# What type of singularity does $\exp(\frac{t}{2} (z - \frac{1}{z}))$ have on $z = 0, \infty$?

What type of singularity does $\exp(\frac{t}{2} (z - \frac{1}{z}))$ have on $z = 0, \infty$ for a fixed $t$? I've concluded this function has simple poles on both $z=0, \infty$ because I thought the function goes to infinity as it approaches to each of those singularity. But I've seen somewhere that they are essential singularities. Which is correct and why?

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– Cameron Buie Nov 9 '12 at 11:24

For the sake of curiosity, in this special case we can give an explicit answer. $$\exp\left[\frac{t}{2} \left(z - \frac{1}{z}\right)\right]=\sum_{n=-\infty}^{\infty}J_n(t)z^n,$$ where $J_n$ is the Bessel function of the first kind.
They are essential singularities. To see this for, for example, the singularity at $z = 0$, you can form the Laurent series expansion there. It will have infinitely many negative-degree terms, thus is cannot be a pole or removable singularity and must be an essential singularity. For the other singularity at $z = \infty$, substitute $z \rightarrow \frac{1}{z}$ and take the Laurent expansion again.
You can also examine the limit behavior along different approaches to the singularities. Consider approaching $z = 0$ along the imaginary axis, for example. The function will oscillate ever more rapidly as the point is approached.
Namely, substitute $z \rightarrow iz$ to get $\exp(\frac{t}{2} (iz - \frac{1}{iz})) = \exp(\frac{t}{2} (iz - -i\frac{1}{z})) = \exp(\frac{t}{2} i(z + \frac{1}{z})) = \cos(\frac{t}{2} (z + \frac{1}{z})) + i \sin(\frac{t}{2} (z + \frac{1}{z}))$ and see where the zeros of cos/sin lie. This will reveal the oscillation. It's just like the $\sin(\frac{1}{x})$ example from intro calc.