# How $m^4+4 = 0 \Rightarrow m = 1 \pm i,-1\pm i$?

This is given in my module as a part of a problem's solution:

$$m^4 + 4 = 0$$ $$\Rightarrow m = 1 \pm i,-1\pm i$$

I am not getting how this conversion is taking place,could somebody explain?

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It follows from Euler's formula. –  Qiaochu Yuan Feb 20 '11 at 12:59
Well, a polynomial of degree $4$ over the complex numbers has at most 4 roots; these four numbers are roots, so they are *all the roots... –  Arturo Magidin Feb 20 '11 at 23:30

One way to look at it is like this: $$m^4+4=0\implies m^4=-4\implies m^2=\pm 2i.$$ But $$m^2=2i\implies m=\pm\sqrt{2}\sqrt{i}.$$ Also, $$m^2=-2i\implies m=\pm i\sqrt{2}\sqrt{i}.$$ Now you can use that fact that there are two square roots of $i$, $$\frac{1+i}{\sqrt{2}}\ \text{and}\ \frac{1+i}{-\sqrt{2}}$$ to simplify each possibility, and get the $4$ values of $m$ you mentioned.

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Here $m = (-4)^{1/4}$.

Now $-4 = 4(\cos(\theta) + i\sin(\theta))$ in polar form where $\theta = (2n-1)\pi$.

By de Moivre's theorem

$m = (-4)^{1/4} = 4^{1/4}(\cos(\frac{\theta}{4}) + i\sin(\frac{\theta}{4}))$

For $n = 0$, $\displaystyle m = \sqrt{2}\left(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) = 1 - i$,

For $n = 1$, $\displaystyle m = \sqrt{2}\left(\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\right) = 1 + i$,

For $n = 2$, $\displaystyle m = \sqrt{2}\left(-\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\right) = -1 + i$,

For $n = 3$, $\displaystyle m = \sqrt{2}\left(-\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) = -1 - i$

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You can also factor $x^4+4=(x^2-2x+2)(x^2+2x+2)$.

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BTW, this also proves that $n^4+4$ is never a prime, for $n$ an integer. See math.stackexchange.com/questions/21146/… –  lhf Feb 20 '11 at 15:31

HINT $\rm\ -4 = m^4\ \Rightarrow\ \pm 2\ i = m^2 = (a+b\ i)^2 = a^2-b^2 + 2\:a\:b\ i\ \Rightarrow\ |a| = |b| = 1\:$.

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