$4^{th}$ root of $i$ in $\mathbb{C}$

Question

I am apparently a little bit confused about some basic calculations in $\mathbb{C}$: given the equation $w^4=(\sqrt{2}e^{i\frac{\pi}{4}})$, I know that there must be 4 solutions in $\mathbb{C}$, but I can't understand why the solutions are $w_1=\sqrt{2}e^{i\frac{\pi}{4}}, w_2=\sqrt{2}e^{i\frac{3\pi}{4}}, w_3=\sqrt{2}e^{i\frac{5\pi}{4}}, w_4=\sqrt{2}e^{i\frac{7\pi}{4}}$.

Answer 1

That is wrong. The $4$^th roots of $\sqrt2e^{i\frac\pi4}$ are$$\sqrt[4]{\sqrt2}e^{i\frac\pi{16}},\ \sqrt[4]{\sqrt2}e^{i\left(\frac\pi{16}+\frac\pi4\right)},\ \sqrt[4]{\sqrt2}e^{i\left(\frac\pi{16}+\frac\pi2\right)}\text{ and },\ \sqrt[4]{\sqrt2}e^{i\left(\frac\pi{16}+\frac{3\pi}4\right)}.$$

Answer 2

Note that $z=\sqrt{2}e^{i\pi/4}=2^{1/2}\big(\cos \tfrac{\pi}{4}+i\sin \tfrac{\pi}{4}\big)$. So, the four solutions to the equation $w^4=z$ are given by $$\begin{align} w_{k+1}&=(2^{1/2})^{1/4}\left[\cos \left(\frac{\tfrac{\pi}{4}+2\pi k}{4}\right)+i\sin \left(\frac{\tfrac{\pi}{4}+2\pi k}{4}\right)\right] \\ &=2^{1/8}\left[\cos \left(\frac{(8k+1)\pi}{16}\right)+i\sin \left(\frac{(8k+1)\pi}{16}\right)\right] \end{align}$$ for $k=0,1,2,3$.

Answer 3

These are wrong. To accurately calculate, them, use Euler's Identity: $$re^{i\theta}=r(\cos\theta+i\sin\theta)$$

and the corollary De Moivre's Theorem: $$[r(\cos\theta+i\sin\theta)]^n=r^n(\cos(n\theta)+i\sin(n\theta))$$

hence:

$$[2^\frac12(\cos\frac\pi4+i\sin\frac\pi4)]^\frac14=2^\frac18(\cos\frac{\pi}{16}+i\sin\frac{\pi}{16})$$ is the base root. Now you need to account for the other three, by noticing that $$\forall k\in \Bbb Z; e^{i\theta}=e^{i\theta+2k\pi}$$

Answer 4

As far as the title (which is different from what you did in the body), you'd want the four solutions to $z^4=i$. One is $z=e^{\frac {\pi i}8}$.

The other three can be gotten by multiplying the first by $\omega, \omega ^2$ and $\omega ^3$, where $\omega$ is a primitive fourth root of unity. So $\omega =e^{\frac{\pi i k}2}$, for $k=1$ or $3$.