singularities and zeros (roots) of a function

Question

Given

$\frac{e^{z^2} -1}{z \sin(z)}$

I have to determine the order of the zeros and of the singularities. I think I'm confused by the difference between the two things.
One should be $z_k=k\pi$ . I usually derivate the function to obtain the order, I'm not sure how to proceed here. Plus, I found different definitions in different sites. Here what my book says for order of zeros: $f(z_0)=f'(z_0)=f''(z_0)=...=f^{(n-1)}(z_0) = 0$ and $f^{(n)} \neq0$;

Answer 1

$z=0$ is a removable singularity since

$$ \lim_{z\to\ 0} \frac{e^{z^2}-1}{z\sin{z}} = 1 $$ For zeros,

$e^{z^2}-1=0$ $\implies$ $e^{z^2}=1$ $\implies$ $e^{z^2} = e^{2n\pi i},$ $\space where\space n=0,1,2,.....$

$\implies$ $z^2 = 2n\pi i \implies z = \sqrt{2n\pi i} $

Note that, $n=0$ gives $z=0$ which is a removable singularity of the given function. So, $z_n=\sqrt{2n\pi i} , where\space n=1,2,3,...$ are the zeros of the given function.

Now, let $f(z) = e^{z^2} -1 $. Then,

$$\begin{align}\frac{e^{z^2} - 1}{z\sin{z}} & = \frac{f(z)}{z\sin{z}} \\ & = \frac{f(z_n) + \frac{(z-z_n)}{1!}f'(z_n) + \frac{(z-z_n)}{2!}f"(z_n) + ....}{z\sin{z}} \end{align} $$

[Exapanding f(z) by Taylor's series about the point $z=z_n$]

Since, $f(z_n)=0$ and $f'(z_n)\neq 0 , f"(z_n) \neq 0$ $$\begin{align}\frac{e^{z^2} - 1}{z\sin{z}} & = (z-z_n)\biggl[\frac{\frac{f'(z_n)}{1!} + (z-z_n) \frac{f"(z_n)}{2!} + .....}{z\sin{z}}\biggr] \\ & = (z-z_n) \phi (z) \end{align}$$

Where $\phi (z_n) =\frac{f'(z_n)}{z_n\sin{z_n}} \neq 0$.

Therefore, by the definition, $z=z_n$ are zeros of order 1 ,i.e., simple zeros.

For poles,

$z\sin{z} = 0 \implies z=z_k = k\pi ,\\ where\space k=1,2,3,...$

So $z=z_k$ are simple poles.

Answer 2

To find the poles and their order, I would take the Laurent series around each point $z=k\pi$ and look at the $z^{-n}$ terms.

For $z=0$, this isn't too hard to find. Start with the Taylor series for $e^z$ about $z=0$:

$$e^z=\sum_{k=0}^{\infty}\frac{z^k}{k!}$$

and then build it up, first subbing in $z^2$, subtracting $1$, and dividing through by $1/z$. You'll then need the expansion for $\frac{1}{\sin(z)}=\csc(z)$, which is

$$\csc(z)=\sum_{k=0}^{\infty}\frac{(-1)^k\cdot2(2^{2k-1}-1)B_{2k}}{(2k)!}z^{2k-1}$$

Multiplying the two series gives $$\frac{e^{z^2}-1}{z\sin (z)}=\sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{(-1)^k\cdot2(2^{2k-1}-1)B_{2k}}{(2k)!\cdot(n-k+1)!}z^{2k-1}\cdot z^{2n-2k+1}$$

which looks messy but it is easy to run through the first few terms and see that there are no negative power $z$ terms. This means that $z=0$ isn't a pole, and must be a removable singularity. Taking $$\lim_{z \to 0}\frac{e^{z^2}-1}{z\sin (z)}=\lim_{z \to 0}\frac{z\sin(z)}{e^{z^2}-1}=1$$ (via L'Hospital's rule) verifies this.

To find about $z=k\pi$ for $k\neq0$ is messier, but it is not necessary to find the full Laurent series. Observing the case for $z=0$, the first term of the expansion for $\frac{e^{z^2}-1}{z}$ was $z$ ; i.e. the constant term vanished. This does not occur for $z=\pm\pi,\pm2\pi,\pm3\pi...$ The first term will be $\frac{e^{k^2\pi^2}-1}{k\pi}$, which when multiplied with the first term of the $\csc(z)$ expansion ($\frac{\pm1}{z-k\pi}$) will mean the full Laurent series has a $z^{-1}$ term. Thus each $z=k\pi$ for $k\in\mathbb Z\setminus\{0\}$ is a simple pole.

To get the zeros, you will want $$e^{z^2}-1=0$$ s.t. $$z\sin(z)\neq0$$

You already know that $0$ is a removable singularity, so which other values of $z$ satisfy $e^{z^2}=1$? Once you find them, sub them into $z\sin(z)$ to verify that the function is defined at these points. You can then use your book's rule to find their order.