Find all three complex solutions of the equation $z^3=-10+5i$ Let $z\in \mathbb{C}$. I want to calculate the three solutions of the equation $z^3=-10+5i$. Give the result in cartesian and in exponential representation.
Let $z=x+yi $.
Then we have $$z^2=(x+yi)^2 =x^2+2xyi-y^2=(x^2-y^2)+2xyi$$
And then $$z^3=z^2\cdot z=[(x^2-y^2)+2xyi]\cdot [x+yi ] =(x^3-xy^2)+2x^2yi+(x^2y-y^3)i-2xy^2=(x^3-3xy^2)+(3x^2y-y^3)i$$
So we get
$$z^3=-10+5i \Rightarrow (x^3-3xy^2)+(3x^2y-y^3)i=-10+5i \\ \begin{cases}x^3-3xy^2=-10 \\ 3x^2y-y^3=5\end{cases} \Rightarrow \begin{cases}x(x^2-3y^2)=-10 \\ y(3x^2-y^2)=5\end{cases}$$
Is everything correct so far? How can we calculate $x$ and $y$ ? Or should we do that in an other way?

There is another way
$$z^3+10-5i=0$$
Let's replace the complex number with a constant $a$
$$z^3+a=0$$
$$z^3+(a^{1/3})^3=0$$
Using the Sum of Cubes
$$(z+a^{1/3})(z^2-a^{1/3}z+a^{2/3})=0$$
We get $z=-a^{1/3}$ as a solution
$$z^2-a^{1/3}z+a^{2/3}=0$$
$$z=\frac{-(-a^{1/3})±\sqrt{(-a^{1/3})^2-4(1)(a^{2/3})}}{2(1)}=\frac{a^{1/3}±\sqrt{-3a^{2/3}}}{2}=\frac{a^{1/3}±\sqrt3ia^{1/3}}{2}$$
$$=\left(\frac{1±i\sqrt3}{2}\right)a^{1/3}=a^{1/3}e^{±\pi i/3}$$
Now we find the cube root of $a$ using polar form
$$(10-5i)^{1/3}=\left(\sqrt{10^2+5^2}\exp\left(i\arctan\left(-\frac{5}{10}\right)\right)\right)^{1/3}$$$$=\left(5\sqrt5\exp\left(-i\arctan\left(\frac{1}{2}\right)\right)\right)^{1/3}=\sqrt{5}\exp\left(\frac{-i}{3}\arctan{\frac{1}{2}}\right)$$
WE GET THE SOLUTION
$${z=\sqrt{5}\exp\left(\frac{-i}{3}\arctan{\frac{1}{2}}\right)e^{k\pi i/3}}, k=-1,0,1$$
NOTE:
Instead of using the sum of cubes, we can notice the following
$$z^3=-a$$
$$z^3=ae^{k\pi i}, k=-1,1,3$$
Take the cube root on both sides
$$z=(ae^{k\pi i})^{1/3}=a^{1/3}e^{k\pi i/3}, k=-1,1,3$$
Then find the cube root and finish
