-i d|}\left[\because\left|z^2\right|=|z|^2 \text { and }\left|\frac{z_1}{z_2}\right|=\frac{\left|z_1\right|}{\left|z_2\right|}\right] \\
& \Rightarrow \quad\left(\sqrt{x^2+y^2}\right)^2=\frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}} \quad\left[\because|x+i y|=\sqrt{x^2+y^2}\right] \\
& \Rightarrow \quad x^2+y^2=\frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}}
\end{aligned}
$

Squaring both sides gives us
$
\left(x^2+y^2\right)^2=\frac{a^2+b^2}{c^2+d^2}
$

Alternative Approach
Given, $x-i y=\sqrt{\frac{a-i b}{c-i d}}$
Squaring both sides gives us
$
(x-i y)^2=\frac{a-i b}{c-i d}
$

Taking the conjugate of both sides of (i) (i.e., replacing $i$ with $-i$),
$
(x+i y)^2=\frac{a+i b}{c+i d}
$






Multiplying equations (i) and (ii) together,
$
\begin{aligned}
& (x-i y)^2(x+i y)^2=\left(\frac{a-i b}{c-i d}\right)\left(\frac{a+i b}{c+i d}\right) \\
& \text { or } \left.(x-i y)(x+i y)^2=\frac{a^2-i^2 b^2}{c^2-i^2 d^2} \quad \right\rvert\, \because \mathrm{A}^2 \mathrm{~B}^2=(\mathrm{AB})^2 \\
& \text { or }\left(x^2-i^2 y^2\right)^2=\frac{a^2+b^2}{c^2+d^2} \\
& \text { or }\left(x^2+y^2\right)^2=\frac{a^2+b^2}{c^2+d^2}
\end{aligned}
$

Sol. (i) $\frac{1+7 i}{(2-i)^2}=\frac{1+7 i}{4+i^2-4 i}=\frac{1+7 i}{3-4 i} \quad\left[\because \quad i^2=-1\right]$
$
\begin{aligned}
& =\frac{1+7 i}{3-4 i} \times \frac{3+4 i}{3+4 i} \\
& =\frac{3+4 i+21 i+28 i^2}{9-16 i^2}=\frac{3+25 i-28}{9+16} \\
& =\frac{-25+25 i}{25}=-1+i
\end{aligned}
$

The polar form can now be found using the standard procedure for $z = -1 + i$.
(ii) $\frac{1+3 i}{1-2 i}=\frac{1+3 i}{1-2 i} \times \frac{1+2 i}{1+2 i}$
$
\begin{aligned}
& =\frac{1+2 i+3 i+6 i^2}{1-4 i^2}=\frac{1+5 i-6}{1+4} \\
& =\frac{-5+5 i}{5}=-1+i
\end{aligned}
$

Again, the polar form is found using the same method as for $z = -1 + i$.






Solve each of the equations in Exercises 6 to 9.

Sol. For the equation $3 x^2-4 x+\frac{20}{3}=0$
We identify the coefficients:
$
\begin{aligned}
a & =3, \quad b=-4, \quad c=\frac{20}{3} \\
\mathrm{D} & =b^2-4 a c=(-4)^2-4 \times 3 \times \frac{20}{3} \\
& =16-80=-64<0
\end{aligned}
$
$
\begin{aligned}
& \therefore \quad x=\frac{-b \pm \sqrt{D}}{2 a}=\frac{4 \pm \sqrt{-64}}{6} \\
& =\frac{4 \pm \sqrt{64} i}{2 \times 3}=\frac{4 \pm 8 i}{6}=\frac{2}{3}(1 \pm 2 i) .
\end{aligned}
$

Sol. Considering the equation $x^2-2 x+\frac{3}{2}=0$
We identify the coefficients: $\quad a=1, b=-2, c=\frac{3}{2}$
$
\begin{aligned}
\mathrm{D} & =b^2-4 a c=(-2)^2-4 \times 1 \times \frac{3}{2} \\
& =4-6=-2 \\
\therefore \quad x & =\frac{-b \pm \sqrt{\mathrm{D}}}{2 a}=\frac{2 \pm \sqrt{-2}}{2 \times 1} \\
& =\frac{2 \pm \sqrt{2} i}{2}=1 \pm \frac{\sqrt{2}}{2} i
\end{aligned}
$

