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Let $a$ be some complex number. I have to solve equation

$$z^4=a^{16}$$

One is tempted to "simplify" it to $z=a^4$, so it is the solutions. But somebody told me, there are more solutions than that. Is it true and why? It seems counterintuitive.

Micah
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adm34
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  • Hint: even within real numbers there would be two square roots. Add to that that square roots of negative numbers are now allowed within complex numbers – Miguel Sep 11 '15 at 16:04
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    $(z-a^4)(z+a^4)(z^2+a^8)$ –  Sep 11 '15 at 16:07

3 Answers3

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Yes, there are more solutions than that. To see that concretely, suppose $a=1$. Then your original equation is $$ z^4 = 1 $$ which has the solutions $\{1,-1,i,-i\}$, but if you rewrite it to $z=1$, you will have lost three of the solutions.

What you can do is rewrite $z^4=a^{16}$ to $$ z=ka^4 \text{ for some }k\text{ with }k^4=1\text{, that is, }k\in\{1,-1,i,-i\} $$ which gives you four separate equations to solve as you plug the possible $k$s in.

hmakholm left over Monica
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Note that every $a \in \mathbb{C}$ can be written as $a = re^{i\varphi}$, hence your equation delivers

$$z^4 = r^{16} e^{16i\varphi}$$

The complex roots are given by

$$z^n = re^{i\varphi} \implies z_k = r^{\frac{1}{n}}e^{i\left( \frac{\varphi}{n} + \frac{2\pi k}{n}\right)} \qquad k=0,\dots,n-1$$

So you have $n$ solutions for the $n^{\text{th}}$ complex root. In your case

\begin{align*} z_0 &= r^4 e^{i\left( \frac{16\varphi}{4} + 0 \right)} = \left( r, \varphi \right) = \left( r^4, 4\varphi \right)\\ z_1 &= r^4 e^{i\left( \frac{16\varphi}{4} + \frac{2\pi}{4} \right)} = \left( r, \varphi \right) = \left( r^4, 4\varphi + \frac{\pi}{2} \right)\\ z_2 &= r^4 e^{i\left( \frac{16\varphi}{4} + \frac{4\pi}{4} \right)} = \left( r, \varphi \right) = \left( r^4, 4\varphi + \pi \right)\\ z_3 &= r^4 e^{i\left( \frac{16\varphi}{4} + \frac{6\pi}{4} \right)} = \left( r, \varphi \right) = \left( r^4, 4\varphi + \frac{3\pi}{2} \right)\\ \end{align*}

root
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We see that $16/4 = 4$, so we can substitute $t = a^4$, then we can rewrite $$z^4 = a^{16} \Leftrightarrow z^4 = t^4$$ This has the solutions Henning described above, the four roots of unity: $\{1,i,-1,-i\} $ times $t$. Just to give some clarity as why it works.

mathreadler
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