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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:… –  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

5 Answers 5

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

Then to find a fourth root...

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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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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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