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

$=\frac{1+3 i}{1-2 i} \times \frac{1+2 i}{1+2 i}$

$=\frac{1+6 i^{2}+5 i}{1-4 i^{2}}$

$=\frac{5 i-5}{5}$

$=i-1$

Let $Z=1-i=r(\cos \theta+i \sin \theta)$

Now , separating real and complex part , we get

$-1=\operatorname{rcos} \theta \ldots \ldots \ldots . . e q .1$

1 = rsinθ …………eq.2

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

$2=r^{2}$

Since r is always a positive no., therefore,

$r=\sqrt{2}$

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

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

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

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

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

$\operatorname{Tan} \theta=-1$, therefore $\theta=\frac{3 \pi}{4}$

Representing the complex no. in its polar form will be

$\mathrm{Z}=\sqrt{2}\left\{\cos \left(\frac{3 \pi}{4}\right)+i \sin \left(\frac{3 \pi}{4}\right)\right\}$