Given that, $\left(\frac{1+i}{1-i}\right)^3-\left(\frac{1-i}{1+i}\right)^3=x+i y\quad \text{... (i)}$

$$\begin{aligned} \therefore \quad\left(\frac{1+i}{1-i}\right)^3 & =\frac{1+i^3+3 i(1+i)}{1-i^3-3 i(1-i)}=\frac{1-i+3 i+3 i^2}{1+i-3 i+3 i^2} \\ & =\frac{2 i-2}{-2 i-2}=\frac{i-1}{-i-1}=\frac{1-i}{1+i} \\ & =\frac{(1-i)}{(1+i)} \frac{(1-i)}{(1-i)}=\frac{1+i^2-2 i}{1+1}=\frac{1-1-2 i}{2} \\ \Rightarrow \quad\left(\frac{1+i}{1-i}\right)^3 & =-i\quad \text{... (ii)} \end{aligned}$$

Similarly, $\quad\left(\frac{1-i}{1+i}\right)^3=\frac{-1}{i}=\frac{i^2}{i}=i\quad \text{... (iii)}$

Using Eqs. (ii) and (iii) in Eq. (i), we get

$$\begin{aligned} -i-i & =x+i y \\ -2 i & =x+i y \end{aligned}$$

On comparing real and imaginary part of complex number, we get

$x=0$ and $y=-2$

So, $$(x, y)=(0,-2)$$

$$\begin{array}{ll} \text { Given that, } & \frac{(1+i)^2}{2-i}=x+i y \\ \Rightarrow & \frac{\left(1+i^2+2 i\right)}{2-i}=x+i y \Rightarrow \frac{2 i}{2-i}=x+i y \\ \Rightarrow & \frac{2 i(2+i)}{(2-i)(2+i)}=x+i y \Rightarrow \frac{4 i+2 i^2}{4-i^2}=x+i y \end{array}$$

$$\begin{aligned} &\Rightarrow \quad \frac{4 i-2}{4+1}=x+i y \Rightarrow \frac{-2}{5}+\frac{4 i}{5}=x+i y\\ &\text { On comparing both sides, we get }\\ &x=-2 / 5 \Rightarrow y=4 / 5\\ &\Rightarrow \quad x+y=\frac{-2}{5}+\frac{4}{5}=2 / 5 \end{aligned}$$

$$\begin{aligned} & \text { Given that, }\left(\frac{1-i}{1+i}\right)^{100}=a+i b \\ & \Rightarrow \quad\left[\frac{(1-i)}{(1+i)} \cdot \frac{(1-i)}{(1-i)}\right]^{100}=a+i b \Rightarrow\left(\frac{1+i^2-2 i}{1-i^2}\right)^{100}=a+i b \\ & \Rightarrow \quad\left(\frac{-2 i}{2}\right)^{100}=a+i b \quad [\because i^2=-1]\\ & \Rightarrow \quad\left(i^4\right)^{25}=a+i b \Rightarrow 1=a+i b \\ & \text { Then, } \\ & a=1 \text { and } b=0 \quad [\because i^4=1]\\ & \therefore \quad(a, b)=(1,0) \end{aligned}$$

Given that, $\quad a=\cos \theta+i \sin \theta$

$$\begin{aligned} \therefore \quad \frac{1+a}{1-a} & =\frac{1+\cos \theta+i \sin \theta}{1-\cos \theta-i \sin \theta} \\ & =\frac{1+2 \cos ^2 \theta / 2-1+2 i \sin \theta / 2 \cdot \cos \theta / 2}{1-1+2 \sin ^2 \theta / 2-2 i \sin \theta / 2 \cdot \cos \theta / 2}=\frac{2 \cos \theta / 2(\cos \theta / 2+i \sin \theta / 2)}{2 \sin \theta / 2(\sin \theta / 2-i \cos \theta / 2)} \\ & =-\frac{2 \cos \theta / 2(\cos \theta / 2+i \sin \theta / 2)}{2 i \sin \theta / 2(\cos \theta / 2+i \sin \theta / 2)}=-\frac{1}{i} \cot \theta / 2 \\ & =\frac{+i^2}{i} \cot \theta / 2=i \cot \theta / 2 \quad\left[\because \frac{-1}{i}=\frac{i^2}{i}\right] \end{aligned}$$

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.