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?

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