Sol. For the equation $27 x^2-10 x+1=0$,
We identify the coefficients:
$
\begin{aligned}
a & =27, b=-10, c=1 \\
\mathrm{D} & =b^2-4 a c=(-10)^2-4 \times 27 \times 1 \\
& =100-108=-8
\end{aligned}
$
$
\begin{aligned}
\therefore \quad x & =\frac{-b \pm \sqrt{\mathrm{D}}}{2 a}=\frac{10 \pm \sqrt{-8}}{2 \times 27} \\
& =\frac{10 \pm 2 \sqrt{2} i}{54}=\frac{5}{27} \pm \frac{\sqrt{2}}{27} i
\end{aligned}
$

Sol. For the equation $21 x^2-28 x+10=0$,
We identify the coefficients:
$
\begin{aligned}
a & =21, b=-28, c=10 \\
\mathrm{D} & =b^2-4 a c=(-28)^2-4 \times 21 \times 10 \\
& =784-840=-56 \\
x & =\frac{-b \pm \sqrt{\mathrm{D}}}{2 a}=\frac{28 \pm \sqrt{-56}}{2 \times 21}=\frac{28 \pm \sqrt{4(14)(-1)}}{42} \\
& =\frac{28 \pm 2 \sqrt{14} i}{42}=\frac{28}{42} \pm \frac{2 \sqrt{14}}{42} i \\
& =\frac{2}{3} \pm \frac{\sqrt{14}}{21} i
\end{aligned}
$

Sol. $\quad\left|\frac{z_1+z_2+1}{z_1-z_2+1}\right|=\left|\frac{(2-i)+(1+i)+1}{(2-i)-(1+i)+1}\right|$
$
\begin{aligned}
& =\left|\frac{4}{2-2 i}\right|=\left|\frac{2}{1-i}\right| \\
& =\frac{|2|}{|1-i|}=\frac{|2+i 0|}{|1-i|} \quad\left[\because\left|\frac{z_1}{z_2}\right|=\frac{\left|z_1\right|}{\left|z_2\right|}\right] \\
& =\frac{\sqrt{4+0}}{\sqrt{1+1}}=\frac{2}{\sqrt{2}}=\sqrt{2} .
\end{aligned}
$

Sol. We are given that $\quad a+i b=\frac{(x+i)^2}{2 x^2+1}$
Taking the modulus of both sides,
$
\begin{array}{lrl}
\Rightarrow & |a+i b|=\left|\frac{(x+i)^2}{2 x^2+1}\right| & \\
\Rightarrow & \sqrt{a^2+b^2}=\frac{\left|(x+i)^2\right|}{\left|2 x^2+1\right|} & {\left[\because\left|\frac{z_1}{z_2}\right|=\frac{\left|z_1\right|}{\left|z_2\right|}\right]}
\end{array}
$




$
=\frac{|x+i|^2}{2 x^2+1}
$
$\left[\because\left|z^2\right|=|z|^2\right.$ and $2 x^2+1$ is a positive real number $]$
$
=\frac{\left(\sqrt{x^2+1}\right)^2}{2 x^2+1}
$
$
\Rightarrow \quad \sqrt{a^2+b^2}=\frac{x^2+1}{2 x^2+1}
$

Squaring both sides gives us
$
a^2+b^2=\frac{\left(x^2+1\right)^2}{\left(2 x^2+1\right)^2} .
$

Alternative Approach
Given, $\quad a+i b=\frac{(x+i)^2}{2 x^2+1}$

Taking the conjugate of both sides of (i) (i.e., replacing $i$ with $-i$),
$
a-i b=\frac{(x-i)^2}{2 x^2+1}
$

