How can I find the order of the pole z=π2 for f(z)=1(2log⁡(z))(1−sin⁡(z))? I know the answer should be 2, but I cant

First, note that 2log⁡(z) does not vanish at z=π2, so we can ignore it; it does not contribute to the pole. So consider simply 11−sin⁡(z). We want to write things in terms of w=z−π2, so let sin⁡(z)=sin⁡(w+π2)=cos⁡(w), where at the end I have used a simple trig formula. Then we can expand cos⁡(w)=1−w22+w424−⋯, so 11−cos⁡(w)=1w22−w424+⋯=1w2(12−w224+⋯) Thus, we can see that the order of the pole is 2.

you have the function h(z)=log⁡(z)(1−sin⁡z)then h(π2)=0,h′(z)=1−sin⁡zz−log⁡zcos⁡z then h′(π2)=0h″(z)=−zcos⁡z−(1−sin⁡z)z2−cos⁡zz+log⁡zsin⁡z thenh″(π2)≠0 hence π2 is a zero of order two for h, therefore is a pole for 1h(z) of order two.

First, note that 2log⁡(z) does not vanish at z=π2, so we can ignore it; it does not contribute to the pole. So consider simply 11−sin⁡(z). We want to write things in terms of w=z−π2, so let sin⁡(z)=sin⁡(w+π2)=cos⁡(w), where at the end I have used a simple trig formula. Then we can expand cos⁡(w)=1−w22+w424−⋯, so 11−cos⁡(w)=1w22−w424+⋯=1w2(12−w224+⋯) Thus, we can see that the order of the pole is 2.

you have the function h(z)=log⁡(z)(1−sin⁡z)then h(π2)=0,h′(z)=1−sin⁡zz−log⁡zcos⁡z then h′(π2)=0h″(z)=−zcos⁡z−(1−sin⁡z)z2−cos⁡zz+log⁡zsin⁡z thenh″(π2)≠0 hence π2 is a zero of order two for h, therefore is a pole for 1h(z) of order two.

Key to "How can I find the order of the..."

Mabel Breault

Answered question

2022-01-14

How can I find the order of the pole

z = \frac{π}{2} for f (z) = \frac{1}{(2 \log (z)) (1 - \sin (z))}

? I know the answer should be 2, but I cant

Answer & Explanation

Durst37

Beginner2022-01-15Added 37 answers

First, note that

2 \log (z)

does not vanish at

z = \frac{π}{2}

, so we can ignore it; it does not contribute to the pole. So consider simply

\frac{1}{1 - \sin (z)}

. We want to write things in terms of

w = z - \frac{π}{2}

,
so let

\sin (z) = \sin (w + \frac{π}{2}) = \cos (w)

, where at the end I have used a simple trig formula. Then we can expand

\cos (w) = 1 - \frac{w^{2}}{2} + \frac{w^{4}}{24} - \dots

, so

\frac{1}{1 - \cos (w)} = \frac{1}{\frac{w^{2}}{2} - \frac{w^{4}}{24} + \dots} = \frac{1}{w^{2} (\frac{1}{2} - \frac{w^{2}}{24} + \dots)}

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

kaluitagf

Beginner2022-01-16Added 38 answers

you have the function

h (z) = \log (z) (1 - \sin z)

then

h (\frac{π}{2}) = 0

h^{'} (z) = \frac{1 - \sin z}{z} - \log z \cos z then h^{'} (\frac{π}{2}) = 0

h^{″} (z) = \frac{- z \cos z - (1 - \sin z)}{z^{2}} - \frac{\cos z}{z} + \log z \sin z then h^{″} (\frac{π}{2}) \neq 0 hence \frac{π}{2}

is a zero of order two for h, therefore is a pole for

\frac{1}{h (z)}

of order two.

alenahelenash

Expert2022-01-24Added 556 answers

Write

f (z) = \frac{1 / \log z^{2}}{(1 - \sin z)} = \frac{g (z)}{h (z)}

Observe that g(z) is analytic and non- zero at

z = \frac{π}{2}

hence the order of zero of h(z) there determines the order of pole for rational function f(z).

h (\frac{π}{2}) = h^{'} (\frac{π}{2}) = 0 and h^{″} (\frac{π}{2}) \neq 0

suggests h(z) has zero of order 2 at

z = \frac{π}{2}

so f(z) has pole of order 2 there.

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