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

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

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

$=\frac{1+i}{\sqrt{2}}$

$=\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}$

Let $Z=\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}=r(\cos \theta+i \sin \theta)$

Now, separating real and complex part, we get

$\frac{1}{\sqrt{2}}=r \cos \theta$ ……….eq.1

$\frac{1}{\sqrt{2}}=r \sin \theta$ …………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,

$\frac{r \sin \theta}{r \cos \theta}=\frac{\frac{i}{\sqrt{2}}}{\frac{i}{\sqrt{2}}}$

tanθ = 1

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

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

Representing the complex no. in its polar form will be

$\mathrm{Z}=1\left\{\cos \left(\frac{\pi}{4}\right)+i \sin \left(\frac{\pi}{4}\right)\right\}$