Multiplying equations (i) and (ii) together,
$
\begin{aligned}
(a+i b)(a-i b) & =\frac{(x+i)^2}{2 x^2+1} \frac{(x-i)^2}{2 x^2+1} \\
\text { or } \mathrm{a}^2-i^2 b^2 & \left.=\frac{\left[(x+i)(x-i)^2\right]}{\left(2 x^2+1\right)^2} \right\rvert\, \because \mathrm{A}^2 \mathrm{~B}^2=(\mathrm{AB})^2 \\
\text { or } \mathrm{a}^2+b^2 & =\frac{\left(x^2-i^2\right)^2}{\left(2 x^2+1\right)^2} \\
\text { or } \mathrm{a}^2+b^2 & =\frac{\left(x^2+1\right)^2}{\left(2 x^2+1\right)^2}
\end{aligned}
$

Sol. (i)
$
\begin{aligned}
& \frac{z_1 z_2}{\bar{z}_1}=\frac{(2-i)(-2+i)}{(2-i)} \\
&=\frac{-4+2 i+2 i-i^2}{2+i}=\frac{-4+4 i+1}{2+i} \\
&=\frac{-3+4 i}{2+i} \times \frac{2-i}{2-i} \quad \text { (Rationalising the denominator) } \\
&=\frac{-6+3 i+8 i-4 i^2}{4-i^2} \\
&=\frac{-6+11 i+4}{4+1}=\frac{-2+11 i}{5} \\
&=-\frac{2}{5}+\frac{11}{5} i . \\
& \therefore \quad \mathrm{R}_e\left(\frac{z_1 z_2}{\bar{z}_1}\right)=\mathrm{R}_e\left(-\frac{2}{5}+\frac{11}{5} i\right)=-\frac{2}{5} . \\
&(i i) \quad \begin{aligned}
\frac{1}{z_1 \bar{z}_1} & =\frac{1}{(2-i)(2+i)}=\frac{1}{4-i^2} \\
& =\frac{1}{4+1}=\frac{1}{5}=\frac{1}{5}+0 i \\
\therefore \quad \operatorname{Im}\left(\frac{1}{z_1 \bar{z}_1}\right) & =\operatorname{Im}\left(\frac{1}{5}+0 i\right)=0 .
\end{aligned}
\end{aligned}
$

Sol. $\quad \frac{1+2 i}{1-3 i}=\frac{1+2 i}{1-3 i} \times \frac{1+3 i}{1+3 i}$
$
\begin{aligned}
& =\frac{1+3 i+2 i+6 i^2}{1-9 i^2} \\
& =\frac{1+5 i-6}{1+9}=\frac{-5+5 i}{10} \\
& =-\frac{1}{2}+\frac{1}{2} i .=x+i y
\end{aligned}
$




This corresponds to the point $\mathrm{P}\left(-\frac{1}{2}, \frac{1}{2}\right)$ in the second quadrant.

We write
$
\begin{aligned}
z & =\frac{1+2 i}{1-3 i}=-\frac{1}{2}+\frac{1}{2} i \\
& =r(\cos \theta+i \sin \theta)
\end{aligned}
$

This gives $r \cos \theta=-\frac{1}{2}=x$ and $r \sin \theta=\frac{1}{2}=y$
Squaring and adding these two equations, we obtain
$
\begin{aligned}
& r^2\left(\cos ^2 \theta+\sin ^2 \theta\right)=\frac{1}{4}+\frac{1}{4} \\
& \Rightarrow \quad r^2=\frac{1}{2} \quad \therefore \quad r=\frac{1}{\sqrt{2}} \\
& \left(\text { or } \quad r=|z|=\sqrt{x^2+y^2}=\sqrt{\left(-\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2}\right. \\
& \left.\quad=\sqrt{\frac{1}{4}+\frac{1}{4}}=\sqrt{\frac{2}{4}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}\right)
\end{aligned}
$

Since the point $\mathrm{P}\left(\frac{-1}{2}, \frac{1}{2}\right)$ lies in the second quadrant, the argument $\arg z$ takes the form $\pi – \theta$.
$
\begin{aligned}
& \tan \theta=\frac{y}{x}=-1=-\tan \frac{\pi}{4}=\tan \left(\pi-\frac{\pi}{4}\right)=\tan \frac{3 \pi}{4} \\
& \Rightarrow \quad \theta=\frac{3 \pi}{4} \\
& \text { Hence, } \quad|z|=r=\frac{1}{\sqrt{2}} \quad \text { and } \quad \arg z=\theta=\frac{3 \pi}{4}
\end{aligned}
$

