Answer :

$\left[i^{18}+\left(\dfrac{1}{i}\right)^{25}\right]^{3}$ $=\left[i^{4 \times 4+2}+\dfrac{1}{i^{4 \times 6+1}}\right]^{3}$

$\hspace{2.5cm} =\left[\left(i^{4}\right)^{4} \cdot i^{2}+\dfrac{1}{\left(i^{4}\right)^{6} \cdot i}\right]^{3}$

$\hspace{2.5cm} =\left[i^{2}+\dfrac{1}{i}\right]^{3} \qquad\left[\because \ \ i^{4}=1\right]$

$\hspace{2.5cm} =\left[-1+\dfrac{1}{i} \times \dfrac{i}{i}\right]^{3} \qquad\left[\because \ \ i^{2}=-1\right]$

$\hspace{2.5cm} =\left[-1+\dfrac{i}{i^{2}}\right]^{3}$

$\hspace{2.5cm} =\left[-1-i\right]^{3}$

$\hspace{2.5cm} =\left(-1\right)^{3}\left[1+i\right]^{3}$

$\hspace{2.5cm} =-\left[1^{3}+i^{3}+3 \cdot 1 \cdot i\left(1+i\right)\right]$

$\hspace{2.5cm} =-\left[1+i^{3}+3 i+3 i^{2}\right]$

$\hspace{2.5cm} =-\left[1-i+3 i-3\right]$

$\hspace{2.5cm} =-\left[-2+2 i\right]$

$\hspace{2.5cm} =2-2 i$

Answer :

Let $z_1=x_1+i y_1$ and $z_2=x_2+i y_2$

$\therefore \ \ z_1 z_2=\left(x_1+i y_1\right)\left(x_2+i y_2\right)$

$\hspace{1.2cm}=x_1\left(x_2+i y_2\right)+i y_1\left(x_2+i y_2\right)$

$\hspace{1.2cm}=x_1 x_2+i x_1 y_2+i y_1 x_2+i^{2} y_1 y_2$

$\hspace{1.2cm}=x_1 x_2+i x_1 y_2+i y_1 x_2-y_1 y_2 \qquad\left[\because \ \ i^{2}=-1\right]$

$\hspace{1.2cm}=\left(x_1 x_2-y_1 y_2\right)+i\left(x_1 y_2+y_1 x_2\right)$

$\Rightarrow Re\left(z_1 z_2\right)=x_1 x_2-y_1 y_2$

$\Rightarrow Re\left(z_1 z_2\right)=Re z_1 Re z_2-Im z_1 Im z_2$

Hence, proved.

Answer :

$\left(\dfrac{1}{1-4 i}-\dfrac{2}{1+i}\right)\left(\dfrac{3-4 i}{5+i}\right) =\left[\dfrac{\left(1+i\right)-2\left(1-4 i\right)}{\left(1-4 i\right)\left(1+i\right)}\right]\left[\dfrac{3-4 i}{5+i}\right]$

$\hspace{4.4cm} =\left[\dfrac{1+i-2+8 i}{1+i-4 i-4 i^{2}}\right]\left[\dfrac{3-4 i}{5+i}\right]$

$\hspace{4.4cm} =\left[\dfrac{-1+9 i}{5-3 i}\right]\left[\dfrac{3-4 i}{5+i}\right]$

$\hspace{4.4cm} =\left[\dfrac{-3+4 i+27 i-36 i^{2}}{25+5 i-15 i-3 i^{2}}\right]$

$\hspace{4.4cm} =\dfrac{33+31 i}{28-10 i}=\dfrac{33+31 i}{2\left(14-5 i\right)}$

$\hspace{4.4cm} =\dfrac{\left(33+31 i\right)}{2\left(14-5 i\right)} \times \dfrac{\left(14+5 i\right)}{\left(14+5 i\right)} \quad \big[ $ On multiplying numerator and denominator by $\left(14+5 i\right)\big]$

$\hspace{4.4cm} =\dfrac{462+165 i+434 i+155 i^{2}}{2\left[\left(14\right)^{2}-\left(5 i\right)^{2}\right]}$

$\hspace{4.4cm} =\dfrac{307+599 i}{2\left(196-25 i^{2}\right)}$

$\hspace{4.4cm} =\dfrac{307+599 i}{2\left(221\right)}$

$\hspace{4.4cm} =\dfrac{307+599 i}{442}$

$\hspace{4.4cm} =\dfrac{307}{442}+\dfrac{599 i}{442}$

This is the required standard form.

