How to find the solutions to $z^4 = -16$

Question

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.

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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 $-4i.$

Answer 6

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: $$\{\sqrt2+\sqrt2i,\sqrt2-\sqrt2i,-\sqrt2+\sqrt2i,-\sqrt2-\sqrt2i\}$$