a while ago we had the solution for a complex number task about the nth-root of unity in the complex. But now I am having some difficulties to fully understand it:
The task was to find all complex solutions to $z^4=-1$
We used polar coordinate and the De Moivre formula:
$\varphi \in \{-\pi, \pi\}$
$(r*e^{i\varphi})^4 = e^{-i\pi}$
The following two steps are the steps I don't understand:
Question: How do I get here from the previous steps? I know that just put the exponent inside and get $r^4$ and $e^{4i\varphi}$ but how did we get from this to $r=1$?
$\Leftrightarrow r^4=1$ and $e^{4i\varphi}=e^{-i\pi}$
$\Leftrightarrow r=1$ and $\underbrace{4\varphi \in \pi+2\pi z}_*$
$* \Leftrightarrow \varphi \in \frac{\pi}{4}+ \frac{1}{2}\pi z = \left\{\pm \frac{\pi}{4};\pm \frac{3\pi}{4};\pm \frac{5\pi}{4}\right\}$
Due to $\varphi \in \{-\pi, \pi\}$ we said that $\varphi \in \left\{\pm \frac{\pi}{4}; \pm \frac{3\pi}{4}\right\}$ and that gives the solution
$z_{1/2}=e^{\pm \frac{i\pi}{4}}, z_{3/4}=e^{\pm \frac{i3\pi}{4}}$
The last steps are diving by 4 to get $\varphi$ and then using the periocidcity to go from $z=0,1..2$ I get the solutions adn the corresponding negative ones.
Regards, Christoph
You get $r=1$ from the equation $r^4\cdot e^{4i\varphi} = 1\cdot e^{-i\pi}$. From this equation, if you take the absolute values of both sides, you get $r^4=1$, and since $x\mapsto x^4$ is a monotonous function, it follows from $r^4=1$ that $r=1$.