Answer :

$x-i y=\sqrt{\dfrac{a-i b}{c-i d}}$

$\qquad\quad =\sqrt{\dfrac{a-ib}{c-id} \times \dfrac{c+id}{c+id}}\qquad \big[$ On multiplying numerator and deno min ator by $\left(c+id\right)\big]$

$\qquad\quad =\sqrt{\dfrac{\left(ac+bd\right)+i\left(ad-bc\right)}{c^{2}+d^{2}}}$

$\therefore \ \ \left(x-i y\right)^{2}=\dfrac{\left(a c+b d\right)+i\left(a d-b c\right)}{c^{2}+d^{2}}$

$\Rightarrow x^{2}-y^{2}-2 ixy=\dfrac{\left(ac+bd\right)+i\left(ad-bc\right)}{c^{2}+d^{2}}$

On comparing real and imaginary parts, we obtain

$ \ \ x^{2}-y^{2}=\dfrac{a c+b d}{c^{2}+d^{2}},$ $\quad-2 x y=\dfrac{a d-b c}{c^{2}+d^{2}}$

$\left(x^{2}+y^{2}\right)^{2}=\left(x^{2}-y^{2}\right)^{2}+4 x^{2} y^{2}$

$\hspace{1.7cm} =\left(\dfrac{a c+b d}{c^{2}+d^{2}}\right)^{2}+\left(\dfrac{a d-b c}{c^{2}+d^{2}}\right)^{2} \quad\left[\text{Using} \ \left(1\right)\right]$

$\hspace{1.7cm} =\dfrac{a^{2} c^{2}+b^{2} d^{2}+2 a c b d+a^{2} d^{2}+b^{2} c^{2}-2 a d b c}{\left(c^{2}+d^{2}\right)^{2}}$

$\hspace{1.7cm} =\dfrac{a^{2} c^{2}+b^{2} d^{2}+a^{2} d^{2}+b^{2} c^{2}}{\left(c^{2}+d^{2}\right)^{2}}$

$\hspace{1.7cm} =\dfrac{a^{2}\left(c^{2}+d^{2}\right)+b^{2}\left(c^{2}+d^{2}\right)}{\left(c^{2}+d^{2}\right)^{2}}$

$\hspace{1.7cm} =\dfrac{\left(c^{2}+d^{2}\right)\left(a^{2}+b^{2}\right)}{\left(c^{2}+d^{2}\right)^{2}}$

$\hspace{1.7cm} =\dfrac{a^{2}+b^{2}}{c^{2}+d^{2}}$

Hence, proved.

Answer :

$ \begin{aligned} & z_1=2-i, z_2=1+i \\ \\ & \therefore \ \ \bigg|\dfrac{z_1+z_2+1}{z_1-z_2+1}\bigg|=\bigg|\dfrac{\left(2-i\right)+\left(1+i\right)+1}{\left(2-i\right)-\left(1+i\right)+1}\bigg| \\ \\ & \hspace{2.5cm} =\bigg|\dfrac{4}{2-2 i}\bigg| =\bigg|\dfrac{4}{2\left(1-i\right)}\bigg| \\ \\ & \hspace{2.5cm} =\bigg|\dfrac{2}{1-i} \times \dfrac{1+i}{1+i}\bigg| =\bigg|\dfrac{2\left(1+i\right)}{1^{2}-i^{2}}\bigg| \\ \\ & \hspace{2.5cm} =\bigg|\dfrac{2\left(1+i\right)}{1+1}\bigg| \qquad\left[\because \ \ i^{2} =-1\right] \\ \\ & \hspace{2.5cm} =\bigg|\dfrac{2\left(1+i\right)}{2}\bigg| =|1+i|=\sqrt{1^{2}+1^{2}}=\sqrt{2} \end{aligned} $

Thus, the value of $\bigg|\dfrac{z_1+z_2+1}{z_1-z_2+1}\bigg|$ is $\sqrt{2}$.

