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(cosθ + isinθ)

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

$1=r^{2}$

Since r is always a positive no., therefore,

r = 1,

Hence its modulus is 1

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

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

$\tan \theta=\infty$

Since $\cos \theta=0, \sin \theta=1$ and $\tan \theta=\infty$. Therefore the $\theta$ lies in first quadrant.

$\tan \theta=\infty$, therefore $\theta=\frac{\pi}{2}$

Representing the complex no. in its polar form will be

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