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

$=\mathrm{i}^{75}$

$=i^{4 n+3}$ where $n=18$

Since $i^{4 n+3}=-i$

$\mathrm{i}^{75}=-\mathrm{i}$

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

Now , separating real and complex part , we get

0 = rcosθ ……….eq.1

-1 = rsinθ …………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

$\frac{r \sin \theta}{r \cos \theta}=\frac{-1}{0}$

$\tan \theta=-\infty$

Since $\cos \theta=0, \sin \theta=-1$ and $\tan \theta=-\infty$. therefore the $\theta$ lies in fourth quadrant.

$\operatorname{Tan} \theta=-\infty$, therefore $\theta=-\frac{\pi}{2}$

Representing the complex no. in its polar form will be

$\mathrm{Z}=1\left\{\cos \left(-\frac{\pi}{2}\right)+i \sin \left(-\frac{\pi}{2}\right)\right\}$