Solving imaginary equation $z^3 = 5i + 5$ I need help solving the equation : $z^3 = 5i + 5$. I'm basically just starting to learn imaginary numbers and how to solve them. 
Straight forward solution is a pain ( was trying to figure out with step-by-step solution from wolframalpha ). After that I was reading wikipedia about them and came across with the De Moivre's formula that states $z^n = |z|^n (\cos(n\phi) + i\sin(n\phi))$ but I have no clue how to use it with this example.
I would greatly appreciate some help on this. Maybe it is possible to do it another way? Thanks in advance.
 A: $$z^3=5(1+i)$$
$$=5\sqrt{2}(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}})$$
$$=5\sqrt{2}(\cos(\frac{\pi}{4})+i\sin(\frac{\pi}{4}))$$
Do you now see some thing worthy???
A: Alternatively, you can write $$5+5i=\sqrt{50}\operatorname{cis}\frac\pi4,$$ so putting $$\zeta=\sqrt[6]{50} \operatorname{cis} \frac\pi{12},$$ we have $$z^3=\zeta^3\\z^3-\zeta^3=0\\(z-\zeta)(z^2+\zeta z+\zeta^2)=0,$$ and so the solutions are $z=\zeta$, and the solutions to the quadratic equation $z^2+\zeta z+\zeta^2=0.$ As yet another alternative, put $\omega= \operatorname{cis} \frac{2\pi}3,$ so that $1,\omega,\omega^2$ are the complex cube roots of $1$, and so $\zeta,\zeta\omega,\zeta\omega^2$ are the solutions to the equation  $z^3=\zeta^3.$
Regardless you should find solutions $$z=\sqrt[6]{50} \operatorname{cis}\left(\frac\pi{12}+\frac{2\pi k}3\right),$$ for $k=0,1,2.$
A: Here's an approach that doesn't use any trig.  Assume that $z=a+bi$ with real values for $a$ and $b$.  Expanding $(a+bi)^3$ and separating real and imaginary parts leads to the pair of equations
$$a^3-3ab^2=5$$
and
$$3a^2b-b^3=5$$
This implies $a^3-3ab^2=3a^2b-b^3$, which implies
$$(a+b)(a^2-4ab+b^2)=0$$
All three solutions for $z$ have to come from this.  I'll do the easy one, corresponding to $a+b=0$.  Plugging this back into either of the pair of equations gives $-2a^3=5$, or $a=-\sqrt[3]{5/2}$ and $b=-a=\sqrt[3]{5/2}$.  So one solution is
$$z=-\sqrt[3]{5\over2}+i\sqrt[3]{5\over2}$$
To get the other two solutions, it's probably best to multiply this one by ${1\over2}\pm i{\sqrt3\over2}$ (i.e., the complex cube roots of unity).
