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I am just beginning a Complex Analysis course and we are asked to find the complex solutions for $z^6=-9$ and I don't know where to start.

I previously solved the equation $z^4=-16$, which was more straight forward because we could take the $4^{\text{th}}$ root of $-16$ in the complex plane, and I am unsure of how to do this for the question at hand.

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2 Answers

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There is a standard way which you handle this type of equations. I will elaborate it for you :

Let's assume the problem : Find all the solutions of the equation $z^n = w$, where $n \in \mathbb N^*$ and $w \in \mathbb C.$

Solution :

Let $ w=|w|e^{iφ} = |w|\cosφ + i|w|\sin φ, φ=\arg w$

If $ z=|z|e^{iθ} = |z|\cosθ + i|z|\sin θ, θ=\arg z$

Then, $|z|^ne^{inθ}=|w|e^{iφ} \Leftrightarrow |z|^n(|z|\cosθ + i|z|\sin θ)=|w|(\cosφ + i\sin φ)$

Obviously, $|z|^n=|w| \Leftrightarrow |z| = \sqrt[n]{|w|}$

We get :

$|w|e^{inθ} = |w|e^{iφ} \Leftrightarrow |w|(\cos nθ + i\sin nθ) = |w|(\cosφ + i\sin φ) \Rightarrow \{\cos nθ = \cos φ , \sin nθ = \sin φ \} \Rightarrow \{nθ = 2κπ + φ \} \Leftrightarrow \{ θ = \frac{2κπ +φ}{n} \}$

Eventually, the solutions are :

$ |z| = \sqrt[n]{|w|}[\cos (\frac{2κπ +φ}{n}) + i\sin (\frac{2κπ +φ}{n})] = \sqrt[n]{|w|}e^{\frac{2κπ +φ}{n}i} , κ \in \mathbb Z, κ=0,1,...,n-1$ or $κ=1,2,...n $

So, in your particular problem the solutions are :

$z_κ = \sqrt[6]{-1}\sqrt[3]{3}e^{\frac{2π}{3}i}$ with $κ\in \mathbb Z, κ= 0,1,...,5$

I skipped the part of the calculations since I provided the theoritical answer.

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$-9=9e^{i\pi}$. So $z=9^{\frac{1}{6}}e^{i\frac{1}{6}(\pi+2k\pi)}=9^{\frac{1}{6}}\Big[\cos\Big(\frac{(2k+1)\pi}{6}\Big)+i\sin\Big(\frac{(2k+1)\pi}{6}\Big)\Big]$

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