# Question 56a98

Sep 10, 2016

$- \left(2 + 2 \sqrt{3} i\right) , \mathmr{and} , = - 2 \left(1 + \sqrt{3} i\right)$

#### Explanation:

To convert $z = x + i y \in \mathbb{C}$ into polar /trigo. form, we have to find

$r > 0 , \mathmr{and} \theta \in \left(- \pi , \pi\right]$ such that, $x = r \cos \theta , y = r \sin \theta$.

Condsider, ${z}_{1} = \sqrt{3} + i$

$\therefore r \cos \theta = x = \sqrt{3} , r \sin \theta = y = 1 \Rightarrow {x}^{2} + {y}^{2} = {r}^{2} = 3 + 1 = 4$

$\therefore r = 2.$

"Then, rcostheta=2costheta=sqrt3 rArr costheta=sqrt3/2 >0, and,

sintheta=1/2 >0". We conclude that, "theta in (0,pi/2), &, theta=pi/6.

Altogether, ${z}_{1} = 2 \left(\cos \left(\frac{\pi}{6}\right) + i \sin \left(\frac{\pi}{6}\right)\right) \ldots \ldots \ldots \ldots \ldots \ldots . . \left(1\right)$.

Similarly, we can show that, ${z}_{2} = 2 \left(\cos \left(- 5 \frac{\pi}{6}\right) + i \sin \left(- 5 \frac{\pi}{6}\right)\right) \ldots \ldots \ldots \left(2\right)$.

Now, it can easily proved that,

${z}_{j} = {r}_{1} \cos {\theta}_{j} + i \sin {\theta}_{j} , j = 1 , 2 ,$

$\Rightarrow {z}_{1} \cdot {z}_{2} = {r}_{1} \cdot {r}_{2} \left\{\cos \left({\theta}_{1} + {\theta}_{2}\right) + i \sin \left({\theta}_{1} + {\theta}_{2}\right)\right\}$.

Accordingly, in our case, we have,

${z}_{1} \cdot {z}_{2} = 2 \cdot 2 \left\{\cos \left(\frac{\pi}{6} - 5 \frac{\pi}{6}\right) + i \sin \left(\frac{\pi}{6} - 5 \frac{\pi}{6}\right)\right\}$

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

$= 4 \left(\cos \left(2 \frac{\pi}{3}\right) - i \sin \left(2 \frac{\pi}{3}\right)\right)$

=4(-cos(pi/3)-isinsin(pi/3)#

$= 4 \left(- \frac{1}{2} - i \frac{\sqrt{3}}{2}\right)$

$= - \left(2 + i 2 \sqrt{3}\right)$

$= - 2 \left(1 + i \sqrt{3}\right)$.

We can verify that, ${z}_{\cdot} {z}_{2} = \left(\sqrt{3} + i\right) \left(- \sqrt{3} - i\right) = - {\left(\sqrt{3} + 1\right)}^{2}$

$= - \left(3 + 2 \sqrt{3} i + {i}^{2}\right) = - \left(2 + 2 \sqrt{3} i\right) = - 2 \left(1 + \sqrt{3} i\right)$

Enjoy maths.!