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Question Number 144094 by mohammad17 last updated on 21/Jun/21

Commented by mohammad17 last updated on 21/Jun/21

$$helpmesirplease$$

Answered by Olaf_Thorendsen last updated on 21/Jun/21

$$\mathrm{Laurent}\mathrm{serie}\mathrm{of}f\mathrm{in}a:$$$$f\left(z\right)=\underset{n=0}{\sum ^{\mathrm{\infty }}}{a}_n{(z-a)}^n$$$$$$$$f\left(z\right)=2z\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n{z}^{2n}$$$$f\left(z\right)=2\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n{z}^{2n+1}$$$$\mathrm{\forall }n,a_{2n}=0\mathrm{and}{a}_{2n+1}=2{(-1)}^n$$

Commented by mohammad17 last updated on 21/Jun/21

$$thankyousir$$

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