Find the modulus of each of the following complex numbers and hence

Solution:

Let $Z=2-2 i=r(\cos \theta+i \sin \theta)$

Now , separating real and complex part , we get

$2=r \cos \theta \ldots \ldots \ldots .$ eq. 1

$-2=r \sin \theta \ldots \ldots \ldots \ldots e q \cdot 2$

Squaring and adding eq.1 and eq.2, we get

$8=r^{2}$

 Since r is always a positive no. therefore,

$r=2^{\sqrt{2}}$

Hence its modulus is $2 \sqrt{2}$.

Now, dividing eq.2 by eq.1 , we get

$\frac{r \sin \theta}{r \cos \theta}=\frac{-2}{2}$

$\operatorname{Tan} \theta=-1$

Since $\cos \theta=\frac{1}{\sqrt{2}}, \sin \theta=-\frac{1}{\sqrt{2}}$ and $\tan \theta=-1$. Therefore the $\theta$ lies in the fourth quadrant.

Tan $\theta=-1$, therefore $\theta=-\frac{\pi}{4}$

Representing the complex no. in its polar form will be

$\mathrm{Z}=2 \sqrt{2}\left\{\cos \left(-\frac{\pi}{4}\right)+\mathrm{i} \sin \left(-\frac{\pi}{4}\right)\right\}$