Find the modulus and the arguments of each of the complex numbers
1. $z=-1-i\sqrt{3}$
$|z|=r=\sqrt{(1)^{2}+(\sqrt{3})^{2}}$
= $\sqrt{1+3}$
= $\sqrt{4}$
$r=2$
Let $r\cos \theta =-1$ and $r\sin \theta =-\sqrt{3}$
$\cos \theta =\frac{-1}{r}$  and $\sin \theta =\frac{-\sqrt{3}}{r}$
$\cos \theta =\frac{-1}{2}$  and $\sin \theta =\frac{-\sqrt{3}}{2}$
Both $\sin \theta$  and $\cos \theta$  are negative in IIIrd quadrant
∴ $arg(z)=-(\pi -\frac{\pi }{3})$ $[-\pi<\theta\leq \pi ]$
$arg(z)=-\frac{2\pi}{3}$
2.  $z=-\sqrt{3}+i$ 
$|z|=r=\sqrt{(\sqrt{3})^{2}+(1)^{2}}$
= $\sqrt{3+1}$
= $\sqrt{4}$
$r=2$
Let $r\cos \theta =-\sqrt{3}$ and $r\sin \theta =1$
$\cos \theta =\frac{-\sqrt{3}}{r}$  and $\sin \theta =\frac{1}{r}$
$\cos \theta =\frac{-\sqrt{3}}{2}$  and $\sin \theta =\frac{1}{2}$
$\sin \theta$ is positive and $\cos \theta$  is negative in IInd quadrant
∴ $arg(z)=(\pi -\frac{\pi }{6})$
$arg(z)=\frac{5\pi}{6}$
Convert each of the complex numbers given in the polar form.
3. $1-i$
Let $r\cos \theta =1$ and $r\sin \theta =-1$
On squaring and adding,
$r^{2}\cos^{2}\theta +r^{2}\sin^{2}\theta$ $=1^{2}+(-1)^{2}$
$r^{2}(\cos^{2}\theta +\sin^{2}\theta)$ $=2$
$r^{2}$ $=2$
$r=\sqrt{2}$
$\cos \theta =\frac{1}{r}$  and $\sin \theta =\frac{-1}{r}$
$\cos \theta =\frac{1}{\sqrt{2}}$  and $\sin \theta =\frac{-1}{\sqrt{2}}$
$\sin \theta$ is negative and $\cos \theta$  is positive in IVth quadrant
$\theta =-\frac{\pi}{4}$
$1-i=r\cos\theta+ir\sin\theta$
= $\sqrt{2}\left (\cos(-\frac{\pi}{4})+i\sin(-\frac{\pi}{4}) \right )$
4. $-1+i$
Let $r\cos \theta =-1$ and $r\sin \theta =1$
On squaring and adding,
$r^{2}\cos^{2}\theta +r^{2}\sin^{2}\theta$ $=(-1)^{2}+1^{2}$
$r^{2}(\cos^{2}\theta +\sin^{2}\theta)$ $=2$
$r^{2}$ $=2$
$r=\sqrt{2}$
$\cos \theta =\frac{-1}{r}$  and $\sin \theta =\frac{1}{r}$
$\cos \theta =\frac{-1}{\sqrt{2}}$  and $\sin \theta =\frac{1}{\sqrt{2}}$
$\sin \theta$ is positive and $\cos \theta$  is negative in IInd quadrant
$\theta =\pi-\frac{\pi}{4}=\frac{3\pi}{4}$
$-1+i=r\cos\theta+ir\sin\theta$
= $\sqrt{2}\left (\cos(\frac{3\pi}{4})+i\sin(\frac{3\pi}{4}) \right )$
5. $-1-i$
Let $r\cos \theta =-1$ and $r\sin \theta =-1$
On squaring and adding,
$r^{2}\cos^{2}\theta +r^{2}\sin^{2}\theta$ $=(-1)^{2}+(-1)^{2}$
$r^{2}(\cos^{2}\theta +\sin^{2}\theta)$ $=2$
$r^{2}$ $=2$
$r=\sqrt{2}$
$\cos \theta =\frac{-1}{r}$  and $\sin \theta =\frac{-1}{r}$
$\cos \theta =\frac{-1}{\sqrt{2}}$  and $\sin \theta =\frac{-1}{\sqrt{2}}$
Both $\sin \theta$ and $\cos \theta$  are negative in IIIrd quadrant
$\theta =-(\pi-\frac{\pi}{4})=-\frac{3\pi}{4}$
$-1-i=r\cos\theta+ir\sin\theta$
= $\sqrt{2}\left (\cos(\frac{-3\pi}{4})+i\sin(\frac{-3\pi}{4}) \right )$
6.  $-3$
Let $r\cos \theta =-3$ and $r\sin \theta =0$
On squaring and adding,
$r^{2}\cos^{2}\theta +r^{2}\sin^{2}\theta$ $=(-3)^{2}$
$r^{2}(\cos^{2}\theta +\sin^{2}\theta)$ $=9$
$r^{2}$ $=9$
$r=3$
$\cos \theta =\frac{-3}{r}$  and $\sin \theta =\frac{0}{r}$
$\cos \theta =\frac{-3}{3}$  and $\sin \theta =0$
$\cos \theta =-1$  and $\sin \theta =0$
$\theta =\pi$
$-3=r\cos\theta+ir\sin\theta$
= $3\left (\cos \pi+i\sin \pi \right )$
7. $\sqrt{3}+i$
Let $r\cos \theta =\sqrt{3}$ and $r\sin \theta =1$
On squaring and adding,
$r^{2}\cos^{2}\theta +r^{2}\sin^{2}\theta$ $=(\sqrt{3})^{2}+(1)^{2}$
$r^{2}(\cos^{2}\theta +\sin^{2}\theta)$ $=4$
$r^{2}$ $=4$
$r=2$
$\cos \theta =\frac{\sqrt{3}}{r}$  and $\sin \theta =\frac{1}{r}$
$\cos \theta =\frac{\sqrt{3}}{2}$  and $\sin \theta =\frac{1}{2}$
$\theta =\frac{\pi}{6}$
$\sqrt{3}+i=r\cos\theta+ir\sin\theta$
= $2\left (\cos(\frac{\pi}{6})+i\sin(\frac{\pi}{6}) \right )$
8. $i$
Let $r\cos \theta =0$ and $r\sin \theta =1$
On squaring and adding,
$r^{2}\cos^{2}\theta +r^{2}\sin^{2}\theta$ $=(1)^{2}$
$r^{2}(\cos^{2}\theta +\sin^{2}\theta)$ $=1$
$r^{2}$ $=1$
$r=\sqrt{1}=1$
$\cos \theta =\frac{0}{r}$  and $\sin \theta =\frac{1}{r}$
$\cos \theta =0$  and $\sin \theta =\frac{1}{1}$
$\cos \theta =0$  and $\sin \theta =1$
$\theta =\frac{\pi}{2}$
$i=r\cos\theta+ir\sin\theta$
= $\cos(\frac{\pi}{2})+i\sin(\frac{\pi}{2}) $