Question Number 37297 by math khazana by abdo last updated on 11/Jun/18
$${calculate}\:\:\int_{{C}} \:\:\:\:\frac{{z}}{{z}^{\mathrm{2}} \:+\mathrm{1}}{dz}\:\:{with}\:{C}=\left\{{z}\in{C}/\mid{z}\mid=\frac{\mathrm{1}}{\mathrm{2}}\right\} \\ $$
Commented by math khazana by abdo last updated on 14/Jun/18
$${let}\:\varphi\left({z}\right)=\frac{{z}}{{z}^{\mathrm{2}} \:+\mathrm{1}}\:\:{the}\:{poles}\:{of}\:\varphi\:{are}\:{i}\:{and}\:−{i} \\ $$$$\left({are}\:{out}\:{of}\:{the}\:{circle}\:\mid{z}\mid=\frac{\mathrm{1}}{\mathrm{2}}\right)\:\Rightarrow\int_{{C}} \varphi\left({z}\right){dz}\:=\mathrm{0}\Rightarrow \\ $$$$\int_{{C}} \:\frac{{z}}{{z}^{\mathrm{2}} \:+\mathrm{1}}{dz}\:=\mathrm{0}\:. \\ $$