$$\begin{aligned} & \text { (i) Given, } \\ & \mathrm{z}=i+\sqrt{3}=r(\cos \theta+\sin \theta) \\ & \because \quad r \cos \theta=\sqrt{3}, r \sin \theta=1 \\ & \Rightarrow \quad r^2=1+3=4 \Rightarrow r=2 \quad [\because r > 0]\\ & \Rightarrow \quad \tan \alpha=\left|\frac{r \sin \theta}{r \cos \theta}\right|=\frac{1}{\sqrt{3}} \\ & \Rightarrow \quad \tan \alpha=\frac{1}{\sqrt{3}} \Rightarrow \alpha=\frac{\pi}{6} \\ & \because \quad x>0, y>0 \\ & \text { and } \quad \arg (z)=\theta=\frac{\pi}{6} \end{aligned}$$

So the polar form of $z$ is $2\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)$.

$$\begin{aligned} &\text { (ii) Given that, }\\ &\begin{aligned} z & =-1+\sqrt{-3}=-1+i \sqrt{3} \\ \therefore \quad \tan \alpha & =\left|\frac{\sqrt{3}}{-1}\right|=\sqrt{3} \\ \Rightarrow\quad \tan \alpha & =\tan \frac{\pi}{3} \Rightarrow \alpha=\frac{\pi}{3} \end{aligned} \end{aligned}$$

$$\begin{array}{ll} \because & x<0, y>0 \\ \Rightarrow & \theta=\pi-\alpha=\pi-\frac{\pi}{3}=\frac{2 \pi}{3} \end{array}$$

(iii) Given that, $$|z+2|=|z-2|$$

$$\begin{aligned} \Rightarrow \quad & |x+2+i y| =|x-2+i y| \\ \Rightarrow \quad & (x+2)^2+y^2 =(x-2)^2+y^2 \\ \Rightarrow \quad & x^2+4 x+4 =x^2-4 x+4 \Rightarrow 8 x=0 \\ \therefore \quad & x =0 \end{aligned}$$

It is a straight line which is a perpendicular bisector of segment joining the points $(-2,0)$ and $(2,0)$.

(iv) Given that, $$|z+2 i|=|z-2 i|$$

$$\begin{array}{lrl} \Rightarrow \quad & |x+i(y+2)| =|x+i(y-2)| \\ \Rightarrow \quad & x^2+(y+2)^2 =x^2+(y-2)^2 \\ \Rightarrow \quad & 4 y =0 \Rightarrow y=0 \end{array}$$

It is a straight line, which is a perpendicular bisector of segment joining $(0,-2)$ and $(0,2)$.

(v) Given that,

$$\begin{aligned} |z+4 i| \geq 3 & =|x+i y+4 i| \geq 3 \\ & =|x+i(y+4)| \geq 3 \\ & =\sqrt{x^2+(y+4)^2} \geq 3 \\ & =x^2+(y+4)^2 \geq 9 \\ & =x^2+y^2+8 y+16 \geq 9 \\ & =x^2+y^2+8 y+7 \geq 0 \end{aligned}$$

Which represent a circle. On or outside having centre $(0,-4)$ and radius 3 .

(vi) Given that, $$|z+4| \leq 3$$

$$\begin{array}{lr} \Rightarrow & |x+i y+4| \leq 3 \\ \Rightarrow & |x+4+i y| \leq 3 \\ \Rightarrow & \sqrt{(x+4)^2+y^2} \leq 3 \\ \Rightarrow & (x+4)^2+y^2 \leq 9 \\ \Rightarrow & x^2+8 x+16+y^2 \leq 9 \\ \Rightarrow & x^2+8 x+y^2+7 \leq 0 \end{array}$$

It represent the region which is on or inside the circle having the centre $(-4,0)$ and radius 3 .

