# How do you use demoivre's theorem to simplify 2(sqrt3+i)^2?

Sep 11, 2017

$2 {\left(\sqrt{3} + i\right)}^{2} = 4 + 4 \sqrt{3} i$

#### Explanation:

First convert your $z = \sqrt{3} + i$ into polar form by taking its absolute value and argument.

Recall that $| z | = \sqrt{{\left(\sqrt{3}\right)}^{2} + {1}^{2}} = 2$.

And that $A r g \left(z\right) = \arctan \left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$.

$\therefore z = 2 \cdot c i s \left(\frac{\pi}{6}\right)$.

By de Moivre's theorem, ${z}^{n} = {r}^{n} \cdot c i s \left(n \theta\right)$.

$\therefore {z}^{2} = 4 \cdot c i s \left(\frac{\pi}{3}\right)$.

Converting back into Cartesian form, we have $x = r \cdot \cos \left(\theta\right)$ and $y = r \cdot \sin \left(\theta\right)$

$\therefore x = 4 \cdot \cos \left(\frac{\pi}{3}\right)$ and $y = 4 \cdot \sin \left(\frac{\pi}{3}\right)$.

$\therefore {z}^{2} = 2 + 2 \sqrt{3} i$.

$\therefore 2 {z}^{2} = 4 + 4 \sqrt{3} i$.