# Determine all the values of $1^{\sqrt{2}}$

I can't seem to understand how to solve this. I mean, if we weren't dealing with complex numbers, then I suppose it is clearly 1, but I don't know how to approach this. Apparently the answer is $\cos(2\sqrt{2} k \pi) + i\sin (2 \sqrt{2} k \pi)$, but I don't know how to go through with this. Do I begin with setting it to $e^{\sqrt{2}ln(1)}$? Even then, $ln(1) = 0$, and $e^\sqrt{2}$ is just that... I'm not sure how to go about this.

In short, how do I go from $1^{\sqrt{2}}$ to $\cos(2\sqrt{2} k \pi) + i\sin (2 \sqrt{2} k \pi)$?

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I see an answer here, but what is the question? – Anderson Green Sep 12 '13 at 1:51
The question is how to get from $1^{\sqrt{2}}$ to the answer shown. – David Sep 12 '13 at 1:51
$\ln(1) = 0$ for the principal branch of the logarithm. But for other choices of branch cut you get $\ln(1) = 2\pi i k$ for $k \in \Bbb Z$. – Henry T. Horton Sep 12 '13 at 1:54
Solve $x^{1/\sqrt{2}}=1\;(=e^{2i k\pi})$ – L. F. Sep 12 '13 at 1:54
How do I know all of my branches? How could I see this? – David Sep 12 '13 at 1:55

Hint: Recall that $e^{x+iy}$ where $x,y\in\mathbb{R}$ is equal to $e^x(\cos(y)+i\sin(y))$. What complex values $z$ give $e^z=1$?
Expanding the hint: We start with $e^{\sqrt{2}\log(1)}$. Solving the above equation, we see that $e^{z}$ is one at precisely $2\pi ik$ for $k\in\mathbb{Z}$, so $\log(1)=\{2\pi i k|k\in\mathbb{Z}\}$. Plugging in, we have the set $e^{2\sqrt{2}\pi i k}$ for $k\in\mathbb{Z}$. Expanding according to the above, we have $e^{2\sqrt{2}\pi i k}=\cos(2\sqrt{2}\pi k)+i\sin(2\sqrt{2}\pi k)$ and we arrive at the answer.
The punchline here is that $\log(z)$ is still the inverse of $e^z$, but $e^z$ is no longer 1-1 and therefore $\log(z)$ can only be a local inverse, and there's some choice involved in which branch to pick.
$x\neq 1$. Solve the equation! What value of $x,y$ can you plug in to get 1? – KReiser Sep 12 '13 at 2:08
Whoops, you're right, I didn't see that. x clearly has to be 0, and then we're left with figuring out when cos(y) = 1, and sin(y) = 0. This happens at $0, \pi, 2\pi, etc.$ – David Sep 12 '13 at 2:28
Though... the answer doesn't make sense to me. $cos(2\sqrt{2}k\pi)$ does not equate to 1. – David Sep 12 '13 at 2:36
I just came back to this one, and I'm wondering: Why are you solving $e^z = 0$ and not $e^z = 1$? – David Sep 12 '13 at 7:07