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})