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

Question

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

Answer 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 $1$.

Answer 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 $k=0,1,2$.

Answer 3

Let $z^n = a$ where $a\in\mathbb{C}\setminus\{0\}$ then $$z = \sqrt[n]{|a|}e^{i\left(\frac{\arg(a)}{n}+p\frac{2\pi}{n}\right)}$$ where $p\in\{0,...,n-1\}$, which gives $n$ solutions. Interestingly enough, the solutions form the vertices of a regular $n$-gon centered around the origin of the complex plane.

If needed you can then use $$e^{i\theta} = \cos\theta+i\sin\theta. $$ The argument of $a$ is given by $$\arg(-26-18i) = \arctan\left(\frac{-18}{-26}\right)-\pi = \arctan\left(\frac{9}{13}\right)-\pi.$$

Answer 4

A systematic way of attempting this is to factor $N=-26-18i$ into irreducibles (primes) in the ring $\mathbb Z[i]$ of Gaussian integers. The norm of our number is $(-26)^2 + (-18)^2 = 1000$. That’s the field-theoretic norm I’m talking about, not the absolute value. Since the $2$-part of the norm is $2^3$, we should divide by $(1+i)^2$ right off the bat, to get $N/(1+i)^3=2+11i$. The norm here is, unsurprisingly, $125=5^3$, and since $5$ does not divide $2+11i$, this number must be divisible by either $2+i$ or $2-i$, and this to the third power. You can try dividing $2+11i$ by $2-i$, and you immediately see that that’s no go. So divide by $(2+i)^3$ and lo and behold! you get a quotient of $1$. And there you are, the only cube root of $-26-18i$ in $\mathbb Z[i]$ is $(1+i)(2+i)=1+3i$. There are others in the field $\mathbb Q(i,\sqrt3)$, you get them by multiplying your $2+i$ by cube roots of unity.