(i) $$\begin{aligned} &\left|a z_1-b z_2\right|^2+\left|b z_1+a z_2\right|^2 \\ &=\left|a z_1\right|^2+\left|b z_2\right|^2-2 \operatorname{Re}\left(a z_1 \cdot b \bar{z}_2\right)+\left|b z_1\right|^2+\left|a z_2\right|^2+2 \operatorname{Re}\left(a z_1 \cdot b \bar{z}_2\right) \\ &=\left(a^2+b^2\right)\left|z_1\right|^2+\left(a^2+b^2\right)\left|z_2\right|^2 \\ &=\left(a^2+b^2\right)\left(\left|z_1\right|^2+\left|z_2\right|^2\right) \end{aligned}$$

(ii) $\sqrt{-25} \times \sqrt{-9}=i \sqrt{25} \times i \sqrt{9}=i^2(5 \times 3)=-15$

$$\text { (iii) } \begin{aligned} \frac{(1-i)^3}{1-i^3} & =\frac{(1-i)^3}{(1-i)\left(1+i+i^2\right)} \\ & =\frac{(1-i)^2}{i}=\frac{1+i^2-2 i}{i}=\frac{-2 i}{i}=-2 \end{aligned}$$

(iv) $i+i^2+i^3+\ldots$ upto 1000 terms $=i+i^2+i^3+i^4+\ldots i^{1000}=0$ $\left[\because i^n+i^{n+1}+i^{n+2}+i^{n+3}=0\right.$, where $n \in$ Ni.e., $\left.\sum_{n=1}^{1000} i^n=0\right]$

(v) Multiplicative inverse of $1+i=\frac{1}{1+i}=\frac{1-i}{1-i^2}=\frac{1}{2}(1-i)$

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

$z_1+z_2=\left(x_1+x_2\right)+i\left(y_1+y_2\right)$, which is real.

If $z_1+z_2$ is real, then $y_1+y_2=0$

$$\begin{array}{ll} \Rightarrow & y_1=-y_2 \\ \because & z_2=x_2-i y_1 \\ \Rightarrow & z_2=\bar{z}_1\quad [\text{when } x_1=x_2] \end{array}$$

(vii) $$\begin{aligned} & \arg (z)+\arg (\bar{z}),(\bar{z} \neq 0) \\ & \Rightarrow \quad \theta+(-\theta)=0 \end{aligned}$$

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

For the greatest value of $|z+1|$.

$$\begin{aligned} \Rightarrow \quad |z+1| & =|z+4-3| \leq|z+4|+|-3| \\ & =|z+4-3| \leq 3+3 \\ & =|z+4-3| \leq 6 \end{aligned}$$

So, greatest value of $|z+1|=6$

For, now, least value of $|z+1|$, we know that the least value of the modulus of a complex number is zero. So, the least value of $|z+1|$ is zero.

$$\begin{aligned} & \text { (ix) Given that, } \\ & \left|\frac{z-2}{z+2}\right|=\frac{\pi}{6} \\ & \Rightarrow \quad \frac{|x+i y-2|}{|x+i y+2|}=\frac{\pi}{6} \Rightarrow \frac{|x-2+i y|}{|x+2+i y|}=\frac{\pi}{6} \\ & \Rightarrow \quad 6|x-2+i y|=\pi|x+2+i y| \\ & \Rightarrow \quad 6 \sqrt{(x-2)^2+y^2}=\pi \sqrt{(x+2)^2+y^2} \\ & \Rightarrow \quad 36\left[x^2+4-4 x+y^2\right]=\pi^2\left[x^2+4 x+4+y^2\right] \\ & \Rightarrow\left(36-\pi^2\right) x^2+\left(36-\pi^2\right) y^2-\left(144+4 \pi^2\right) x+144+4 \pi^2=0 \text {, which is a circle. } \end{aligned}$$

(x) Given that, $|z|=4$ and $\arg (z)=\frac{5 \pi}{6}$

$$\begin{aligned} & \text { Let } \quad z=x+i y=r(\cos \theta+i \sin \theta) \\ & \Rightarrow \quad|z|=r=4 \text { and } \arg (z)=\theta \\ & \because \quad \tan \theta=\frac{5 \pi}{6} \\ & \Rightarrow \quad z=4\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)=4[\cos (\pi-\pi / 6)+i \sin (\pi-\pi / 6)] \\ & =4\left[-\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right]=4\left[\frac{-\sqrt{3}}{2}+i \frac{1}{2}\right]=-2 \sqrt{3}+2 i \end{aligned}$$