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?
What type of singularity does $\exp(\frac{t}{2} (z - \frac{1}{z}))$ have on $z = 0, \infty$?
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1Related: http://math.stackexchange.com/questions/206245/singularities-of-ez-frac1z/206251#206251 – 2012-11-09
2 Answers
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.
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.