# What are the solutions to $z^4+1=0$?

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 $z^4 = -1$?

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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: math.stackexchange.com/questions/233999/… – 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.

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

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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 $z_3$...

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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...

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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.

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

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