I am getting two answers for the fourth root of i. What am I doing wrong?

Question

Using Algebra:

$\sqrt[4]{i}=\sqrt[16]{i^4}$

$=\sqrt[16]{i^2i^2}$

$=\sqrt[16]{(-1)(-1)}$

$=\sqrt[16]{1}$

$=1$

Using Euler's Identity:

$\sqrt[4]{j}=\sqrt[4]{e^{i\pi/2}}$

$=(e^{i\pi/2})^{1/4}$

$=e^{i\pi/8}$

$=\cos{(\frac{\pi}{8})}+i\sin{(\frac{\pi}{8})}$

$=0.9239+i0.3827$

Answer 1

There are four different $4$th roots of $i$, so finding multiple roots isn't what you're doing wrong. What you ARE doing wrong is saying that $1^4 = i$ though.

Answer 2

Note that any nonzero complex number has $4$ distinct 4th roots; as such, there isn't inherently a problem with finding different solutions. However, as noted, your first answer is incorrect:

What happens is that you're introducing extraneous solutions in the first case. By exponentiating both sides, you generate an equation that has more roots than the original. You then only find one such solution, which happens to be extraneous.

An analogous example would be the equation $x = y$; clearly any solution of this has the form $(x,x)$. Squaring both sides gives you $x^2 = y^2$, which introduces extra solutions of the form $(x,-x)$.

Answer 3

The 16th root of 1 are 1, i^(1/4), ..., i^(15/4), where the 1st, 5th, 9th and 13th are 4th root of i