How do you solve $z^4 = 2(1+i\sqrt{3})$

Solve $z^4 = 2(1+i\sqrt{3})$ in the form $r(\cos\alpha+i\sin\alpha)$ where $r>0$ and $0\le\alpha<2\pi$

I know you have to find $\arctan(\frac{\sqrt{3}}{1})=\frac{\pi}{3}$ and that is $\alpha$? I am not really sure how to go about doing this.

• hint: $1+i\sqrt{3}=\sqrt{1^2+\sqrt{3}^2}\left(\frac 12+i\frac{\sqrt{3}}2\right)=2\,e^{i\pi/3}$ – Raymond Manzoni Apr 27 '13 at 10:59
• One huge contributor to the community recently answered a question that described the process of solving this kind of problems. I can't remember who it was. Hopefully someone will know what I'm talking about and provide the link to that answer. Despite this I'm sure there are other similar questions here and this should probably be marked as a duplicate. – Git Gud Apr 27 '13 at 10:59
• $z^4=4(\frac{1}{2}+i\frac{\sqrt3}{2})$ – maxmitch Apr 27 '13 at 11:03
• $z=\sqrt2(\frac{1}{2}+i\frac{\sqrt3}{2})$ – maxmitch Apr 27 '13 at 11:04
• Where did he claim to understand the equivalence: $z^4 = 4e^{i\pi/3}$ ? I really think he needs to take this slower and that the existing answers aren't of much help to him. But perhaps it's the wrong forum too. – roliu Apr 27 '13 at 11:30

$$cos(\alpha)+isin(\alpha)=e^{i\alpha}$$ $$z^{4}=re^{i\alpha}\implies z=(r)^{1/4}e^{i\alpha/4}=r^{1/4}[cos(\alpha /4)+isin(\alpha/4)]$$
Here r=4 and $\alpha=\dfrac{\pi}{3}$
$$(1)\;\;\;w:=2+2\sqrt 3\,i\implies |w|=\sqrt{4+12}=4\;,\;\;\arg w=\arctan\frac{2\sqrt 3}{2}=\frac{\pi}{3}\;\implies$$
$$(2)\;\;z^4=\left(re^{i\phi}\right)^4=w=4e^{\frac{\pi i}{3}+2k\pi i}\implies z_k=4^{1/4}e^{\frac{\pi i}{12}(1+2k)}\;,\;\;k=0,1,2,3\; (\text{why only these?})$$