Question Number 37298 by math khazana by abdo last updated on 11/Jun/18
$${calculate}\:\:\int_{\gamma} \:\:\:\:\frac{{z}+\mathrm{1}}{{z}\left({z}−\mathrm{1}\right)\left({z}+\mathrm{2}\right)}{dz}\:\:{with}\:\gamma\:{is}\:{the} \\ $$$${circle}\:\gamma\:=\left\{{z}\in{C}/\:\:\mid{z}\mid\:=\frac{\mathrm{3}}{\mathrm{2}}\right\} \\ $$
Commented by prof Abdo imad last updated on 15/Jun/18
$${let}\:\varphi\left({z}\right)\:=\:\frac{{z}+\mathrm{1}}{{z}\left({z}−\mathrm{1}\right)\left({z}+\mathrm{2}\right)}\:\:{tbe}\:{poles}\:{of}\:\varphi\:{are} \\ $$$$\mathrm{0},\mathrm{1},−\mathrm{2}\:\:\left(−\mathrm{2}\:{is}\:{out}\:{of}\:{the}\:{circle}\:\gamma\right) \\ $$$$\int_{\gamma} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:{Res}\left(\varphi,\mathrm{0}\right)\:+{Res}\left(\varphi,\mathrm{1}\right)\right\} \\ $$$${Res}\left(\varphi,\mathrm{0}\right)\:=\:\frac{\mathrm{1}}{−\mathrm{2}}\:=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${Res}\left(\varphi,\mathrm{1}\right)\:=\:\frac{\mathrm{2}}{\mathrm{3}}\:\Rightarrow \\ $$$$\int_{\gamma} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:−\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{2}}{\mathrm{3}}\right\}=\mathrm{2}{i}\pi.\frac{\mathrm{1}}{\mathrm{6}}\:=\frac{{i}\pi}{\mathrm{3}}\:. \\ $$