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?
 A: Hint: Write $-10+5i$ in its polar form, then recognize the exponential has period $2\pi i$. Use this fact to take the cube root of both sides and obtain all solutions. Once you have the solutions, you can find individual real and imaginary parts.
A: The exact algebraic answer, which expressible in terms of real quantities by real-valued radicals, is:
$$\bbox[5px,border:2px solid #C0A000]{\begin{align}&x=-\alpha_k\sqrt [3]{\frac{5}{3\alpha_k^2-1}}\\
&y=-\sqrt [3]{\frac{5}{3\alpha_k^2-1}}\end{align}}$$
where,
$$\begin{align}\alpha_k=-2+2\sqrt{5}\cos\left(\frac{1}{3}\arccos\left(-\frac{2\sqrt 5}{5}\right)-\frac{2\pi k}{3}\right)\end{align}$$
for $k\in\{0,1,2\}\thinspace .$

$\rm Construction:$
In this answer, we find all pairs of $(x,y)\in\mathbb R^{2}$ in terms of real quantities by real-valued radicals.
After expanding the parentheses, we have:
$$\begin{align}&\begin{cases}x(x^2-3y^2)=-10 \\y(3x^2-y^2)=5\end{cases}\\
\implies &\frac xy\left(\frac {\frac {x^2}{y^2}-3}{\frac {3x^2}{y^2}-1}\right)=-2\end{align}$$
Substituting $\dfrac xy=u$, leads to:
$$\begin{align}&\frac{u(u^2-3)}{3u^2-1}=-2\\
\implies &u^3+6u^2-3u-2=0\end{align}$$
We have $3$ distinct real roots, however this is the case of Casus irreducibilis.
Using the key substitution $u=v-2$, yields:
$$v^3-15v+20=0$$
We know that, the depressed monic cubic equation $x^3+px+q=0$ is solved by
$$\begin{align}x_k=2\sqrt{-\frac{p}{3}}\cos\left(\frac{1}{3}\arccos\left(\frac{3q}{2p}\sqrt{\frac{-3}{p}}\right)-\frac{2\pi k}{3}\right) \quad \text{for} \quad k=0,1,2.\end{align}$$
Thus we obtain:
$$\begin{align}u_k=-2+2\sqrt{5}\cos\left(\frac{1}{3}\arccos\left(-\frac{2\sqrt 5}{5}\right)-\frac{2\pi k}{3}\right) \quad \text{for} \quad k=0,1,2.\end{align}$$
Finally, setting $\dfrac xy=\alpha\,$ we get:
$$\begin{align}&y(3y^2\alpha^2-y^2)+5=0\\
\implies &(3\alpha^2-1)y^3+5=0\\
\implies &y^3=-\frac{5}{3\alpha^2-1}\\
\implies &y=-\sqrt [3]{\frac{5}{3\alpha^2-1}}\\
\implies &x=-\alpha\sqrt [3]{\frac{5}{3\alpha^2-1}}\end{align}$$
where $\alpha\in\{u_0,u_1,u_2\}\thinspace.$

Note that, casus irreducibilis cannot be solved in radicals in terms of real quantities, it can be solved trigonometrically in terms of real quantities.
Therefore, we can no longer escape the general trigonometric or another non-algebraic representations.
A: It is correct so far. Note that\begin{align}-10+5i&=5\sqrt5\left(-\frac2{\sqrt5}+\frac i{\sqrt 5}\right)\\&=5\sqrt5\exp\left(\left(\pi-\arctan\left(\frac12\right)\right)i\right).\end{align}Therefore, if $\alpha=\pi-\arctan\left(\frac12\right)$, then $z^3=-10+5i$ if and only if $z$ is one of the numbers$$\sqrt5\exp\left(\frac\alpha3i\right),\ \sqrt5\exp\left(\frac{\alpha+2\pi}3i\right)\text{ or }\sqrt5\exp\left(\frac{\alpha+4\pi}3i\right).$$
A: 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
A: Use a primitive third root of unity,  say $\zeta_3=e^{\frac {2\pi i}3}.$
Take $\omega $ with $w^3=-10+5i$, say $\omega =-\sqrt {125}^{\frac 13}e^{\frac{i\arctan -\frac 12}3}=-\sqrt 5e^{\frac{i\arctan -\frac 12}3}.$
Then the roots are $\{\omega, \zeta_3 \omega, \zeta_3 ^2\omega \}.$
