Compute $z^3 = -26 -18i$

$z^3 = -26 -18i, i =$ imaginary unit.

How do I solve this? So far I've calculated the length of it;

$|z| = \sqrt{(-26)^2+(-18)^2}=10\sqrt10$

I think I'm supposed to use the polar form of it, but I'm not sure how.

Solution 1:

You can see from your norm calculation (Which I believe should say $\left|z^3\right|=10\sqrt{10}$) that $z$ has norm $\sqrt{10}$.

If you have reason to believe that $z$ has integer parts, there aren't that many options. You need $z=a+bi$ with $a^2+b^2=10$. One of them is $\pm1$ and one is $\pm3$. Check the 8 possibilities.

$$\begin{align} (1+3i)^3&=1+9i-27-27i&&=-26-18i\\ (1-3i)^3&=1-9i-27+27i&&\neq-26-18i\\ (-1+3i)^3&=-1+9i+27-27i&&\neq-26-18i\\ (-1-3i)^3&=-1-9i+27+27i&&\neq-26-18i\\ (3+i)^3&=27+27i-9-i&&\neq-26-18i\\ (3-i)^3&=27-27i-9+i&&\neq-26-18i\\ (-3+i)^3&=-27+27i+9-i&&\neq-26-18i\\ (-3-i)^3&=-27-27i+9+i&&\neq-26-18i\\ \end{align}$$

Only the first works. Since you know that there must be three cube roots, the other two are $(1+3i)\omega$ and $(1+3i)\omega^2$ where $\omega=e^{2\pi i/3}=\frac{-1+i\sqrt{3}}{2}$, a noninteger cube root of $1$.

Solution 2:

$\left|-26-18i\right|=10\sqrt{10}=10^{3/2}$ and $\text{Arg}\left(-26-18i\right)=\arctan\left(18/26\right)-\pi$ so that \begin{align*} z^{3} & =10^{3/2}e^{\left[\arctan\left(18/26\right)-\pi+2\pi k\right]i}\\ z & =10^{1/2}e^{\left[\arctan\left(18/26\right)-\pi+2\pi k\right]i/3} \end{align*} for $k=0,1,2$.