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I would like to know the most efficient way to solve (in $\mathbb{C}$ of course):

$$z^4+1 = 0$$

I already tried:

$$\Longleftrightarrow z^2 = \pm \sqrt {-1} = \pm i$$ $$\Longleftrightarrow z = \pm \sqrt{\pm i}$$ which is quite confusing. I also tried:

$$\rho \exp{(4i\theta)} = -1$$ $$\Longleftrightarrow 4i\theta = \ln \left(\frac{-1}{\rho}\right)$$ which is also confusing, as $\ln(z\le 0)$ does not exist...

  • 4
    $z^4=-1\implies z^8=1$. Take the "odd" roots of unity. – Dr. Mathva Nov 18 '22 at 18:54
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    As Dr. Mathva says. Alternately: $$e^{i\theta} = \cos\theta + i \sin \theta$$ so when $\theta = (2n+1)\pi$ for any integer $n$, you have $$e^{i(2n+1)\pi} = -1$$ But you need $$e^{4i\theta} = -1$$ so $$4\theta = (2n+1)\pi$$ and find values for $\theta$ in $[0,2\pi)$ to get all four values. – SlipEternal Nov 18 '22 at 18:56

4 Answers4

3

You should understand what modulus is:

$z=re^{i\theta}$ then $|z|=\sqrt{z\bar{z}}=\sqrt{re^{i\theta}re^{-i\theta}}=r$

now $z^4=-1\implies r^4e^{4i\theta}=-1$ so the modulus is $r=1$ and we are left with

$e^{4i\theta}=-1\implies 4\theta=\pi+2n\pi\implies\\ \theta=\frac{\pi}{4}+\frac{n}{2}\pi$

ryaron
  • 1,231
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One can rewrite $z^4$ as $(z^2)^2$, then, $(z^2)^2 = -1,$ and $z^2 = \pm i$ Now recall that since i corresponds with $\frac{\pi}{2}$ on the unit circle, the square root must correspond with $\frac{\pi}{4}$ so the solutions must be: $$ z=\pm\frac{1+i}{\sqrt 2},\ \ z=\pm\frac{1-i}{\sqrt 2} $$ And here you have your four roots.

Itamar
  • 130
3

We don't need to appeal to the use of the complex exponential--the polynomial can be factored directly by "completing the square" in a clever way. Observe: $$ z^4 + 1 = (z^4 + 2z^2 + 1) - 2z^2 = (z^2 + 1)^2 - 2z^2 = [(z^2+1)-\sqrt{2}z]\cdot[(z^2+1) + \sqrt{2}z]. $$ In $\mathbb{R}$, this would be as far as we can go in factoring, because the resultant quadratics has discriminant $\Delta = (\pm\sqrt{2})^2 - 4\cdot1\cdot1 = 2-4 = -2 < 0$. But in $\mathbb{C}$, we can just apply the quadratic formula: $$ z^2 \pm \sqrt{2}z + 1 =0 \implies z = \frac{\mp\sqrt{2} \pm \sqrt{-2}}{2} = \frac{1}{\sqrt{2}}(\mp1 \pm i). $$ Hence the four roots are $$ z_1 = \frac{1+i}{\sqrt{2}}, z_2 = -\frac{1+i}{\sqrt{2}}, z_3 = \frac{1-i}{\sqrt{2}}, z_4 = -\frac{1-i}{\sqrt{2}}. $$

This factoring technique can be used to solve any quartic in $\mathbb{C}$ where only even powers show up.

Edit: This technique was given in the linked answer here--more discussion on the same polynomial is given there.

Nick F
  • 1,289
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I just thought about a trick: we know that $1 = \exp{(2ik\pi})$ and $-1 = \exp{(2ik\pi+\pi})$ with $k \in \mathbb{Z}$, so why not writing:

$$\rho\exp{(4i\theta)} = \exp{(2ki\pi+\pi)}$$ $$\Longleftrightarrow \begin{cases}\theta = k\pi/2+\pi/4\\ \rho=1 \end{cases}$$

So the solutions would be:

$$z_k = \exp(ik\pi/2 + i\pi/4)$$