Answer :

$ \begin{aligned} a+i b & =\dfrac{\left(x+i\right)^2}{2 x^2+1} \\ \\ & =\dfrac{x^2+i^2+2 x i}{2 x^2+1} =\dfrac{x^2-1+i 2 x}{2 x^2+1} \\ \\ & =\dfrac{x^2-1}{2 x^2+1}+i\left(\dfrac{2 x}{2 x^2+1}\right) \end{aligned} $

On comparing real and imaginary parts, we obtain

$ \begin{aligned} & a=\dfrac{x^{2}-1}{2 x^{2}+1} \quad \text{ and } \quad b=\dfrac{2 x}{2 x^{2}+1} \\ \\ & \begin{aligned} \therefore \ \ a^{2}+b^{2} & =\left(\dfrac{x^{2}-1}{2 x^{2}+1}\right)^{2}+\left(\dfrac{2 x}{2 x^{2}+1}\right)^{2} \\ \\ & = \dfrac{x^{4}+1-2 x^{2}+4 x^{2}}{\left(2 x+1\right)^{2}} \\ \\ & =\dfrac{x^{4}+1+2 x^{2}}{\left(2 x^{2}+1\right)^{2}} =\dfrac{\left(x^{2}+1\right)^{2}}{\left(2 x^{2}+1\right)^{2}} \\ \\ \therefore \ \ a^{2}+b^{2} \ & =\dfrac{\left(x^{2}+1\right)^{2}}{\left(2 x^{2}+1\right)^{2}} \end{aligned} \end{aligned} $

Hence, proved.

Answer :

$z_1=2-i, z_2=-2+i$

(i) $\quad z_1 z_2=\left(2-i\right)\left(-2+i\right)=-4+2 i+2 i-i^{2}=-4+4 i-\left(-1\right)=-3+4 i$

$ \qquad\quad\overline{z}_1=2+i $

$\therefore \ \ \dfrac{z_1 z_2}{\overline{z}_1}=\dfrac{-3+4 i}{2+i}$

On multiplying numerator and denominator by $\left( 2 - i\right)$, we obtain

$ \begin{aligned} \dfrac{z_1 z_2}{\bar{z}_1} & =\dfrac{\left(-3+4 i\right)\left(2-i\right)}{\left(2+i\right)\left(2-i\right)} \\ \\ & =\dfrac{-6+3 i+8 i-4 i^{2}}{2^{2}+1^{2}} \\ \\ & =\dfrac{-6+11 i-4\left(-1\right)}{2^{2}+1^{2}} \\ \\ & =\dfrac{-2+11 i}{5}=\dfrac{-2}{5}+\dfrac{11}{5} i \end{aligned} $

On comparing real parts, we obtain

$Re\left(\dfrac{z_1 z_2}{\bar{z}_1}\right)=\dfrac{-2}{5}$

$ \text{(ii)}\quad\dfrac{1}{z_1 \bar{z}_1}=\dfrac{1}{\left(2-i\right)\left(2+i\right)}=\dfrac{1}{\left(2\right)^{2}+\left(1\right)^{2}}=\dfrac{1}{5} $

On comparing imaginary parts, we obtain

$Im\left(\dfrac{1}{z_1 \bar{z}_1}\right)=0$

Answer :

Let $z=\left(x-i y\right)\left(3+5 i\right)$

$\quad \ z=3 x+5 x i-3 y i-5 y i^{2}=3 x+5 x i-3 y i+5 y=\left(3 x+5 y\right)+i\left(5 x-3 y\right)$

