 Mabel Breault

2022-01-14

How can I find the order of the pole ? I know the answer should be 2, but I cant Durst37

Expert

First, note that $2\mathrm{log}\left(z\right)$ does not vanish at $z=\frac{\pi }{2}$, so we can ignore it; it does not contribute to the pole. So consider simply $\frac{1}{1-\mathrm{sin}\left(z\right)}$. We want to write things in terms of $w=z-\frac{\pi }{2}$,
so let $\mathrm{sin}\left(z\right)=\mathrm{sin}\left(w+\frac{\pi }{2}\right)=\mathrm{cos}\left(w\right)$, where at the end I have used a simple trig formula. Then we can expand $\mathrm{cos}\left(w\right)=1-\frac{{w}^{2}}{2}+\frac{{w}^{4}}{24}-\cdots$, so
$\frac{1}{1-\mathrm{cos}\left(w\right)}=\frac{1}{\frac{{w}^{2}}{2}-\frac{{w}^{4}}{24}+\cdots }=\frac{1}{{w}^{2}\left(\frac{1}{2}-\frac{{w}^{2}}{24}+\cdots \right)}$
Thus, we can see that the order of the pole is 2. kaluitagf

Expert

you have the function $h\left(z\right)=\mathrm{log}\left(z\right)\left(1-\mathrm{sin}z\right)$
then $h\left(\frac{\pi }{2}\right)=0$,

is a zero of order two for h, therefore is a pole for $\frac{1}{h\left(z\right)}$ of order two. alenahelenash

Expert

Write $f\left(z\right)=\frac{1/\mathrm{log}{z}^{2}}{\left(1-\mathrm{sin}z\right)}=\frac{g\left(z\right)}{h\left(z\right)}$Observe that g(z) is analytic and non- zero at $z=\frac{\pi }{2}$ hence the order of zero of h(z) there determines the order of pole for rational function f(z). suggests h(z) has zero of order 2 at $z=\frac{\pi }{2}$ so f(z) has pole of order 2 there.