Linear Algebra - args complex number question

Question

I need to solve this problem : $$z^3-(2+2i)^2=0$$

This is what I did :

$$z^3 = (2+2i)^2$$ $$z^3 = 8i$$

The formula for args is : $$\tan(args)=\frac{b}{a}$$ in this case its clear that the args is $\frac{\pi}{2}$ since $z^3=0+8i$ however if I use the formula I get $\tan(args)=\frac{8}{0}$ But it impossible to divide by zero, in these case should I rely on the graph?

Any ideas? Thanks!

Answer 1

Notice that $$ i = \cos (\tfrac{\pi}{2}) + i\sin(\tfrac{\pi}{2}) = e^{i \tfrac{\pi}{2}}. $$ Hence $$8i = 8e^{i \tfrac{\pi}{2}}.$$ Moreover, for all complex numbers $w$, there exists a unique $r \geq 0$ and a unique $\varphi \in \mathbb{R}$ (modulo $2\pi$) such that $w = re^{i\varphi}$, where $\varphi$ is the argument of $w$ and $r$ is the $lenght$ of $w$. Thus the $z$ in our equation can be written as $z=|\,z\,|e^{i\arg(z)}$. Hence, $$z^3 = \left(|\,z\,|e^{i\arg(z)}\right)^3 = |\,z\,|^3(e^{i\arg(z)})^3 = |\,z\,|^3e^{3i\arg(z)}, $$ by some standard laws. We have $$ |\,z\,|^3e^{3i\arg(z)} = 8e^{i \tfrac{\pi}{2}}. $$ Which holds if and only if $|\,z\,|^3 = 8 (\,\Leftrightarrow |\,z\,| = 2 \,$) and $3\arg(z) = \tfrac{\pi}{2} + k2\pi$, $k \in \mathbb{Z}$ ($\, \Leftrightarrow \arg(z) = \tfrac{\pi}{6} + k\tfrac{2\pi}{3} $, $k \in \mathbb{Z}$\,).

Hence $$ z = 2e^{i(\tfrac{\pi}{6} + k\tfrac{2\pi}{3})}\,,k \in \mathbb{Z}, $$ which of course is covered by $k=0,1,2$, so we have either $z = \sqrt 3 + i$, $z = -\sqrt 3 +i$ or $z = -i$.

Answer 2

$2+2i=2(1+i)=2e^{i\pi/2}$ then, $$z^3=(2+2i)\iff z^3=4e^{i\pi}$$ and so $$z=\sqrt[3]{4}e^{\frac{i\pi}{k}}$$ with $k=0,1,2$

Answer 3

One Idea: Note that $i = e^{i(\frac{\pi}{2} + 2\pi k)}$ for integer $k$. Writing $z = re^{i \theta}$, then $z^3 = r^3 e^{i\theta}$. So, $$ z^3 = 8i \implies r^3 e^{i 3\theta} = 8 e^{i(\frac{\pi}{2} + 2\pi k)}. $$ Now, solve $r^3 = 8$ for a real number $r$, and $\theta = \frac{1}{3} (\frac{\pi}{2} + 2\pi k)$ ...