$\therefore \ \ \bar{{}z}=\left(3 x+5 y\right)-i\left(5 x-3 y\right)$

It is given that, $\bar{{}z}=-6-24 i$

$\therefore \ \ \left(3 x+5 y\right)-i\left(5 x-3 y\right)=-6-24 i$

Equating real and imaginary parts, we obtain

$3 x+5 y=-6\qquad \ldots(i)$

$5 x-3 y=24\qquad \ldots(ii)$

Multiplying equation $(i)$ by 3 and equation $\left(ii\right)$ by 5 and then adding them, we obtain

$ \begin{aligned} & 9x + 15y = -18 \\ & \underline{25x - 15y = 120} \\ & \qquad\quad 34x = 102 \\ \\
& \quad\therefore \ \ \qquad x = \dfrac{102}{34} = 3 \end{aligned} $

Putting the value of $x$ in equation $\left(i\right),$ we obtain

$ \begin{aligned} & 3\left(3\right)+5 y=-6 \\ \\ & \Rightarrow 5 y=-6-9=-15 \\ \\ & \Rightarrow y=-3 \end{aligned} $

Thus, the values of $x$ and $y$ are $3$ and $- 3$ respectively.

Answer :

$ \begin{aligned} \dfrac{1+i}{1-i}-\dfrac{1-i}{1+i} & =\dfrac{\left(1+i\right)^{2}-\left(1-i\right)^{2}}{\left(1-i\right)\left(1+i\right)} \\ \\ & =\dfrac{1+i^{2}+2 i-1-i^{2}+2 i}{1^{2}+1^{2}} \\ \\ & =\dfrac{4 i}{2}=2 i \\ \\ \therefore \ \ \bigg|\dfrac{1+i}{1-i}-\dfrac{1-i}{1+i}\bigg| & =|2 i|=\sqrt{2^{2}}=2 \end{aligned} $

Answer :

$ \begin{aligned} & \left(x+i y\right)^{3}=u+i v \\ \\ & \Rightarrow x^{3}+\left(i y\right)^{3}+3 \cdot x \cdot i y\left(x+i y\right)=u+i v \\ \\ & \Rightarrow x^{3}+i^{3} y^{3}+3 x^{2} y i+3 x y^{2} i^{2}=u+i v \\ \\ & \Rightarrow x^{3}-i y^{3}+3 x^{2} y i-3 x y^{2}=u+i v \\ \\ & \Rightarrow\left(x^{3}-3 x y^{2}\right)+i\left(3 x^{2} y-y^{3}\right)=u+i v \end{aligned} $

On equating real and imaginary parts, we obtain $u=x^{3}-3 x y^{2}, v=3 x^{2} y-y^{3}$

$ \begin{aligned} \therefore \ \ \ \dfrac{u}{x}+\dfrac{v}{y} & =\dfrac{x^{3}-3 x y^{2}}{x}+\dfrac{3 x^{2} y-y^{3}}{y} \\ \\ & =\dfrac{x\left(x^{2}-3 y^{2}\right)}{x}+\dfrac{y\left(3 x^{2}-y^{2}\right)}{y} \\ \\ & =x^{2}-3 y^{2}+3 x^{2}-y^{2} \\ \\ & =4 x^{2}-4 y^{2} \\ \\ & =4\left(x^{2}-y^{2}\right) \\ \\ \therefore \ \ \dfrac{u}{x}+\dfrac{v}{y} & =4\left(x^{2}-y^{2}\right) \end{aligned} $

Hence, proved.

Answer :

