Answer
The possible solutions for $x$ are:
$cos~22.5^{\circ}+i~sin~22.5^{\circ}$
$cos~112.5^{\circ}+i~sin~112.5^{\circ}$
$cos~202.5^{\circ}+i~sin~202.5^{\circ}$
$cos~292.5^{\circ}+i~sin~292.5^{\circ}$
Work Step by Step
$x^4-i = 0$
$x^4 = i$
$x = (i)^{1/4}$
We need to find the fourth roots of $i$.
Let $z = 0 + i$
$z = cos~90^{\circ}+i~sin~90^{\circ}$
$r = 1$ and $\theta = 90^{\circ}$
We can use this equation to find the fourth 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:
$z^{1/4} = 1^{1/4}~[cos(\frac{90^{\circ}}{4}+\frac{(360^{\circ})(0)}{4})+i~sin(\frac{90^{\circ}}{4}+\frac{(360^{\circ})(0)}{4})]$
$z^{1/4} = 1~[cos~22.5^{\circ}+i~sin~22.5^{\circ}]$
$z^{1/4} = cos~22.5^{\circ}+i~sin~22.5^{\circ}$
When k = 1:
$z^{1/4} = 1^{1/4}~[cos(\frac{90^{\circ}}{4}+\frac{(360^{\circ})(1)}{4})+i~sin(\frac{90^{\circ}}{4}+\frac{(360^{\circ})(1)}{4})]$
$z^{1/4} = 1~[cos~112.5^{\circ}+i~sin~112.5^{\circ}]$
$z^{1/4} = cos~112.5^{\circ}+i~sin~112.5^{\circ}$
When k = 2:
$z^{1/4} = 1^{1/4}~[cos(\frac{90^{\circ}}{4}+\frac{(360^{\circ})(2)}{4})+i~sin(\frac{90^{\circ}}{4}+\frac{(360^{\circ})(2)}{4})]$
$z^{1/4} = 1~[cos~202.5^{\circ}+i~sin~202.5^{\circ}]$
$z^{1/4} = cos~202.5^{\circ}+i~sin~202.5^{\circ}$
When k = 3:
$z^{1/4} = 1^{1/4}~[cos(\frac{90^{\circ}}{4}+\frac{(360^{\circ})(3)}{4})+i~sin(\frac{90^{\circ}}{4}+\frac{(360^{\circ})(3)}{4})]$
$z^{1/4} = 1~[cos~292.5^{\circ}+i~sin~292.5^{\circ}]$
$z^{1/4} = cos~292.5^{\circ}+i~sin~292.5^{\circ}$