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

$=\frac{1-i}{1+i} \times \frac{1-i}{1-i}$

$=\frac{1+i^{2}-2 i}{1-i^{2}}$

$=-\frac{2 i}{2}$

$=-i$

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

Now, separating real and complex part, we get

0 = rcosθ……….eq.1

-1 = rsinθ …………eq.2

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

r = 1,

Hence its modulus is 1.

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

$\frac{r \sin \theta}{r \cos \theta}=\frac{-1}{0}$

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

Since $\cos \theta=0, \sin \theta=-1$ and $\tan \theta=-\infty$, therefore the $\theta$ lies in fourth quadrant.

$\operatorname{Tan} \theta=-\infty$, therefore $\theta=-\frac{\pi}{2}$

Representing the complex no. in its polar form will be

$\mathrm{Z}=1\left\{\cos \left(-\frac{\pi}{2}\right)+i \sin \left(-\frac{\pi}{2}\right)\right\}$