Solve the equation $z^3=\sqrt3 +i$

Question

I am trying to solve the equation: $$z^3=\sqrt3 +i,$$ where $z$ is a complex number.

I have already found that $$ \sqrt3+i=2(\cos(\pi/6)+i \sin(\pi/6))$$

How can I solve the equation above? Thank you very much.

Answer 1

One cube root of $2\exp\frac{i\pi}{6}$ is $\sqrt[3]{2}\exp\frac{i\pi}{18}$. Multiplying by $\exp\frac{2i\pi}{3}$ gives another, $\sqrt[3]{2}\exp\frac{13i\pi}{18}$. We multiply by $\exp\frac{2i\pi}{3}$ again to get the final cube root, $\sqrt[3]{2}\exp\frac{25i\pi}{18}$.

Answer 2

v$$z^3=2 e^{i\pi/6} =2 e^{i(2n+1/6)\pi} \implies z=(2)^{1/3} e^{i(2n+1/6)\pi}, n=0,1,2.$$ So there three roots three roots $(n-0,1,2 <3)$: $$ 2^{1/3} e^{i\pi/6},~ 2^{1/3} e^{13 i\pi/18}, ~2^{1/3} e^{25i\pi/18}$$

Answer 3

Correct me if I am wrong but I believe you are not familiar with the form $z=re^{i\theta}$.

Are you familiar with $cis$? Basically, $cis\theta=cos\theta+i*sin\theta$.

Notice that $z^3=\sqrt3+i=2cis\frac{\pi}{6}+2\pi k$

That means that $z=2^\frac{1}{3}cis\frac{\pi}{18}+\frac{2}{3}\pi k$

Therefore $z=2^\frac{1}{3}cis\frac{\pi}{18},2^\frac{1}{3}cis\frac{13\pi}{18},2^\frac{1}{3}cis\frac{25\pi}{18}$ are all solutions.