Sol. We are given that $(x-i y)(3+5 i)=\overline{-6-24 i}$
$
=-6+24 i
$




$
\begin{aligned}
\Rightarrow \quad x-i y & =\frac{-6+24 i}{3+5 i} \\
& =\frac{-6+24 i}{3+5 i} \times \frac{3-5 i}{3-5 i} \quad \text { (Rationalising the denominator) } \\
& =\frac{-18+30 i+72 i-120 i^2}{9-25 i^2} \\
& =\frac{-18+102 i+120}{9+25} \\
& =\frac{102+102 i}{34}=\frac{34(3+3 i)}{34} \\
\Rightarrow \quad x-i y & =3+3 i
\end{aligned}
$

Comparing the real and imaginary parts on both sides, we get
$
\Rightarrow \quad \begin{array}{ll}
x=3 & \text { and }-y=3 \\
\Rightarrow & x=3 \\
\text { and } & y=-3 .
\end{array}
$

Sol. Let $z=\frac{1+i}{1-i}-\frac{1-i}{1+i}$
Finding the common denominator,
$
\begin{aligned}
& =\frac{(1+i)^2-(1-i)^2}{(1-i)(1+i)}=\frac{\left(1+i^2+2 i\right)-\left(1+i^2-2 i\right)}{1-i^2} \\
& =\frac{1-1+2 i-(1-1-2 i)}{1+1} \\
& =\frac{2 i+2 i}{2}=\frac{4 i}{2}=2 i=0+2 i \\
\therefore \quad|z| & =\sqrt{0+4}=2 .
\end{aligned}
$

Sol. We are given that $u+i v=(x+i y)^3$
$
\begin{aligned}
& =x^3+3 x^2 \cdot i y+3 x \cdot(i y)^2+(i y)^3 \\
\mid \because \quad(a+b)^3 & =a^3+b^3+3 a b(a+b) \\
& \left.=a^3+3 a^2 b+3 a b^2+b^3\right]
\end{aligned}
$



$
\begin{aligned}
& =x^3+3 i x^2 y+3 i^2 x y^2+i^3 y^3 \\
& =x^3+3 i x^2 y-3 x y^2-i y^3 \\
& {\left[\because \quad i^2=-1, i^3=-i\right] } \\
\Rightarrow \quad u+i v & =\left(x^3-3 x y^2\right)+i\left(3 x^2 y-y^3\right)
\end{aligned}
$

Comparing the real and imaginary parts on both sides, we get
$
\begin{aligned}
& \Rightarrow \\
& \Rightarrow \\
& \Rightarrow
\end{aligned} \quad \begin{aligned}
u & =x^3-3 x y^2 \text { and } v=3 x^2 y-y^3 \\
& u=x\left(x^2-3 y^2\right) \text { and } v=y\left(3 x^2-y^2\right) \\
\text { Adding } & \frac{u}{x}=x^2-3 y^2 \text { and } \frac{v}{y}=3 x^2-y^2 \\
& \frac{u}{x}+\frac{v}{y}=4 x^2-4 y^2=4\left(x^2-y^2\right) .
\end{aligned}
$

Sol.
$
\begin{array}{ll}
& |\beta|=1 \\
\therefore & |\beta|^2=1 \text { or } \beta \bar{\beta}=1
\end{array}
$
$\left[\because|z|^2=z \bar{z}\right.$ for every complex number $z$.]
$
\therefore \quad \begin{aligned}
\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right| & =\left|\frac{\beta-\alpha}{\beta \bar{\beta}-\bar{\alpha} \beta}\right| \quad \text { [Putting } 1=\beta \bar{\beta} \text { from (ii)] } \\
& =\left|\frac{\beta-\alpha}{\beta(\bar{\beta}-\bar{\alpha})}\right|=\left|\frac{\beta-\alpha}{\beta(\overline{\beta-\alpha)}}\right| \\
& =\frac{|\beta-\alpha|}{|\beta||(\overline{\beta-\alpha})|} \\
& {\left[\because\left|\frac{z_1}{z_2}\right|=\frac{\overline{z_2}}{\left|z_1\right|} \bar{z}_1-z_2\right] } \\
& =\frac{|\beta-\alpha|}{|\beta||(\beta-\alpha)|} \\
& =\frac{1}{|\beta|}=\frac{1}{1} \\
& =1 .
\end{aligned}
$

