Answer
The two possible values of $x$ are:
$cos~135^{\circ}+i~sin~135^{\circ}$
$cos~315^{\circ}+i~sin~315^{\circ}$
Work Step by Step
$x^2+i = 0$
$x^2 = -i$
$x = (-i)^{1/2}$
$x = [1(0-~i)]^{1/2}$
Let $z = 1(0-~i)$
$z = 1(cos~270^{\circ}+i~sin~270^{\circ})$
$r = 1$ and $\theta = 270^{\circ}$
We can use this equation to find the square roots:
$z^{1/n} = r^{1/n}~[cos(\frac{\theta}{n}+\frac{360^{\circ}~k}{n})+i~sin(\frac{\theta}{n}+\frac{360^{\circ}~k}{n})]$, where $k \in \{0, 1, 2,...,n-1\}$
When k = 0:
$z^{1/2} = 1^{1/2}~[cos(\frac{270^{\circ}}{2}+\frac{(360^{\circ})(0)}{2})+i~sin(\frac{270^{\circ}}{2}+\frac{(360^{\circ})(0)}{2})]$
$z^{1/2} = cos~135^{\circ}+i~sin~135^{\circ}$
When k = 1:
$z^{1/2} = 1^{1/2}~[cos(\frac{270^{\circ}}{2}+\frac{(360^{\circ})(1)}{2})+i~sin(\frac{270^{\circ}}{2}+\frac{(360^{\circ})(1)}{2})]$
$z^{1/2} = cos~315^{\circ}+i~sin~315^{\circ}$
