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

$=\frac{5-i}{2-3 i} \times \frac{2+3 i}{2+3 i}$

$=\frac{10-3 i^{2}+13 i}{4-9 i^{2}}$

$=\frac{+13 i+13}{13}$

= i + 1

Let Z = 1 + i = r(cosθ + isinθ)

Now , separating real and complex part , we get

1 = rcosθ ……….eq.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 first quadrant.

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

Representing the complex no. in its polar form will be

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