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

Solution:

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

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 fourth quadrant.

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

Representing the complex no. in its polar form will be

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