Question

Calculeaza rădăcina de ordinul 3

Answer 1

 

$\displaystyle\bf\\Z=\sqrt[\b3]{4+4i\sqrt{3}}\\\\ \textbf{Ne ocupam de numarul complex de sub radical.}\\\\z=4+4i\sqrt{3}\\\\\textbf{Il scriem pe z sub forma trigonometrica.}\\\\|z|=r=\sqrt{4^2+\Big(4\sqrt{3}\Big)^2}=\sqrt{16+16\times3}=\sqrt{16+48}=\sqrt{64}=\boxed{\bf8}\\\\cos\,\varphi=\frac{4}{r}=\frac{4}{8}=\boxed{\bf\frac{1}{2}}\\\\sin\,\varphi=\frac{4\sqrt{3}}{r}=\frac{4\sqrt{3}}{8}=\boxed{\bf\frac{\sqrt{3}}{2}}\\\\\implies~\varphi=60^o=\boxed{\bf\frac{\pi}{3}}$

.

$\displaystyle\bf\\z=r(cos\,\varphi+i\,sin\,\varphi)\\\\z=8\left(cos\frac{\pi}{3}+i\,sin\frac{\pi}{3}\right)\\\\Z=\sqrt[\b3]{\bf~z}\\\\Z=\sqrt[\b3]{\bf8\left(cos\frac{\pi}{3}+i\,sin\frac{\pi}{3}\right)}\\\\Z=\sqrt[\b3]{\bf8}~\times\sqrt[\b3]{\bf\left(cos\frac{\pi}{3}+i\,sin\frac{\pi}{3}\right)}$

.

$\displaystyle\bf\\\boxed{\bf~Z=2\left(cos\frac{\dfrac{\pi}{3}+2k\pi}{3}+i\,sin\frac{\dfrac{\pi}{3}+2k\pi}{3}\right)}$