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Identify singularities of the function $f(z)=\frac{1}{\cos{z^2}}$ and classify them. Find the residue of the function that the point $z_0=\sqrt{\frac{\pi}{2}i}$ .

I am hoping to find a clear solution to this problem.


My thoughts: I know that the singularities of a rational functions are the zeros of the denominator. I think the singularities are $z=0$ and $z=n\pi$. I'm not quite sure on how to classify them. I think that 0 is a pole of order 2 because the angle is squared. However I also think it might be a simple pole because the denominator itself isn't raised to a power.

Also if I'm being asked to find the residue, doesn't that mean that the point $z_0$ itself is a singularity? I'm generally feeling confused and would appreciate a clear solution to this.

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We know the singularities of $\frac{1}{\cos(z^2)}$ are $\{z | z^2 = (n+1/2)\pi\}$ for integer n. Now let's classify the singularities. We expand $\cos(z^2)$ and its derivative is $-2z\sin(z^2)$ which is not zero at any of the points, so all of the poles are of order 1.

Now, because we are dealing with a simple pole, at $z_0 = i\sqrt{\pi/2}$, $Res(f,z_0) = \frac{1}{-2z\sin(z^2)} = \frac{1}{-2i\sqrt{\pi/2}\sin(-\pi/2)} = \frac{i}{2 \sqrt{\pi/2}}$

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