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What are possible solutions to $$z^4 = w$$ where $z \in \mathbb{C} $ and $w \in \mathbb{R}$?

Here is my attempt:

Write both numbers in polar form: $r^4(\cos 4 \theta + i\sin 4 \theta) = w(\cos 0 + i \sin 0) $.

Working with the conventions of $r \geq 0$ and $0 \leq \theta < 2\pi$ I get the following:

$r^4 = w \iff r = w^\frac{1}{4}$

and

$4\theta = n2\pi$ where $n \in \mathbb{Z}$.

$\Rightarrow \theta = \frac{n}{2}\pi$

Thus we get four possible solutions:

$z = w^\frac{1}{4}(\cos 0 + i \sin 0) = w^\frac{1}{4}$

$z = w^\frac{1}{4}(\cos \frac{\pi}{2} + i \sin \frac{\pi}{2})= iw^\frac{1}{4}$

$z = w^\frac{1}{4}(\cos \pi + i \sin \pi) = -w^\frac{1}{4}$

$z = w^\frac{1}{4}(\cos \frac{3\pi}{2} + i \sin \frac{3\pi}{2}) = -iw^\frac{1}{4}$

Is this legit, something missing or very much not rigorous?

Travis Willse
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snoram
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    Looks good to me, for positive $w$. When you've studied a little more you will realize that you can just multiply the real fourth root of $w$ by each of the four complex fourth roots of $1$ - no need to work those out from scratch. – Ethan Bolker Dec 01 '15 at 15:32
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    There is one special case that may be worth noting separately. If $w=0$ then your 4 possible solutions collapse into one solution though that may be getting a bit nit-picky here. – JB King Dec 01 '15 at 15:33
  • Thanks. I should have been more specific about possible values of $w$. – snoram Dec 01 '15 at 15:40

2 Answers2

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If $w < 0$, then the equation $r = w^{1 / 4}$ is not even sensible, as any fourth root of $w$ is nonreal (and anyway in this case there is not a preferred choice among these) but $r$ is a real variable. The fix is to write any $w < 0$ in polar form as $$w = |w|e^{\pi i} = |w|(\cos \pi + i \sin \pi).$$ Then, regardless of the sign of $w$ we have $r = |w|^{1 / 4}$, and when $w < 0$ our equation in $\theta$ becomes $4 \theta = (2n + 1) \pi$.

Travis Willse
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Assume $z\in\mathbb{C}$ and $w\in\mathbb{R}$:

$$z^4=w\Longleftrightarrow$$ $$z^4=|w|e^{\arg(w)i}\Longleftrightarrow$$ $$z^4=we^{0i}\Longleftrightarrow$$ $$z=\left(we^{2\pi ki}\right)^{\frac{1}{4}}\Longleftrightarrow$$ $$z=\sqrt[4]{w}e^{\frac{2\pi ki}{4}}\Longleftrightarrow$$ $$z=\sqrt[4]{w}e^{\frac{\pi ki}{2}}$$

With $k\in\mathbb{Z}$ and $k:0-3$

So the solutions are:

$$z_0=\sqrt[4]{w}e^{\frac{\pi\cdot 0i}{2}}=\sqrt[4]{w}e^{0}=\sqrt[4]{w}$$ $$z_1=\sqrt[4]{w}e^{\frac{\pi\cdot 1i}{2}}=\sqrt[4]{w}e^{\frac{\pi i}{2}}=i\sqrt[4]{w}$$ $$z_2=\sqrt[4]{w}e^{\frac{\pi\cdot 2i}{2}}=\sqrt[4]{w}e^{\pi i}=-\sqrt[4]{w}$$ $$z_3=\sqrt[4]{w}e^{\frac{\pi\cdot 3i}{2}}=\sqrt[4]{w}e^{\frac{3\pi i}{2}}=-i\sqrt[4]{w}$$

So we can conclude that your solutions are:

$$z=\pm\sqrt[4]{w} \space\space\vee\space\space z=\pm i\sqrt[4]{w}$$

Jan Eerland
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