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,

$\mathrm{r}=\sqrt{2}$

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

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

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

Tanθ = -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

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