Let $\alpha=a+i b $ and $ \beta=x+i y $

It is given that, $\big|\beta\big|=1$

$\therefore \ \ \sqrt{x^{2}+y^{2}}=1$

$\Rightarrow x^{2}+y^{2}=1$

$ \begin{aligned} \bigg|\dfrac{\beta-\alpha}{1-\bar{{}\alpha} \beta}\bigg| & =\bigg|\dfrac{\left(x+i y\right)-\left(a+i b\right)}{1-\left(a-i b\right)\left(x+i y\right)}\bigg| \\ \\ & = \bigg|\dfrac{\left(x-a\right)+i\left(y-b\right)}{1-\left(a x+a i y-i b x+b y\right)}\bigg| \\ \\ & = \bigg|\dfrac{\left(x-a\right)+i\left(y-b\right)}{\left(1-a x-b y\right)+i\left(b x-a y\right)}\bigg| \\ \\ & = \dfrac{\bigg|\left(x-a\right)+i\left(y-b\right)\bigg|}{|\left(1-a x-b y\right)+i\left(b x-a y\right) \mid} \quad\left[\because \ \ \bigg|\dfrac{z_1}{z_2}\bigg|=\dfrac{\big|z_1\big|}{\big|z_2\big|}\right] \\ \\ & = \dfrac{\sqrt{\left(x-a\right)^{2}+\left(y-b\right)^{2}}}{\sqrt{\left(1-a x-b y\right)^{2}+\left(b x-a y\right)^{2}}} \\ \\ & = \dfrac{\sqrt{x^{2}+a^{2}-2 a x+y^{2}+b^{2}-2 b y}}{\sqrt{1+a^{2} x^{2}+b^{2} y^{2}-2 a x+2 a b x y-2 b y+b^{2} x^{2}+a^{2} y^{2}-2 a b x y}} \\ \\ & = \dfrac{\sqrt{\left(x^{2}+y^{2}\right)+a^{2}+b^{2}-2 a x-2 b y}}{\sqrt{1+a^{2}\left(x^{2}+y^{2}\right)+b^{2}\left(y^{2}+x^{2}\right)-2 a x-2 b y}} \\ \\ & = \dfrac{\sqrt{1+a^{2}+b^{2}-2 a x-2 b y}}{\sqrt{1+a^{2}+b^{2}-2 a x-2 b y}} =1 \\ \\ \therefore \quad \bigg|\dfrac{\beta-\alpha}{1-\bar{{}\alpha} \beta}\bigg| & =1 \end{aligned} $

Answer :

$ \begin{aligned} & \big|1-i\big|^{x}=2^{x} \\ \\ & \Rightarrow\left(\sqrt{1^{2}+\left(-1\right)^{2}}\right)^{x}=2^{x} \\ \\ & \Rightarrow\left(\sqrt{2}\right)^{x}=2^{x} \\ \\ & \Rightarrow 2^{\frac{x}{2}}=2^{x} \\ \\ & \text{On comparing both sides.} \\ \\ & \Rightarrow \dfrac{x}{2}=x \\ \\ & \Rightarrow x=2 x \\ \\ & \Rightarrow 2 x-x=0 \\ \\ & \Rightarrow x=0 \end{aligned} $

Thus, $0$ is the only integral solution of the given equation. Therefore, the number of non-zero integral solutions of the given equation is $0 .$

Answer :

$\left(a+i b\right)\left(c+i d\right)\left(e+i f\right)\left(g+i h\right)=A+i B$

$\therefore \ \ \big|\left(a+i b\right)\left(c+i d\right)\left(e+i f\right)\left(g+i h\right)\big|=\big|A+i B\big|$

$\Rightarrow\big|\left(a+i b\right)\big| \times\left(c+i d\right)\big|\times\big|\left(e+i f\right)\big|\times\big|\left(g+i h\right)\big|=\big| A+i B \mid \qquad\left[\because \ \ \big|z_1 z_2\big|=\big|z_1\big|\big|z_2\big|\right]$

$\Rightarrow \sqrt{a^{2}+b^{2}} \times \sqrt{c^{2}+d^{2}} \times \sqrt{e^{2}+f^{2}} \times \sqrt{g^{2}+h^{2}}=\sqrt{A^{2}+B^{2}}$

On squaring both sides, we obtain

$\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)\left(e^{2}+f^{2}\right)\left(g^{2}+h^{2}\right)=A^{2}+B^{2}$

Hence, proved.

Answer :

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

$\therefore \ \ m=4 k$, where $k$ is some integer.

Therefore, the least positive integer is $1.$

Thus, the least positive integral value of $m$ is $4\left(=4 \times 1\right)$