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Question Number 34116 by tanmay.chaudhury50@gmail.com last updated on 30/Apr/18

Commented by abdo mathsup 649 cc last updated on 02/May/18

$$letputf\left(x\right)=\frac{1-x^2}{x^2-2xcos\theta +1}thepolesoffare$$$$x_1=\frac{1-x^2}{(x-e^{i\theta })(x-e^{-i\theta })}$$$$=(1-x^2)(\frac{1}{x-e^{i\theta }}-\frac{1}{x-e^{-i\theta }})\frac{1}{2isin\theta }$$$$=\frac{1-x^2}{2isin\theta }(-\frac{1}{e^{i\theta }-x}+\frac{1}{e^{-i\theta }-x})$$$$=\frac{1-x^2}{2isin\theta }(\frac{e^{i\theta }}{1-x{e}^{i\theta }}-\frac{e^{-i\theta }}{1-x{e}^{-i\theta }})$$$$=\frac{1-x^2}{2isin\theta }(e^{i\theta }\sum _{n=0}^{\mathrm{\infty }}{x}^n{e}^{in\theta }-e^{-i\theta }\sum _{n=0}^{\mathrm{\infty }}{x}^n{e}^{-in\theta })$$$$=\frac{1-x^2}{2isin\theta }(\sum _{n=0}^{\mathrm{\infty }}{x}^n{e}^{i(n+1)\theta }-\sum _{n=0}^{\mathrm{\infty }}{x}^n{e}^{-i(n+1)\theta })$$$$=\frac{1-x^2}{2isin\theta }\left(\sum _{n=0}^{\mathrm{\infty }}{x}^n\left(2isin(n+1)\theta \right)\right)$$$$=\frac{1-x^2}{sin\theta }\sum _{n=0}^{\mathrm{\infty }}\left(sin(n+1)\theta \right)x^n$$$$=\frac{1}{sin\theta }\sum _{n=0}^{\mathrm{\infty }}sin(n+1)\theta .x^n-\frac{1}{sin\theta }\sum _{n=0}^{\mathrm{\infty }}sin(n+1)\theta .x^{n+2}$$$$=....$$$$$$

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