Toronto Math Forum
MAT3342020F => MAT334Tests and Quizzes => Test 4 => Topic started by: Xinxuan Lin on December 01, 2020, 01:42:57 PM

f(z) = (z$^4$ $\pi^4$)tan$^2$($\frac{z}{2}$)
Part b of this question is asking to determine the types of the singular point.
In solution, it says z=2n$\pi$ with n$\neq$ $\pm$ 1 are double zeros; z=(2n+1)$\pi$ with n$\neq$ $\pm$ 1 are double poles.
Could anyone explain why n$\neq$ $\pm$ 1 here? Why is not n$\neq$ 1, 0?
Thanks in advanced!

What I got on these poles were the same as you did $n \neq 0\ \& 1$. I believe that was a typo.

I got that there is no restriction on the n when z=2n𝜋, and n ≠ 0 & 1 when z=(2n+1)𝜋