(vii) Given that,

$$\begin{aligned} z & =\frac{1+2 i}{1-i}=\frac{(1+2 i)(1+i)}{(1-i)(1+i)} \\ & =\frac{1+2 i+i+2 i^2}{1-i^2}=\frac{1-2+3 i}{1+1}=\frac{-1+3 i}{2} \\ \bar{z} & =\frac{-1}{2}-\frac{3 i}{2} \end{aligned}$$

Hence, $\left(\frac{-1}{2}, \frac{-3}{2}\right)$ lies in third quadrant.

(viii) Given that, $z=1-i$

$$\therefore \quad \frac{1}{z}=\frac{1}{1-i}=\frac{1+i}{(1-i)(1+i)}=\frac{1+i}{1-i^2}=\frac{1}{2}(1+i)$$

Hence, $\left(\frac{1}{2}, \frac{1}{2}\right)$ lies in first quadrant.

Hence, the correct matches are (a) $\rightarrow$ (v), (b) $\rightarrow$ (iii), (c) $\rightarrow$ (i), (d) $\rightarrow$ (iv), (e) $\rightarrow$ (ii), (f) $\rightarrow$ (vi), (g) $\rightarrow$ (viii), (h) $\rightarrow$ (vii)

Given that,

$$ \begin{aligned} z & =\frac{2-i}{(1-2 i)^2}=\frac{2-i}{1+4 i^2-4 i} \\ & =\frac{2-i}{1-4-4 i}=\frac{2-i}{-3-4 i} \\ & =\frac{(2-i)}{-(3+4 i)}=-\left[\frac{(2-i)(3-4 i)}{(3+4 i)(3-4 i)}\right] \\ & =-\left(\frac{6-8 i-3 i+4 i^2}{9+16}\right)=-\frac{(-11 i+2)}{25} \\ & =\frac{-1}{25}(2-11 i) \Rightarrow z=\frac{1}{25}(-2+11 i) \\ \therefore \quad\bar{z} & =\frac{1}{25}(-2-11 i)=\frac{-2}{25}-\frac{11}{25} i \end{aligned}$$

Given that, $$\left|z_1\right|=\left|z_2\right|$$

Let $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$

$$\begin{aligned} \Rightarrow \quad & \left|x_1+i y_1\right| =\left|x_2+i y_2\right| \\ \Rightarrow \quad & x_1^2+y_1^2 =x_2^2+y_2^2 \\ \Rightarrow \quad & x_1^2 =x_2^2, y_1^2=y_2^2 \\ \Rightarrow \quad & x_1 = \pm x_2, y_1= \pm y_2 \\ \Rightarrow \quad & z_1 =x_1+i y_1 \text { or } z_1= \pm x_2 \pm i y_2 \end{aligned}$$

Hence, it is not neccessary that $z_1=z_2$.

Given that, $\quad \frac{\left(a^2+1\right)^2}{2 a-i}=x+i y \Rightarrow \frac{\left(a^2+1\right)^2}{(2 a-i)}=x+i y$

$\begin{aligned} & \Rightarrow \quad \frac{\left(a^2+1\right)^2(2 a+i)}{(2 a-i)(2 a+i)}=x+i y \\ & \Rightarrow \quad \frac{\left(a^2+1\right)^2(2 a+i)}{4 a^2+1}=x+i y\end{aligned}$

$\begin{aligned} \Rightarrow \quad x & =\frac{2 a\left(a^2+1\right)^2}{4 a^2+1} \text { and } y=\frac{\left(a^2+1\right)^2}{4 a^2+1} \\ \therefore \quad x^2+y^2 & =4 a^2\left[\frac{\left(a^2+1\right)^2}{4 a^2+1}\right]^2+\left[\frac{\left(a^2+1\right)^2}{4 a^2+1}\right]^2 \\ & =\frac{\left(4 a^2+1\right)\left(a^2+1\right)^4}{\left(4 a^2+1\right)^2}=\frac{\left(a^2+1\right)^4}{\left(4 a^2+1\right)}\end{aligned}$