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

Solution:

Let $Z=3 \sqrt{2} i-3 \sqrt{2}=r\left(\cos ^{\theta}+i \sin \theta\right)$

Now, separating real and complex part, we get

$-3 \sqrt{2}=r \cos \theta$

$3 \sqrt{2}=r \sin \theta$

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

$36=r^{2}$

Since r is always a positive no., therefore,

r = 6

Hence its modulus is 6.

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

$\frac{r \sin \theta}{r \cos \theta}=\frac{3 \sqrt{2}}{-3 \sqrt{2}}$

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

Since $\cos \theta=-\frac{1}{\sqrt{2}}, \sin \theta=\frac{1}{\sqrt{2}}$ and $\tan \theta=-1$. therefore the $\theta$ lies in secothe nd quadrant.

$\operatorname{Tan} \theta=-1$, therefore $\theta=\frac{3 \pi}{4}$

Representing the complex no. in its polar form will be

$\mathrm{Z}=6\left\{\cos \left(\frac{3 \pi}{4}\right)+i \sin \left(\frac{3 \pi}{4}\right)\right\}$