How can I solve the complex equation $$z^3=1+i \sqrt3$$
Splitting $z$ into $z=a+bi$ gives me the mess
$$a^3+3a^2b^2i-3ab^2-b^3i-i \sqrt3-1=0$$
where I dont know how to continue. I have never really worked with complex numbers, so I am not familiar with the "tricks" or steps that have to be done.
On
Note that $$a^3-3ab^2-1=0$$ and $$3a^2b^2-b^3-\sqrt{3}=0$$ must hold. So we obtain $$b=3\,{\frac {a \left( {a}^{4}-\sqrt {3}-a \right) }{{a}^{3}-1}}$$ and then you can compute $$a$$.
On
More generally, by using the exponential form (instead of the cartesian one) $$1+i \sqrt3=2\exp(i\pi/3),$$ it is easier to solve the equation $z^n=1+i \sqrt3$ where $n$ is a positive integer. There are $n$ distinct solutions given by $$z_k=2^{1/n}\exp\left(i\frac{\pi/3+2k\pi}{n}\right)\quad \text{for $k=0,1,2,\dots,n-1$}.$$ As a reference take a look at $n$-roots of a complex number.
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For products of complex numbers it's often easiest to use the polar form, $z = r e^{i \phi}$, which yields $z^3 = r^3 e^{3 i \phi}$. Converting the right-hand side of the equation to this form yields $1 + i \sqrt{3} = 2 e^{i \frac{\pi}{3}}$ (cf. the hint of José Carlos Santos). Now you compare both sides to obtain two easily solvable equations for $r$ and $\phi$.
You approach this using the fact that$$1+i\sqrt3=2\left(\cos\left(\frac\pi3\right)+i\sin\left(\frac\pi3\right)\right)$$and using de Moivre's formula.