Finding the limit of a function with a complex exponent

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Homework Statement



\lim_{z \to 0} (\frac{sinz}{z})^{1/z^2}

where z is complex

Homework Equations



The standard definition of a limit
L'hopital's rule?

The Attempt at a Solution


I'm quite stumped by this one. There doesn't seem to be a way to break it down into different limits or even to manipulate it much algebraically. Wolfram says the answer is e^{-1/6} but I'm not sure how to arrive at it. I tried starting by proving that that is the limit if z is real and then extending it to the complex plane, but I can't even solve it for that case. I feel like there's an obvious theorem or something along those lines that I am forgetting. Any hints?
 
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1) log it
2) find the limit. It helps to know what (sinz)/z approaches as z approaches 0. Can we apply l'hopital's rule?
3) reverse the log.
 
It worked! Thanks, that was a really clever and elegant solution
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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