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I want to identify and classify the singularities of $\frac{\sin(z)}{1-\tan(z)}$. There are obvious singularities where $\tan(z)=1$. I have two problems:

  1. I don't know how to classify $z = \frac{1}{2}(n+1)\pi$ where $\tan$ is not defined, are those points even singularities?
  2. How do I find the Laurent series? I don't know if it can even be found directly. Of course, I know the power series for sin and tan, but there is division involved. Is there a smarter way to classify the singularities?
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Note that $\displaystyle \lim_{z \to \frac{(n+1)\pi}{2}} \frac{\sin z}{1-\tan z} = \lim_{z \to \frac{(n+1)\pi}{2}} \frac{\sin z \cos z}{\cos z-\sin z} = 0$. So those points are removable singularities.

You can check that $\displaystyle \tan z = 1 \iff z = \frac{\pi}{4} + n\pi, \, n\in\mathbb{Z}$ and those points are simple poles. Hence Laurent series becomes $\displaystyle \frac{a_{-1}}{z-w_n} + a_0 + a_1(z-w_n) + a_2(z-w_n)^2 + \cdots$ where $\displaystyle w_n = \frac{\pi}{4} + n\pi$. This gives:

$\displaystyle \frac{\sin z}{1-\tan z} = \frac{a_{-1}}{z-w_n} + a_0 + a_1(z-w_n) + a_2(z-w_n)^2 + \cdots$

$\displaystyle \sin z = (1-\tan z) \left( \frac{a_{-1}}{z-w_n} + a_0 + a_1(z-w_n) + a_2(z-w_n)^2 + \cdots \right)$. Use this equality to find Laurent coefficients by considering Taylor series of $\sin z$ and $1-\tan z$ at $\displaystyle z=w_n$.

For example, let $n=0$ so $w_0 =\dfrac{\pi}{4}$

$\displaystyle \sin z = \frac{1}{\sqrt{2}} + \frac{x-\frac{\pi}{4}}{\sqrt{2}} - \frac{\left( x-\frac{\pi}{4} \right)^2}{2\sqrt{2}} - \frac{\left( x-\frac{\pi}{4} \right)^3}{6\sqrt{2}} + \frac{\left( x-\frac{\pi}{4} \right)^4}{24\sqrt{2}} + \frac{\left( x-\frac{\pi}{4} \right)^5}{120\sqrt{2}} - \cdots $

$\displaystyle = \left( -2\left( x-\frac{\pi}{4} \right) - 2\left( x-\frac{\pi}{4} \right)^2 - \frac{8}{3}\left( x-\frac{\pi}{4} \right)^3 + \cdots \right) \left( \frac{a_{-1}}{z-\frac{\pi}{4}} + a_0 + a_1\left(z-\frac{\pi}{4}\right) + a_2\left(z-\frac{\pi}{4}\right)^2 + \cdots \right)$

$\displaystyle \implies -2a_{-1} = \frac{1}{\sqrt{2}}, \, -2a_0 -2a_{-1} = \frac{1}{\sqrt{2}}, \, -2a_1 -2a_0 -\frac{8}{3}a_{-1} = -\frac{1}{2\sqrt{2}}, \, \dots$ by multiplying and comparing the terms of same degree.

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