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I'm having trouble finding the solutions to $z^4 = -16$

At first I did $\sqrt {z^4} = \sqrt {-16}$

Then I'd have $z^2 = 4i$ but this doesn't seem right or I just don't know what to do afterwords.

José Carlos Santos
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Andrew Quinonez
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6 Answers6

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Take $z=re^{i\theta}$ so $$r^4e^{4i\theta}=16e^{i\pi}\to\\r=2\\4\theta=(2k+1)\pi\to\\\theta=\frac{2k+1}{4}\pi\qquad,\qquad k=0,1,2,3\to \\z=2e^{\dfrac{2k+1}{4}\pi}\qquad,\qquad k=0,1,2,3$$

Mostafa Ayaz
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Instead of $z^2 = 4i,$ you actually need $z^2 = \pm4i.$

You have $$ 4i = 4(\cos90^\circ + i\sin90^\circ). $$ Therefore $$ \pm\sqrt{4i} = \pm 2(\cos45^\circ + i\sin45^\circ) = \pm2\left( \frac 1 {\sqrt2} + i\frac1 {\sqrt2} \right) = \pm (\sqrt 2 + i\sqrt2) $$ and then proceed similarly with $-4i.$

Michael Hardy
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It's $$z^4+16=0$$ or $$z^4+8z^2+16-8z^2=0$$ or $$(z^2-2\sqrt2z+4)(z^2+2\sqrt2z+4)=0,$$ which gives the answer: $$\{\sqrt2+\sqrt2i,\sqrt2-\sqrt2i,-\sqrt2+\sqrt2i,-\sqrt2-\sqrt2i\}$$

Michael Rozenberg
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You just calculate the (real) $4$th root of $16$, and multiply by each of the (complex) $4$-th roots of $-1=\mathrm e^{i\pi}$.

Bernard
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Here is how to find one. You have $$ - 16 = 16e^{i\pi} = (2e^{\pi/4})^4.$$ Can you find the others? Four exist.

ncmathsadist
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Let $\,z = r e^{i \varphi}\,$ in polar form, then solve $\,r^4 e^{4 i \varphi}= 2^4 e^{i \pi}\,$.

dxiv
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  • So from this method, do we arrive at $r^4=2^4e^{i(\pi - 4\varphi)}$ then take a fourth root and obtain $r=2e^{i(\frac{\pi}{4}-\varphi)}$ – homosapien Jun 26 '22 at 22:17
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    @HossienSahebjame What you get is $,r^4 = 2^4,$ and $,e^{4i\varphi} = e^{i\pi},$, and from there $,r=2,$ and $,\varphi = \frac{1}{4}(\pi + 2k\pi),$, which gives the $4$ roots for $k=0,1,2,3$. – dxiv Jun 26 '22 at 22:35
  • Is the reason $k =0,1,2,3$ that, after that you get repeated roots ? @dxiv – homosapien Jun 26 '22 at 22:45
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    @HossienSahebjame Yes, because $,\frac{\pi + 2k\pi}{4} \equiv \frac{\pi + 2(k+4)\pi}{4}\pmod{2\pi},$. – dxiv Jun 26 '22 at 22:55