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

Question

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?

Answer 1

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.

Answer 2

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

Answer 3

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

Answer 4

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