[tex]f=\frac{z}{1-\cos z}[/tex](adsbygoogle = window.adsbygoogle || []).push({});

the singular points are z=2pik and zero

i solved for z=2pik

and poles because there limit is infinity

now i want to determine te power of the pole

g=1/f=[tex]\frac{1-\cos z}{z}[/tex]

[tex]g'=\frac{(-\sin z)z-(1-\cos z)}{z^2}[/tex]

[tex]g'(0)=0/0[/tex]

[tex]g''=\frac{-\sin z z^2 -(cos z -1)2z}{z^4}[/tex]

[tex]g''(0)=0/0[/tex]

the book says that its a first order pole

it should differ zero in order to be pole

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# Homework Help: Power of poles

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