Question

use demoivres theorem (2sqrt3+2i)^2

Answer 1

\[(2\sqrt3+2i)^2\]
First convert to polar form. If \(z=x+iy\), then \(z=r(\cos \theta+i\sin\theta)\) where \(r=|z|\), \(\cos\theta=\dfrac{x}{r}\), and \(\sin\theta=\dfrac{y}{r}\).
\[\begin{cases}x=2\sqrt3\\
y=2\\
|z|=\sqrt{(2\sqrt3)^2+2^2}=\sqrt{16}=4\\
\cos\theta=\dfrac{2\sqrt3}{4}=\dfrac{\sqrt3}{2}~~\Rightarrow~~\theta=\dfrac{\pi}{6}
\end{cases}\]
So, converting polar coordinates and applying DeMoivre's Thereom, you get
\[(2\sqrt3+2i)^2=\left(4\left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right)\right)^2=16\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)=8+8i\sqrt3\]