Find the nth root of i I saw a video some time ago showing a nice method for showing that $\sqrt{i} = \pm\left(\frac1{\sqrt2} + i \frac1{\sqrt2}\right)$. The teacher assumes that $\sqrt{i}$ can be equated to some complex number $a +bi$, and  solves accordingly:

$$\sqrt{i} = a+bi$$
$$\iff i = a^2-b^2+ i 2ab$$
$$\implies 
\begin{cases}
a^2-b^2=0 \\
2ab=1
\end{cases}
$$
Solving this latter system of equations yields $a = \pm\frac1{\sqrt2}$ and $b = \pm\frac1{\sqrt2}$.

Seeing this made me curious about the general case, $\sqrt[n]i = a + bi$. I first tried isolating $a$ and $b$ by using the teacher’s method:
$$\sqrt[n]{i} = a+bi \\ 
\iff i = (a+bi)^n \\
 = a^n + i{n\choose 1}a^{n-1}b + i^2 {n\choose 2}a^{n-2}b^2 + … + i^{n-2} {n\choose n-2}a^2b^{n-2} + i^{n-1} {n\choose n-1}ab^{n-1} + i^nb^n
$$
But this quickly becomes complicated.
The next step would be to group terms with respect to whether they have a factor of $i$ or not, but that would require knowing the divisibility of $n$. Otherwise, we wouldn’t be able to know what $i^n$ is, or any subsequent terms for that matter.
Moreover, even if were to isolate certain divisibility cases for $n$, this would yield an at-least-nth degree system of equations that would be really hard to solve.
How could I proceed from here?
 A: $n$-th root of $i$ is easy to express using Euler's formula:
$$
i = e^{\frac{i\pi}{2}}
$$
$$
i^{\frac{1}{n}} = e^{\frac{i\pi(1+4k)}{2n}} =
 \cos \frac{\pi(1+4k)}{2n} + i \sin \frac{\pi(1+4k)}{2n}
$$
where $0 \leq k < n$
A: Let me begin by saying that this problem is pretty much always attacked using deMoivre's Theorem as in the other answers, but I thought it was still worth pointing out that your approach to the problem using the binomial expansion can be pushed further due to the cyclic behavior of powers of $i$.
$$
\begin{align}
i^1 &=i \\
i^2 & -1\\
i^3 &=-i\\
i^4 &=1\\
i^5 &=i\\
\end{align}
$$
after which the pattern repeats.  So if we were to write your sum as $\sum_{k=0}^n i^k a^{n-k}b^k$ we can collect terms based on the residue of $k$ modulo $4$
$$
\begin{align}
\sum_{k=0}^n i^k a^{n-k}b^k&=\sum_{0\le k \le n \\k \equiv 0 \bmod 4}^n a^{n-k}b^k+i\sum_{0\le k \le n \\k \equiv 1 \bmod 4}^n a^{n-k}b^k-\sum_{0\le k \le n \\k \equiv 2 \bmod 4}^n a^{n-k}b^k-i\sum_{0\le k \le n \\k \equiv 3 \bmod 4}^n a^{n-k}b^k \\ 
&= \sum_{0\le k \le n \\k \equiv 0 \bmod 2}^n (-1)^{k/2}a^{n-k}b^k+i\sum_{0\le k \le n \\k \equiv 1 \bmod 2}^n (-1)^{(k-1)/2}a^{n-k}b^k \\
&=\sum_{j=0}^{\lfloor{n/2}\rfloor} (-1)^{j}a^{n-2j}b^{2j}+i\sum_{j=1}^{\lfloor{(n+1)/2}\rfloor} (-1)^{(j-1)}a^{n-2j+1}b^{2j-1}
\end{align}
$$
so for a given $n$ we have to solve
$$
\begin{align}
\sum_{j=0}^{\lfloor{n/2}\rfloor} &(-1)^{j}a^{n-2j}b^{2j}=0 \\
\sum_{j=1}^{\lfloor{(n+1)/2}\rfloor} &(-1)^{(j-1)}a^{n-2j+1}b^{2j-1}=1
\end{align}
$$
Taking $n=3$ we get
$$
\begin{align}
a^3-3ab^2&=0 \\
3a^2b-b^3&=1
\end{align}
$$
from which it's not difficult to obtain the solutions $-i, \frac{\sqrt 3 +i}{2}, \frac{-\sqrt 3 +i}{2}$
But for $n=4$ and
$$
\begin{align}
a^4-6a^2b^2+b^4&=0 \\
4a^3b-4ab^3&=1
\end{align}
$$
things start getting harder.  And by the time we get to $n=5$ and
$$
\begin{align}
a^5-10a^3b^2+5ab^4&=0 \\
5a^4b-10a^2b^3+b^5&=1
\end{align}
$$
the solutions are getting difficult without the use of symbolic algebra software.
A: We write $a +bi$ in polar coordinates.  Let $i^{1/n} = a + bi = r(\cos \theta + i \sin \theta)$. Then $i = r^n(\cos\theta + i \sin\theta)^n = r^n(\cos n \theta + i \sin n \theta).$ You can prove the last equality by induction. It's also called deMoivre's formula.
Then we have, $r^n\cos n\theta = 0$ and $r^n \sin n \theta = 1$. All the solutions are counted by letting $r = 1$ and $\theta = \frac{\pi}{2n}+\frac{2\pi k}{n}$ with $k = 0, 1, \dots n-1$.
