If $a=\cos \theta+i \sin \theta$, then find the value of $\frac{1+a}{1-a}$
Given that, $\quad a=\cos \theta+i \sin \theta$
$$\begin{aligned} \therefore \quad \frac{1+a}{1-a} & =\frac{1+\cos \theta+i \sin \theta}{1-\cos \theta-i \sin \theta} \\ & =\frac{1+2 \cos ^2 \theta / 2-1+2 i \sin \theta / 2 \cdot \cos \theta / 2}{1-1+2 \sin ^2 \theta / 2-2 i \sin \theta / 2 \cdot \cos \theta / 2}=\frac{2 \cos \theta / 2(\cos \theta / 2+i \sin \theta / 2)}{2 \sin \theta / 2(\sin \theta / 2-i \cos \theta / 2)} \\ & =-\frac{2 \cos \theta / 2(\cos \theta / 2+i \sin \theta / 2)}{2 i \sin \theta / 2(\cos \theta / 2+i \sin \theta / 2)}=-\frac{1}{i} \cot \theta / 2 \\ & =\frac{+i^2}{i} \cot \theta / 2=i \cot \theta / 2 \quad\left[\because \frac{-1}{i}=\frac{i^2}{i}\right] \end{aligned}$$
If $(1+i) z=(1-i) \bar{z}$, then show that $z=-i \bar{z}$.
We have, $$(1+i) z=(1-i) \bar{z} \Rightarrow \frac{z}{\bar{z}}=\frac{(1-i)}{(1+i)}$$
$$\begin{array}{ll} \Rightarrow & \frac{z}{\bar{z}}=\frac{(1-i)}{(1+i)} \frac{(1-i)}{(1-i)} \Rightarrow \frac{z}{\bar{z}}=\frac{1+i^2-2 i}{1-i^2} \quad [\because i^2=-1]\\ \Rightarrow & \frac{z}{\bar{z}}=\frac{1-1-2 i}{2} \Rightarrow \frac{z}{\bar{z}}=-i \end{array}$$
$\therefore\quad z=-i\bar{z}$
Hence proved.
If $z=x+i y$, then show that $z \bar{z}+2(z+\bar{z})+b=0$, where $b \in R$, represents a circle.
$$\begin{aligned} & \text { Given that, } \quad z=x+i y \\ & \text { Then, } \quad \bar{z}=x-iy\\ & \text { Now, } \quad z \bar{z}+2(z+\bar{z})+b=0 \\ & \Rightarrow \quad(x+i y)(x-i y)+2(x+i y+x-i y)+b=0 \\ & \Rightarrow \quad x^2+y^2+4 x+b=0,\\ &\text {which is the equation of a circle. } \end{aligned}$$
If the real part of $\frac{\bar{z}+2}{\bar{z}-1}$ is 4 , then show that the locus of the point representing $z$ in the complex plane is a circle.
$$\begin{aligned} \text{Let}\quad z & =x+i y \\ \text{Now,}\quad \frac{\bar{z}+2}{\bar{z}-1} & =\frac{x-i y+2}{x-i y-1} \end{aligned}$$
$$\begin{aligned} & =\frac{[(x+2)-i y][(x-1)+i y]}{[(x-1)-i y][(x-1)+i y]} \\ & =\frac{(x-1)(x+2)-i y(x-1)+i y(x+2)+y^2}{(x-1)^2+y^2} \\ & =\frac{(x-1)(x+2)+y^2+i[(x+2) y-(x-1) y]}{(x-1)^2+y^2} \quad\left[\because-i^2=1\right] \end{aligned}$$
Taking real part, $\quad \frac{(x-1)(x+2)+y^2}{(x-1)^2+y^2}=4$
$\Rightarrow \quad x^2-x+2 x-2+y^2=4\left(x^2-2 x+1+y^2\right)$
$\Rightarrow \quad 3 x^2+3 y^2-9 x+6=0$, which represents a circle.
Hence, $z$ lies on the circle.
Show that the complex number $z$, satisfying the condition arg $\left(\frac{z-1}{z+1}\right)=\frac{\pi}{4}$ lies on a circle.
$$\begin{aligned} & \text { Let } \quad z=x+i y \\ & \text { Given that, } \quad \arg \left(\frac{z-1}{z+1}\right)=\pi / 4 \\ & \Rightarrow \quad \arg (z-1)-\arg (z+1)=\pi / 4 \\ & \Rightarrow \quad \arg (x+i y-1)-\arg (x+i y+1)=\pi / 4 \\ & \Rightarrow \quad \arg (x-1+i y)-\arg (x+1+i y)=\pi / 4 \end{aligned}$$
$$\begin{aligned} & \Rightarrow \quad \tan ^{-1} \frac{y}{x-1}-\tan ^{-1} \frac{y}{x+1}=\pi / 4 \\ & \Rightarrow \quad \tan ^{-1}\left[\frac{\frac{y}{x-1}-\frac{y}{x+1}}{1+\left(\frac{y}{x-1}\right)\left(\frac{y}{x+1}\right)}\right]=\pi / 4 \\ & \Rightarrow \quad \frac{y\left[\frac{x+1-x+1}{x^2-1}\right]}{\frac{x^2-1+y^2}{x^2-1}}=\tan \pi / 4 \\ & \Rightarrow \quad \frac{2 y}{x^2+y^2-1}=1 \\ & \Rightarrow \quad x^2+y^2-1=2 y \\ & \Rightarrow \quad x^2+y^2-2 y-1=0 \text {, which represents a circle. } \end{aligned}$$