I have some problems with the following excersice: Find the order of the pole of: $$\frac{1}{(2\cos z -2 + z^2)^2}$$ at $z=0$. I thought it is maybe better to work here with $1/f$ and find the order of the zero at $z=0$. But I don't know how to continue. Thanks in advance!
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1Use the Taylor expansion of $\cos$. Use as many terms as you need to determine the order. – Daniel Fischer Jan 09 '14 at 12:57
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Thank you for your fast reply! – Leslie Jan 09 '14 at 13:10
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The pole is of order $8$. – Mhenni Benghorbal Jan 09 '14 at 13:12
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Thank you for your fast reply! I don't understand how this helps me to get the order, because then I get: (2(1-x^2/2!+x^4/4!-..+..) - 2 + z^2)^2 and I do not see how I – Leslie Jan 09 '14 at 13:19
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@ Mhenni Benghorbal, that is indeed the correct answer, but how do you see that so easy? – Leslie Jan 09 '14 at 13:19
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@Leslie: See this. – Mhenni Benghorbal Jan 09 '14 at 15:57
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$$2\cos z-2+z^2=2\left(1-\frac{z^2}2+\frac{z^4}{24}-\ldots\right)-2+z^2=$$
$$=\frac{z^4}{12}-\frac{2z^6}{6!}+\ldots=z^4\left(\frac1{12}-\frac{2z^2}{6!}+\ldots\right)\implies$$
$$\left(2\cos x-2+z^2\right)^2=z^8\left(\frac1{144}+\frac{z^2}{3\cdot 6!}+\ldots\right)\implies$$
$$\frac1{(2\cos z-2+z^2)^2}=\;\;\ldots\ldots$$
DonAntonio
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Oo thank you now I see it. It is really easy actually. I just had to write it out and forgot about the outer square in the beginning! – Leslie Jan 09 '14 at 13:33
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