Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I can't seem to find the solutions to $z^4+1=0 $. $z$ is in the complex plane.

The solutions show four roots; however, how do I find them once $z^4 = -1$?

share|cite|improve this question
Can you find one? Do you know geometrically (in polar coordinates) what happens to a complex number $z$ when raised to power $n$? In the solution, basically you will draw a square in the unit circle. – Berci Nov 29 '12 at 15:31
Do you know the magic factorization $z^4+4=(z^2-2z+2)(z^2+2z+2)$? From this you can get what you’re seeking. – Lubin Nov 29 '12 at 17:10
After answering, I realized I had already answered the same question, not too long ago. Duplicate of:… – mrf Nov 29 '12 at 22:53
Note: I think it is more appropriate to say "find the solutions to $X$" when $X$ is an equation; and "find the roots of $X$" when $X$ is a polynomial. – Pedro Tamaroff Nov 29 '12 at 23:39

You can write $z^4=-1$ as $(z^2)^2=-1$. The two square roots of $-1$ are $i$ and $-i$, so we get the two equations $z^2=\pm i$.

Since $i$ corresponds to $\pi/2$ on the unit circle, its square root will have to correspond to $\pi/4$ (or use De Moivre if you don't see this). So $$ z=\pm\frac{1+i}{\sqrt 2},\ \ z=\pm\frac{1-i}{\sqrt 2} $$ are the roots.

share|cite|improve this answer

$$z^4=-1=e^{\pi i+2k\pi i}=e^{\pi i(1+2k)}\Longrightarrow z=e^{\frac{\pi i}{4}(1+2k)}\,\,,\,k=0,1,2,3$$

share|cite|improve this answer

Here is how i was taught to find roots. I'll try to give a fully worked out answer, with no shortcuts, for the first 2 roots; then you should be able to do the 2nd two roots on your own.

$1+z^4=0$ gives $z^4=-1$ We know that $-1=e^{i \pi}$ because Euler tells us $e^{i\pi}= \cos \pi + i \sin \pi =-1 +i(0)=-1$ So, in this problem, we can write: $z^4 =e^{i \pi}$ But $e^{i\pi}=e^{i(\pi + 2\pi n)}$ for n=0,1,2,3,... (think of a point on the unit circle and do n complete rotations, for each integer n you end up right back where you started). So let us say that $$z_n^4=-1=e^{i(\pi +2\pi n)}$$ Then we have, $$z_n =[e^{i(\pi + 2 \pi n)}]^{\frac{1}{4}}=e^{i(\frac{\pi}{4}+\frac{\pi}{2}n)}$$ Now we can find the roots for n=0,1,2,3; that is, the 4 desired roots. First, $$z_0 =e^{i(\frac{\pi}4 +\frac{\pi}2(0))}=e^{i\frac{\pi}{4}}=\cos \frac{\pi}4 +i\sin \frac{\pi}4=\frac{1+i}{\sqrt{2}}$$ Second, $$z_1 =e^{i(\frac{\pi}4 +\frac{\pi}2 (1))} =e^{i\frac{3\pi}4}=\cos \frac{3\pi}4 +i \sin \frac{3\pi}4=\frac{-1+i}{\sqrt{2}}$$

Using this method you should be able to work out the next two roots :) Can you guess what $z_4$ would be? Hint: think of a unit circle; you've been working your way around it from $z_0$ to $z_3$...

share|cite|improve this answer
Thanks for the tip, Roland! Edit made. – Iron Charioteer Mar 9 at 17:55

You can write $-i$ in polar form: $$-i = e^{i \cdot 3 \pi /2} $$

Then to find a fourth root...

share|cite|improve this answer

Since we have $i^2=-1$ $$z^4+1=(z^2)^2-(i)^2$$ $a^2-b^2=(a-b)(a+b)$, so we can factor to have $$z^4+1=(z^2)^2-(i)^2=(z^2-i)(z^2+i)$$ It's easy to solve from here on. $$z^4+1=0 \implies \left \{ \begin{align}&z^2-i=0\implies z=\pm\sqrt i \\&z^2+i=0 \implies z=\pm\sqrt{-i}\end{align}\right.$$

Using the properties

  • $i=e^{i(\pi/ 2)}$
  • $-i=e^{i(3\pi/ 2)}$
  • $e^{i\theta}=\cos \theta + i\cdot \sin \theta$

you can express the result in much more interesting forms.

share|cite|improve this answer

$$z^4+1 = (z^2+1)^2-2z^2 = (z^2+1-z\sqrt2)(z^2+1+z\sqrt2)$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.