Find all $z \in \mathbb{C}$ such that $z^4 + c = 0$

Question

Find all $z \in \mathbb{C}$ such that $z^4 + c = 0$ where $c = c' + 0i$ for some $c' \in \mathbb{R}$

How could I solve for $z$ in this case? I could convert $z$ into polar co-ordinates and let $z = re^{i\phi}$ and then I'd have $z^4 = r^4 e^{i 4\phi} = -1$ but in that case I'm still now sure how to solve for $r$ and $\phi$.

What is the best method to solve for $z$ in this case?

Answer 1

The fourth roots of $-c$ are simply made up of one fourth root of $ -c$ multiplied by each of the fourth roots of unity, $1,\,-1,\,i,\,-i$.

Now either $-c>0$ or $-c<0$. In the first case, a fourth root is simply $\sqrt[4]{-c}$, and in the second case, it is $\sqrt[4]{c}\:\mathrm e^{\tfrac{i\pi}4}$.

Answer 2

Let $a=-c$. Then we want to solve $$ z^4=a $$ Now think about what $^4$ does to a complex number: It quadruples the angle, and raises the length to the fourth power. So if $a$ has (positive) length $r$, what length must $z$ have? If $a$ has angle $\phi$, what angles can $z$ have?

Answer 3

Please correct me if I am wrong,

$z^4 = -c$, so $z = c^\frac{1}{4}.(-1)^\frac{1}{4}$ or $z = c^{1/4}.e^{i(\frac{\pi}{4}+2n\pi)}, n \in \Bbb{Z}$.

Answer 4

Let $\operatorname{cis}\theta=\cos\theta+i\sin\theta$. Then, \begin{align} z^4+c&=0 \\ \implies z^4&=-c \\ z^4&=c\operatorname{cis}(\pi+2\pi k) \qquad k\in\mathbb{Z} \\ \implies z&=\sqrt[4]{c}\operatorname{cis}(\pi/4 + \pi k/2) \quad k=0,1,2,3 \end{align}