Spring 2020 Test 2 Monday Sitting Problem 3

[Xinxuan Lin]

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!

[Jiaqi Bi]

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

[Xuefen luo]

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