But can't the $\pi$th root of $-1$ also be $\frac1{e^i}$?

Question

So I was bored, and decided to do some math for fun. I came up with $$\text{Find all solutions to }\sqrt[\pi]{-1}\text{ other than just }e^i$$which I thought that I might be able to do. Here is my attempt at doing so:

Know: $\sqrt[\pi]{-1}=e^i$, however to find another solution (if possible)$$e^i=e^{\left((-1)^\frac12\right)}=x$$$$\implies(-1)^i=\ln(x)$$$$\implies\dfrac12i\pi=\ln(\ln(x))$$$$\implies i\pi=2\ln(\ln(x))$$Setting $\ln(x)=a$,$$i\pi=2\ln(a)$$$$-1=a^2$$$$-(a)^2-1=0$$$$\implies a^2+1=0$$$$\implies a=\pm i$$$$\ln(x)=\pm i$$$$x=e^{\pm i}$$$$\therefore x=e^i,\dfrac1{e^i}$$

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My question

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Am I right in assuming that $\sqrt[\pi]{-1}=e^{\pm i}$, or is there only one solution to $\sqrt[\pi]{-1}=x$?

Answer 1

Such problems may be solved in a systematic way. You have $z=-1=e^{\pi i}$ in standard form and look for (all) complex numbers $w=z^{\frac{1}{\pi}}$.

In general, you employ $e^{2\pi i n}=1$ for all $n\in \mathbb{Z}$ and thus write

$$w=(ze^{2\pi i n})^{\frac{1}{\pi}} =(e^{\pi i}e^{2\pi i n})^{\frac{1}{\pi}} =(e^{\pi i(2n+1)})^{\frac{1}{\pi}} =e^{i(2n+1)} = e^{im} $$ for all odd integer $m$.

So your set of all solutions is $W=\{\ldots ,e^{-3i},e^{-i},e^{i},e^{3i},e^{5i},... \}$

Answer 2

For complex numbers $z$ and $w$ the "function" $z^w$ is multivaluated, and it's defined by $z^w = e^{w\ \text{Log}\ z}$ for all the posible values of $\text{Log}\ z$.

Then

$$(-1)^{1/\pi} = e^{\tfrac{1}{\pi}\text{Log}(-1)} = e^{\tfrac{1}{\pi}(\pi i +2\pi i k)} = e^{(2k+1)i} = \cos(2k+1) + i\sin(2k+1)$$ where $k\in \Bbb Z$.

When $k=0$ we get the principal value $e^i = \cos 1+ i\sin 1$, and for $k=-1$ we get also the value $e^{-i} = \cos 1- i\sin 1$, but those are just two from an infinity of values that the expression takes.