Classify the singularity of $\frac{e^{z^2}-1}{z}$ and $\frac{e^{z^2}-1}{z^k}$ where $k$ is an integer greater then or equal to 2

Question

I was wondering if someone could verify my solution for these two problems. They are based on the textbook Complex Variables by Fisher.

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Let us start with $\frac{e^{z^2}-1}{z}$. We use the fact that $e^z = \sum_{n=0}^\infty z^n/n!$ to obtain: $$\frac{e^{z^2}-1}{z} = \sum_{n=0}^\infty \left(\frac{z^{2n-1}}{n!} \right) - \frac{1}{z} = \sum_{n=1}^\infty \frac{z^{2n-1}}{n!}$$ Since there are no negative exponents, we conclude that it has a removable singularity.

Similarly, we note that $$\frac{e^{z^2}-1}{z^k} = \sum_{n=0}^\infty \left(\frac{z^{2n-k}}{n!} \right) - \frac{1}{z^k} = \sum_{n=1}^\infty \frac{z^{2n-k}}{n!}$$ Now, if $k=2$, we are left with a summation with no negative exponents, hence a removable singularity.

If $k>2$, we are left with negative exponents. Specifically, the largest negative exponent is when $n=1$, and the exponent is thus $2-k$. This means that we have a pole of order $2-k$.

Answer 1

This is perfectly done. A minor mistake in the second equation: LHS should be divided by $z^k$ instead of $z$.

Answer 2

As $\lim_{z\rightarrow 0}\frac{e^{z^2}-1}{z^2}= 1$ so writing the given function as the product $\frac{e^{z^2}-1}{z^2}.\frac{1}{z^{k-2}}$ the factor $\frac{1}{z^{k-2}}$ is responsible to give a pole of order $k-2$, at $z=0$ where $(k>2)$. In case $k\le2$, this factor is analytic at $z=0$ so the former factor is responsible to give a removable singularity at $z=0$.