No Branch Cut Needed for cos(sqrt(z))

NT123
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Homework Statement

Branch cut for cos(sqrt(z)).



Homework Equations





The Attempt at a Solution

Apparently there is no need for a branch cut for this function, but I am not sure why - I heard it has something to do with cos being an even function. Any clarification would be appreciated.
 
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I'm sure you've done an exercise like writing down all of the values of a multivalued function like sqrt(z) in terms of ordinary functions. (e.g. in terms of the principal branch Sqrt(z))

That seems like an obvious place to start to me.
 
Hurkyl said:
I'm sure you've done an exercise like writing down all of the values of a multivalued function like sqrt(z) in terms of ordinary functions. (e.g. in terms of the principal branch Sqrt(z))

That seems like an obvious place to start to me.

You mean write z as r*exp(i(t+2*k*pi)), so sqrt(z) = (r^1/2)*exp(it/2) or (r^1/2)*exp(i(t/2+pi)? = -(r^1/2)*exp(it/2) ?
 
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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