How to find the roots of $x^4-i=0$ I need the manually analysis to calculate the roots without using the numerical methods 
 A: Let $x=re^{i \theta}$ (with $r \ge 0$) to get $r^4 e^{i 4 \theta} = e^{i {\pi \over 2}}$. It follows that $r=1$ and
$4 \theta - { \pi \over 2} \in 2 \pi \mathbb{Z}$.
Compute the unique (modulo $2 \pi$) values of $\theta$.
A: $$x^4-i=0$$
$$x^4=i=e^{i(2k+1)\pi/2}=\cos\left(\frac{(2k+1)\pi}{2}\right)+i\sin\left(\frac{(2k+1)\pi}{2}\right)$$
$$x=i^{1/4}=e^{i(2k+1)\pi/8}=\left[\cos\left(\frac{(2k+1)\pi}{2}\right)+i\sin\left(\frac{(2k+1)\pi}{2}\right)\right]^{1/4}$$
Using De Moivre's formula
$$(\cos\theta+i\sin\theta)^n=\cos(n\theta)+i\sin (n\theta)$$
$$x=e^{i(2k+1)\pi/8}=\cos\left(\frac{(2k+1)\pi}{8}\right)+i\sin\left(\frac{(2k+1)\pi}{8}\right)$$
$$k=0 , 1 , 2 , 3$$
$$x=e^{i\pi/8} ,\ e^{3i\pi/8}, \ e^{5i\pi/8} ,\ e^{7i\pi/8}$$
or 
$$x=\cos\left(\frac{\pi}{8}\right)+i\sin\left(\frac{\pi}{8}\right)$$
$$x=\cos\left(\frac{3\pi}{8}\right)+i\sin\left(\frac{3\pi}{8}\right)$$
$$x=\cos\left(\frac{5\pi}{8}\right)+i\sin\left(\frac{5\pi}{8}\right)$$
$$x=\cos\left(\frac{7\pi}{8}\right)+i\sin\left(\frac{7\pi}{8}\right)$$
A: Hint: $i = \left(e^{\dfrac{i\pi}{8}}\right)^4$
A: Hint: Write $x = r(\cos \theta + i \sin \theta)$ and $i = 1(\cos \pi/2 + i \sin \pi/2)$, and compare $$r^4(\cos 4\theta + i\sin4\theta) = 1(\cos \pi/2 + i\sin\pi/2).$$
