Pole of the function $f(z)=\frac{z}{(1-e^z).\sin z}.$

Question

Let , $f(z)$ be a meromorphic function given by $$f(z)=\frac{z}{(1-e^z).\sin z}.$$Prove or disprove that " $z=0$ is a pole of order $2$ ".

1st Argument :

Firstly we can rewrite the given function as $\displaystyle f(z)=\frac{z}{\sin z}.\frac{1}{1-e^z}$. Now $\displaystyle \frac{z}{\sin z}$ has a removable singularity at $z=0$. Also , $\displaystyle \frac{1}{1-e^z}$ has a pole of order $1$ at $z=0$. So finally the function has a pole of order $1$ at $z=0$.

2nd Argument :

But we know that " A function $f$ has a pole of order $m$ if and only if $(z-a)^mf(z)$ has a removable singularity at $z=a$."

So here , if we can show that $\displaystyle z^2f(z)=\frac{z^3}{(1-e^z).\sin z}=g(z)(say)$ has a removable singularity at $z=0$ then we can show that $f$ has a pole of order $2$ at $z=0$.

Now , as $\displaystyle \lim_{z\to 0}z.g(z)=0$ , so $g$ has a removable singularity at $z=0$ and consequently $f$ has a pole of order $2$ at $z=0$.

I'm confused that which is correct ? I could not find any mistake in both of my arguments. Can anyone detect the fallacy ?

Answer 1

You remember wrongly the result: a holomorphic function $f$ over $B_r(a)\setminus\{a\}$ (the punctured circle of radius $r$ around $a$) has a pole at $a$ if and only if it hasn't a removable singularity at $a$ but $(z-a)^nf(z)$ has a removable singularity at $a$ for some integer $n>0$. The minimum such integer is the order of the pole.

It is easy to see that this is the same as saying that there exists an integer $m$ such that $(z-a)^mf(z)$ has a removable singularity and $$ \lim_{z\to a}(z-a)^mf(z)\ne0 $$ In this case, $m$ is uniquely determined and is the order of the pole.

It can be shown that if $f$ has a pole of order $m$ at $a$, then its representation as a Laurent series $$ f(z)=\frac{c_{-m}}{(z-a)^m}+ \frac{c_{-1}}{(z-a)}+c_0+c_1z+\dotsb $$ holds for every $z\in B_r(a)\setminus\{a\}$.

In your case, the singularity at $0$ is not removable, but $$ zf(z)=\frac{z}{e^z-1}\frac{z}{\sin z} $$ has a removable singularity at $0$. Hence the order of the pole is $1$. Note that $\lim_{z\to0}zf(z)=1\ne0$.

Answer 2

Your first argument is correct. It has a pole of order 1.

For the second argument, where you go wrong is jumping to the conclusion that since $\lim_{z\rightarrow 0}g(z) = 0$ $g(z)$ has a removable singularity and $f(z)$ has a pole order 2. Think about what would happen if you instead let $g(z) = z^{27}f(z).$ You'd still have $g(z)\rightarrow 0.$ Does that mean that $g$ has a 27 'th order pole?