What are the complex solutions of $z^4+16=0$?

Question

What are the complex solutions for $z^4+16=0$? I know that one solution is $z=a+bi=-2$. How can I figure out the other solutions?

No, $(-2)^4+16=32$, and not $0$. – Dietrich Burde Dec 15 '13 at 20:06 — Dietrich Burde, Dec 15 '13 at 20:06

score 3 · Answer 1 · 2013-12-15T20:16:44.670

3

$z=-2$ is not a solution. Plug it in: $(-2)^4+16=32 \neq 0$. What we want to do is factor: $$(z^4+16) = (z^2+4i)(z^2-4i)$$

It's not blatantly obvious how to continue from here, but we know now that the solutions are the solutions to $z^2=4i$ and $z^2=-4i$. What we really need to know here are the solutions to $z^2=i$ and $z^2=-i$ and multiply them by two; but (check!) these are $$\frac{\sqrt{2}}{2}\left(\pm 1 \pm i\right)$$ These are known as the primitive eighth roots of unity.

Multiplying we two, we find that the four solutions to your equation are $\sqrt{2}(\pm 1 \pm i)$.

edited Dec 15 '13 at 20:16

answered Dec 15 '13 at 20:08

3

You really should check that the numbers you propose as solutions are solutions... – Did Dec 15 '13 at 20:09
You're off by a factor of $\sqrt{2}$ Congratulations, you found the solutions to $z^4+4=0$. – Christopher K Dec 15 '13 at 20:11
And after I'd told the OP to check. Fixed. Thanks and sorry. – Dec 15 '13 at 20:17
@Mike: perhaps I came off as a bit righteous there. If so I apologise. – Christopher K Dec 15 '13 at 20:19
@ChrisK No need, you pointed out an error, and I didn't take your comment negatively. But apology accepted anyway :) – Dec 15 '13 at 20:22
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+1 for mathematical correctness. I don't understand the downvoter. – Christopher K Dec 15 '13 at 20:24

score 1 · Answer 2 · answered Dec 15 '13 at 20:11

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Warning: I know nothing about complex numbers.

The most general technique:

Let $p(x)$ be a polynomial and let $p(c)=0$. Then $x-c$ is a factor of $p(x)$, so you can divide $p(x)$ by $x-c$ and look for roots again.

In this case, the polynomial is of a particularly simple form, $z^n=c$. Writing $c=re^{i\theta}$ for $r$ a positive real number and $0\le\theta<2\pi$, $z=r^{1/n}e^{i(\theta+2\pi k)/n}$ is a solution for any integer $k$.

answered Dec 15 '13 at 20:11

dfeuer

9,069

Edited. If there are any errors remaining, I would like the downvoter to explain. – dfeuer Dec 15 '13 at 20:16
It seems all the answers were downvoted (although I can't fathom why). +1 Your answer is concise, readable and mathematically correct. – Christopher K Dec 15 '13 at 20:21

Magdiragdag · Answer 3 · 2013-12-15T20:41:03.697

Once you've realized that $2^4 = 16$, the only further thing you need to find is an element $\zeta$ (well four of them) with $\zeta^4 = -1$, because then $(2\zeta)^4 = -16$.

If you know about roots of unity, you'll know these $\zeta$ lie on the unit circle, but right now, let's do this more elementary.

So it would help to look for a $\zeta$ with $\zeta^2 = i$ ($-i$ will do as well). Write $\zeta = a + bi$ and work out $\zeta^2 = i$. This gives $(a^2 - b^2) + 2abi = i$, so $a^2 = b^2$ and $2ab = 1$. Now solve for $a$ and $b$, which gives two possibilies for $\zeta$ and $z = 2 \zeta$ are two solutions to your original equation.

To find the other two, solve $\zeta^2 = -i$ or just multiply the ones you already found by $i$, which works because $i^4 = 1$.

score 0 · Answer 4 · answered Dec 15 '13 at 20:10

Well... $z^4 = -16$ and so $z = 16^{1/4}\cdot cis(\pi/4 + 2\pi n / 4)$, since $Arg(-16)=\pi$ and $|z|=16$. We then apply de Moivre's Theorem.

So $z=2cis(\pi/4) = 2(1/\sqrt{2}+i/\sqrt{2})$ or $z = 2cis(3\pi/4) = 2(-1/\sqrt{2} + i/\sqrt{2})$, ...

Hence, $z = \sqrt{2}\cdot (\pm 1 \pm i)$. Indeed, $z^4 = 16cis((2n+1)\pi) = -16$. I'll leave it to you to fill in the blanks.

What are the complex solutions of $z^4+16=0$?

4 Answers4