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Question Number 35220 by abdo.msup.com last updated on 16/May/18
$${let}\:{z}\:{from}\:{C}\:{and}\:{f}\left({z}\right)=\:\frac{\mathrm{2}{z}}{\left({z}−\mathrm{1}\right)\left(\mathrm{2}{z}\:+\mathrm{1}\right)} \\ $$$${developp}\:{f}\:{at}\:{integr}\:{serie}. \\ $$
Commented by abdo mathsup 649 cc last updated on 18/May/18
$${let}\:{decompose}\:{f}\left({z}\right)=\: \\ $$$${f}\left({z}\right)=\:\frac{{a}}{{z}−\mathrm{1}}\:+\:\frac{{b}}{\mathrm{2}{z}+\mathrm{1}} \\ $$$${a}\:={lim}_{{z}\rightarrow\mathrm{1}} \left({z}−\mathrm{1}\right){f}\left({z}\right)\:=\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${b}\:={lim}_{{z}\rightarrow−\frac{\mathrm{1}}{\mathrm{2}}} \:\:\:\left(\mathrm{2}{z}+\mathrm{1}\right){f}\left({z}\right)=\:\frac{−\mathrm{1}}{−\frac{\mathrm{3}}{\mathrm{2}}}\:=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${f}\left({z}\right)=\:\frac{\mathrm{2}}{\mathrm{3}}\left\{\:\frac{\mathrm{1}}{{z}−\mathrm{1}}\:\:+\frac{\mathrm{1}}{\mathrm{2}{z}+\mathrm{1}}\right\} \\ $$$$=−\frac{\mathrm{2}}{\mathrm{3}}\:\frac{\mathrm{1}}{\mathrm{1}−{z}}\:\:+\:\frac{\mathrm{2}}{\mathrm{3}}\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}{z}}\:{so}\:{for}\:\mid{z}\mid<\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${f}\left({z}\right)\:=−\frac{\mathrm{2}}{\mathrm{3}}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{z}^{{n}} \:\:\:+\frac{\mathrm{2}}{\mathrm{3}}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\left(−\mathrm{1}\right)^{{n}} \mathrm{2}^{{n}} \:{z}^{{n}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\left\{\left(−\mathrm{1}\right)^{\boldsymbol{{n}}} \:\mathrm{2}^{\boldsymbol{{n}}} \:−\mathrm{1}\right\}\boldsymbol{{z}}^{\boldsymbol{{n}}} \:\:.\boldsymbol{{the}}\:\boldsymbol{{radius}}\:\boldsymbol{{of}} \\ $$$$\boldsymbol{{convergence}}\:\boldsymbol{{is}}\:\boldsymbol{{R}}=\frac{\mathrm{1}}{\mathrm{2}}\:. \\ $$