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