Classifying the singular point $z=\frac{\pi}{2}$ for $f(z)=\frac{z-\frac{\pi}{2}}{1-\sin(z)}$

Question

I am trying to find determine the type of singular point $z=\frac{\pi}{2}$ is for the function, $$f(z)=\frac{z-\frac{\pi}{2}}{1-\sin(z)}.$$

My attempt:

I originally thought that $$\lim_{z\to\frac{\pi}{2}} z-\frac{\pi}{2}$$ was a simple zero and that $$\lim_{z\to\frac{\pi}{2}} 1-\sin(z)$$ was also a simple zero. Hence $z=\frac{\pi}{2}$ is a removable singularity. But the answer states that for $1-\sin(z)$, $z=\frac{\pi}{2}$ is a zero of order $2$. How can this be?

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However, if I use the fact that $z=\frac{\pi}{2}$ is a zero of order $k$ iff $$f(\frac{\pi}{2})=f'(\frac{\pi}{2})=..=f^{k-1}(\frac{\pi}{2})=0, \ \text{where} \ f^k(\frac{\pi}{2})\neq 0,$$ then $$f(\frac{\pi}{2})=1-\sin(\frac{\pi}{2})=0,$$ $$f'(\frac{\pi}{2})=-\cos(\frac{\pi}{2})=0,$$ $$f''(\frac{\pi}{2})=\sin(\frac{\pi}{2})\neq 0.$$ Hence $z=\frac{\pi}{2}$ is a zero of order $k=2$ for $1-\sin(z)$. Is this correct?

Answer 1

We can instead study $g(z):=f(z+\frac\pi2)$ at $z=0$.

$$\begin{align} g(z)&= \frac{z}{1-\cos z}\\ &=\frac{z}{z^2+O(z^3)}\\ &=\frac{z}{z^2\left(1+\frac{O(z^3)}{z^2}\right)}\\ &=\frac{1}z(1+O(z))\qquad{(1)}\\ &=\frac{1}z+O(1) \end{align} $$ $(1)$: by geometric series.

Thus it is a simple pole with residue $1$.

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To see why $\frac1{1-\sin z}$ has a pole of order two at $z=\frac\pi2$, we can instead look at the order of zero of $1-\sin z$ at $z=\frac\pi2$.

$$\sin z=\cos\left(z-\frac{\pi}2\right)=1-\left(z-\frac\pi2\right)^2+\cdots$$

Hence, $$1-\sin z=\left(z-\frac\pi2\right)^2+\cdots$$

Obviously, that is a zero of order two, and its reciprocal will be a pole of order two.

Answer 2

By basic trigonometry $$ 1-\sin z=1-\cos (\pi /2 -z)=$$ $$=2\sin^2 ((\pi /2 -z)/2)= 2g(z)\cdot ((\pi /2-z)/2)^2$$ where $\lim_{z\to \pi /2}\,g(z)=1.$... So when $\sin z \ne 1$ we have $$(\pi /2 -z)/(1-\sin z)=2g(z)/(\pi /2 -z).$$