Given that, $\sum_\limits{n=1}^{13}\left(i^n+i^{n+1}\right), n \in N$

$$\begin{aligned} = & \left(i+i^2+i^3+i^4+i^5+i^6+i^7+i^8+i^9+i^{10}+i^{11}+i^{12}+i^{13}\right) \\ & \quad+\left(i^2+i^3+i^4+i^5+i^6+i^7+i^8+i^9+i^{10}+i^{11}+i^{12}+i^{13}+i^{14}\right) \\ = & \left(i+2 i^2+2 i^3+2 i^4+2 i^5+2 i^6+2 i^7+2 i^8+2 i^9+2 i^{10}+2 i^{11}+2 i^{12}+2 i^{13}+i^{14}\right) \\ = & i-2-2 i+2+2 i+2\left(i^4\right) i^2+2(i)^4 i^3+2\left(i^2\right)^4+2\left(i^2\right)^4 i+2\left(i^2\right)^5 \\ & +2\left(i^2\right)^5 \cdot i+2\left(i^2\right)^6+2\left(i^2\right)^6 \cdot i+\left(i^2\right)^7 \\ = & i-2-2 i+2+2 i-2-2 i+2+2 i-2-2 i+2+2 i-1-1+i \end{aligned}$$

$$\begin{aligned} &\text { Alternate Method }\\ &\begin{aligned} & \sum_{n-1}^{13}\left(i^n+i^{n+1}\right), n \in N=\sum_{n-1}^{13} i^n(1+i) \\ & \quad=(1+i)\left[i+i^2+i^3+i^4+i^5+i^6+i^7+i^8+i^9+i^{10}+i^{11}+i^{12}+i^{13}\right] \\ & =(1+i)\left[i^{13}\right] \quad[\because i^n+i^{n+1}+i^{n+2}+i^{n+3}=0 \text {, where } n \in N i . e ., \sum_{n=1}^{12} i^n=0 \\ & =(1+i) i \quad [\because (i^4)^3.i=i]\\ & =\left(i^2+i\right)=i-1 \end{aligned} \end{aligned}$$

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

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