# Does x^4=-1 have a sollution

## Main Question or Discussion Point

I am trying to find a sollution to this in the complex plane. One that seems to work is sqrt(i), but is this valid or not?

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phyzguy
Yes, it has a solution. In fact, the fundamental theorem of algebra tells you that it has 4 solutions. In this case they are :
$$\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i, -\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i,\frac{\sqrt2}{2}-\frac{\sqrt2}{2}i,-\frac{\sqrt2}{2}-\frac{\sqrt2}{2}i$$

yup,it is valid
x= +/- sqrt(i)
sqrt(i) = +/- (1+i)/sqrt(2)
so you get four solutions

Yes, it has a solution. In fact, the fundamental theorem of algebra tells you that it has 4 solutions. In this case they are :
$$\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i, -\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i,\frac{\sqrt2}{2}-\frac{\sqrt2}{2}i,-\frac{\sqrt2}{2}-\frac{\sqrt2}{2}i$$
We can also write them in exponential form.

$-1 = e^{j180} = e^{j540} = e^{j900} = e^{j1260}$

Solutions are:
$e^{j\frac{180}{4}} = e^{j45} = +\frac{\sqrt2}{2}+j\frac{\sqrt2}{2}$
$e^{j\frac{540}{4}} = e^{j135} = -\frac{\sqrt2}{2}+j\frac{\sqrt2}{2}$
$e^{j\frac{900}{4}} = e^{j225} = e^{-j135} = -\frac{\sqrt2}{2}-j\frac{\sqrt2}{2}$
$e^{j\frac{1260}{4}} = e^{j315} = e^{-j45} = +\frac{\sqrt2}{2}-j\frac{\sqrt2}{2}$