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

Solution:

Let $Z=-2=r(\cos \theta+i \sin \theta)$

Now, separating real and complex part, we get

-2 = rcosθ………. eq.1

0 = rsinθ …………eq.2

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

$4=r^{2}$

Since r is always a positive no, therefore,

r = 2,

Hence its modulus is 2.

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

$\frac{r \sin \theta}{r \cos \theta}=\frac{0}{-2}$

Tanθ = 0

Since $\cos \theta=-1, \sin \theta=0$ and $\tan \theta=0$. Therefore the $\theta$ lies in second quadrant.

$\operatorname{Tan} \theta=0$, therefore $\theta=\pi$

Representing the complex no. in its polar form will be

$Z=2(\cos \pi+i \sin \pi)$