# Understanding the nature of a singularity

I'm trying to find the nature of the point $$z=\pi/2$$ for $$f(z)=\tan^2z$$ (e.g if it's regular, an essential singularity, a pole of degree n). I sought help from this similar post and what I essentially shown is that: $$\tan^2 z = \left(\dfrac{\sin z}{\cos z}\right)^2 = \left(\dfrac{\cos(z-\pi/2)}{\sin(z-\pi/2)}\right)^2 = \left(\dfrac{1}{z-\pi/2}\dfrac{z-\pi/2}{\sin(z-\pi/2)}\cos(z-\pi/2) \right)^2$$ Now, if I understand correctly then showing $$\lim_{z\rightarrow\pi/2}(z-\pi/2)\tan^2z$$ is finite would allow me to deduce that the pole is singular. From my above expansion, it's clear that this limit would not converge since I still have a remaining $$1/(z-\pi/2)$$ term that will blow up.

However, I can see how $$\lim_{z\rightarrow\pi/2}(z-\pi/2)^2\tan^2z$$ would converge instead, and this suggests that I have a pole of degree 2.

My problem with this is that I could have chosen an arbitrarily high $$n$$ such that $$\lim_{z\rightarrow\pi/2}(z-\pi/2)^n\tan^2z$$ would converge (e.g if I decided to guess $$n=5$$, then would I have claimed that the pole is of degree 5)? How do I know that I actually have a pole of degree 2? Isn't it possible that I just did not rewrite $$f(z)$$ correctly such that the $$1/(z-\pi/2)$$ cancels out?

• If you get finite non-zero limit for $(z-\pi /2)^{n}f(z)$ then the order of the pole is exactly $n$. For higher powers you will get the limit as $0$. Commented Jan 23, 2020 at 5:41

The order of a pole(if it has finite order)$$z_0$$ is the unique number $$n$$ such that $$(z-z_0)^n f(z)$$ is holomorphic non-zero in a neighbourhood of $$z_0$$(uniqueness follows from the analyticity property).
With this in mind, one easily sees that for this $$n$$ we must have $$\lim_{z\to z_0} (z-z_0)^n f(z) \neq 0$$ and exists. Thus, locating any one such $$n$$ suffices, since that must be the order of the pole. We have located $$n=2$$ by checking the limit exists and is non-zero. Now, for $$n<2$$ the limit will be infinite, and for $$n>2$$ the limit will be $$0$$ (just by comparison/domination arguments, product of limit etc.), so these cases become trivial.
EDIT : In the above case, one sees that $$\lim_{z \to \frac{\pi}{2}} \left(z - \frac \pi 2\right)^2 \tan^2 z =\lim_{z \to \pi/2} \sin^2 z \lim_{z \to \pi/2} \frac{(z-\pi/2)^2}{\cos^2 z} = 1$$
Therefore, for all $$k>2$$, we have $$\lim_{z \to \frac{\pi}{2}}\left(z - \frac \pi 2\right)^2 \tan^2 z = 0$$ and for $$k=1$$ we have $$\lim_{z \to \frac{\pi}{2}}\left(z - \frac \pi 2\right)^2 \tan^2 z = +\infty$$, and the pole of order $$2$$.