Mabel Breault

Answered

2022-01-14

How can I find the order of the pole $z=\frac{\pi}{2}\text{}\text{for}\text{}f\left(z\right)=\frac{1}{\left(2\mathrm{log}\left(z\right)\right)(1-\mathrm{sin}\left(z\right))}$ ? I know the answer should be 2, but I cant

Answer & Explanation

Durst37

Expert

2022-01-15Added 37 answers

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$\mathrm{sin}\left(z\right)=\mathrm{sin}(w+\frac{\pi}{2})=\mathrm{cos}\left(w\right)$ , 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

$\frac{1}{1-\mathrm{cos}\left(w\right)}=\frac{1}{\frac{{w}^{2}}{2}-\frac{{w}^{4}}{24}+\cdots}=\frac{1}{{w}^{2}(\frac{1}{2}-\frac{{w}^{2}}{24}+\cdots )}$

Thus, we can see that the order of the pole is 2.

so let

Thus, we can see that the order of the pole is 2.

kaluitagf

Expert

2022-01-16Added 38 answers

you have the function $h\left(z\right)=\mathrm{log}\left(z\right)(1-\mathrm{sin}z)$

then$h\left(\frac{\pi}{2}\right)=0$ ,

${h}^{\prime}\left(z\right)=\frac{1-\mathrm{sin}z}{z}-\mathrm{log}z\mathrm{cos}z\text{}\text{then}\text{}{h}^{\prime}\left(\frac{\pi}{2}\right)=0$

$h{}^{\u2033}\left(z\right)=\frac{-z\mathrm{cos}z-(1-\mathrm{sin}z)}{{z}^{2}}-\frac{\mathrm{cos}z}{z}+\mathrm{log}z\mathrm{sin}z\text{}\text{then}h{}^{\u2033}\left(\frac{\pi}{2}\right)\ne 0\text{}\text{hence}\text{}\frac{\pi}{2}$ is a zero of order two for h, therefore is a pole for $\frac{1}{h\left(z\right)}$ of order two.

then

alenahelenash

Expert

2022-01-24Added 366 answers

Write $f(z)=\frac{1/\mathrm{log}{z}^{2}}{(1-\mathrm{sin}z)}=\frac{g(z)}{h(z)}$ 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).$h(\frac{\pi}{2})={h}^{\prime}(\frac{\pi}{2})=0\text{}\text{and}\text{}{h}^{\u2033}(\frac{\pi}{2})\ne 0$ suggests h(z) has zero of order 2 at $z=\frac{\pi}{2}$ so f(z) has pole of order 2 there.

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