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Question-144094




Question Number 144094 by mohammad17 last updated on 21/Jun/21
Commented by mohammad17 last updated on 21/Jun/21
help me sir please
$${help}\:{me}\:{sir}\:{please} \\ $$
Answered by Olaf_Thorendsen last updated on 21/Jun/21
Laurent serie of f in a :  f(z) = Σ_(n=0) ^∞ a_n (z−a)^n     f(z) = 2zΣ_(n=0) ^∞ (−1)^n z^(2n)   f(z) = 2Σ_(n=0) ^∞ (−1)^n z^(2n+1)   ∀n, a_(2n)  = 0 and a_(2n+1)  = 2(−1)^n
$$\mathrm{Laurent}\:\mathrm{serie}\:\mathrm{of}\:{f}\:\mathrm{in}\:{a}\:: \\ $$$${f}\left({z}\right)\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{n}} \left({z}−{a}\right)^{{n}} \\ $$$$ \\ $$$${f}\left({z}\right)\:=\:\mathrm{2}{z}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} {z}^{\mathrm{2}{n}} \\ $$$${f}\left({z}\right)\:=\:\mathrm{2}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} {z}^{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\forall{n},\:{a}_{\mathrm{2}{n}} \:=\:\mathrm{0}\:\mathrm{and}\:{a}_{\mathrm{2}{n}+\mathrm{1}} \:=\:\mathrm{2}\left(−\mathrm{1}\right)^{{n}} \\ $$
Commented by mohammad17 last updated on 21/Jun/21
thank you sir
$${thank}\:{you}\:{sir} \\ $$

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