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

Solution:

Let $Z=4 \sqrt{2} i-4=r(\cos \theta+i \sin \theta)$

Now, separating real and complex part, we get

$-4=\operatorname{rcos} \theta \ldots \ldots \ldots . . \mathrm{eq} .1$

$4 \sqrt{3}=r \sin \theta$

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

$64=r^{2}$

Since r is always a positive no., therefore,

r = 8

Hence its modulus is 8.

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

$\frac{r \sin \theta}{r \cos \theta}=\frac{4 \sqrt{3}}{-4}$

$\operatorname{Tan} \theta=-\frac{\sqrt{3}}{1}$

Since $\cos \theta=-\frac{1}{2} \sin \theta=\frac{\sqrt{3}}{2}$ and $\tan \theta=-\frac{\sqrt{3}}{1}$. Therefore the $\theta$ lies in second the quadrant.

$\operatorname{Tan} \theta=-\sqrt{3}$, therefore $\theta=\frac{2 \pi}{3}$

Representing the complex no. in its polar form will be

$\mathrm{Z}=8\left\{\cos \left(\frac{2 \pi}{3}\right)+i \sin \left(\frac{2 \pi}{3}\right)\right\}$