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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$\begingroup$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.
$\endgroup$ 9 $\begingroup$$e^0 = 1$. However, remember that $e^{2\pi ki}$ for $k \in \mathbb{Z}$ is also $1$. Therefore, you can rewrite $1^{\sqrt{2}}$ as $e^{{(2\pi ki)}^{\sqrt{2}}}$, and simplify using exponent laws to $e^{2\sqrt{2}\pi ki}$. From there, use Euler's identity.
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