Find the solutions to $z^3 = 2 + 11i$. 
Find the solutions to $z^3 = 2 + 11i$.


I wrote $z = x + iy$ to get
$(x + iy)^3 = 2+ 11i$.
Expanding and equating the real and Imaginary parts I got, 
$$x^3 - 3xy^2 = 2$$
$$y^3 - 3x^2y = 11$$
Let $tx = y$, Substituting this I got,
$$x^3 - 3t^2x^3 = 2$$
$$x^3(3t-t^3) = 11$$
Susbtituting for $x^3$ from first equation into second eaution I got,
$$2t^3 + -6t - 33t^2 + 11 = 0$$
I don't know how to solve this cubic, I tried finding roots by putting different numbers but I can't find any of them. 
What are some other ways of solving without running into a cubic and how can I solve this cubic ?   
 A: To avoid cubics, we use trig:  $\theta=\arctan(11/2)$ and $r=\sqrt{11^2+2^2}=5\sqrt5$
The cube roots are then given as follows:
$$z^3=2+11i=r\operatorname{cis}(\theta)=re^{i\theta}$$
$$z=\sqrt[3]{r}\operatorname{cis}\left(\frac{\theta+2\pi k}3\right)=\sqrt[3]{r}e^{i\frac{\theta+2\pi k}3}$$
Where $\operatorname{cis}(\theta)=\cos(\theta)+i\sin(\theta)$ and $k\in\mathbb N$.
In general, the $n$th roots are:
$$z^n=2+11i$$

$$z=\sqrt[n]{r}\operatorname{cis}\left(\frac{\theta+2\pi k}n\right)$$

A: your number has norm 125. The answer has norm 5. If they intended integers, that means either $\pm1 \pm 2 i$ or $\pm 2 \pm i.$ Try them.
As 11 is odd, we need $y = \pm 1,$ so $\pm 2 \pm i.$
A: 
Expanding and equating the real and Imaginary parts I got,
$$x^3 - 3xy^2 = 2 \tag{1}$$
$$y^3 - 3x^2y = 11$$

The second equation lost a sign along the way, it should be:
$$y^3 - 3x^2y = -11 \tag{2}$$
Subtracting $(1)-(2)$ and grouping:
$$x^3-y^3 - 3xy(y-x)=13 \;\iff\; (x-y)(x^2+4 xy+y^2) = 13$$
Looking for "nice" integer solutions leaves only $x-y=\pm 1,\pm 13$ to try.