Sol. We are given that $|1-i|^x=2^x$
$
\begin{array}{rrlrl}
& \Rightarrow & \left(\sqrt{1^2+(-1)^2}\right)^x=2^x & & \\
\Rightarrow & (\sqrt{2})^x=2 & & \Rightarrow & \left(2^{1 / 2}\right)^x=2^x \\
& \Rightarrow & 2^{x / 2}=2^x & & \Rightarrow \\
\Rightarrow & & & \quad \frac{x}{2}=x \\
& & 2 x=x & \Rightarrow & 2 x-x=0
\end{array}
$

⇒ The only solution to equation (i) is $x = 0$.
⇒ Therefore, the given equation has no non-zero integral solution.
∴ Hence, the number of non-zero integral solutions is 0.

Sol. Starting with $\quad(a+i b)(c+i d)(e+i f)(g+i h)=\mathrm{A}+i \mathrm{~B}$
Taking the modulus of both sides,
$
\begin{array}{cc}
\Rightarrow & |(a+i b)(c+i d)(e+i f)(g+i h)|=|\mathrm{A}+i \mathrm{~B}| \\
\Rightarrow & |a+i b||c+i d||e+i f||g+i h|=|\mathrm{A}+i \mathrm{~B}| \\
\Rightarrow & {\left[\because\left|z_1 z_2 \cdots z_n\right|=\left|z_1\right|\left|z_2\right| \cdots\left|z_n\right|\right]} \\
\Rightarrow & \sqrt{a^2+b^2} \cdot \sqrt{c^2+d^2} \cdot \sqrt{e^2+f^2} \cdot \sqrt{g^2+h^2}=\sqrt{\mathrm{A}^2+\mathrm{B}^2}
\end{array}
$

Squaring both sides gives us
$
\left(a^2+b^2\right)\left(c^2+d^2\right)\left(e^2+f^2\right)\left(g^2+h^2\right)=\mathrm{A}^2+\mathrm{B}^2 .
$

Alternative Approach:
We are given that $(a+i b)(c+i d)(e+i f)(g+i h)=\mathrm{A}+i \mathrm{~B}$

Taking conjugates on both sides of ( $i$ ),
$
(a-i b)(c-i d)(e-i f)(g-i h)=\mathrm{A}-i \mathrm{~B}
$

Multiplying equations (i) and (ii) together,
$
\begin{aligned}
& (a+i b)(a-i b)(c+i d)(c-i d)(e+i f)(e-i f)(g+i h) \\
& (g-i h)=(\mathrm{A}+i \mathrm{~B})(\mathrm{A}-i \mathrm{~B}) \\
& \Rightarrow\left(a^2-i^2 b^2\right)\left(c^2-i^2 d^2\right)\left(e^2-i^2 f^2\right)\left(g^2-i^2 h^2\right) \\
& =\mathrm{A}^2-i^2 \mathrm{~B}^2 \\
& \left(a^2+b^2\right)\left(c^2+d^2\right)\left(e^2+f^2\right)\left(g^2+h^2\right)=\mathrm{A}^2+\mathrm{B}^2
\end{aligned}
$

Sol. $\quad\left(\frac{1+i}{1-i}\right)^m=1 \quad \Rightarrow \quad\left(\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right)^m=1$
$
\begin{aligned}
& \Rightarrow\left(\frac{1+i^2+2 i}{1-i^2}\right)^m=1 \\
& \Rightarrow\left(\frac{1-1+2 i}{1+1}\right)^m=1 \\
& \Rightarrow \quad\left(\frac{2 i}{2}\right)^m=1 \quad \Rightarrow i^m=1
\end{aligned}
$

The smallest positive integer $m$ satisfying $i^m = 1$ is $m = 4$
$(\because i^4 = 1)$