Question Number 74914 by arkanmath7@gmail.com last updated on 03/Dec/19
$${Can}\:{someone}\:{solve}\:{this}\:{question}\:{plz}? \\ $$$${solve}\:{the}\:{contour}\:{integral}\: \\ $$$$\int_{{C}} \:\frac{{e}^{{iz}} }{{z}^{\mathrm{3}} }\:{dz}\:{where}\:{C}\:{is}\:{the}\:{circle}\:\mid{z}\mid=\mathrm{2} \\ $$
Answered by mind is power last updated on 05/Dec/19
$$\mathrm{pols}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{z}\right)=\frac{\mathrm{e}^{\mathrm{iz}} }{\mathrm{z}^{\mathrm{3}} }\:\:\mathrm{has}\:\mathrm{pol}\:\mathrm{at}\:\mathrm{z}=\mathrm{0}\:\mathrm{order}\:\mathrm{3} \\ $$$$\int\frac{\mathrm{e}^{\mathrm{iz}} }{\mathrm{z}^{\mathrm{3}} }\mathrm{dz}=\mathrm{2i}\pi\:\mathrm{Res}\left(\mathrm{f},\mathrm{0}\right)=\mathrm{2i}\pi.\frac{\mathrm{1}}{\mathrm{2}!}\frac{\mathrm{d}^{\mathrm{2}} }{\mathrm{dz}^{\mathrm{2}} }.\mathrm{z}^{\mathrm{3}} \mathrm{f}\left(\mathrm{z}\right)\mid_{\mathrm{z}=\mathrm{0}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{2i}\pi\left(−\mathrm{e}^{\mathrm{iz}} \right)\mid_{\mathrm{z}=\mathrm{0}} =−\mathrm{i}\pi \\ $$
Commented by arkanmath7@gmail.com last updated on 04/Dec/19
$${this}\:{qust}.\:{isn}'{t}\:{solved}\:{from}\:{Schoum}'{s}\:{and}\:{the} \\ $$$${answer}\:{is}\:−\pi{i} \\ $$
Commented by mind is power last updated on 05/Dec/19
$$\mathrm{yeah}\:\mathrm{Res}\left(\mathrm{f},\mathrm{0}\right)=\frac{\mathrm{1}}{\mathrm{2}!}\frac{\mathrm{d}^{\mathrm{2}} }{\mathrm{dz}^{\mathrm{2}} }\mathrm{z}^{\mathrm{3}} \mathrm{f}\left(\mathrm{z}\right)\mid_{\mathrm{z}=\mathrm{0}} \\ $$
Commented by arkanmath7@gmail.com last updated on 05/Dec/19
$${thank}\:{you}\:{sir} \\ $$
Commented by mind is power last updated on 05/Dec/19
$$\mathrm{y}'\mathrm{re}\:\mathrm{welcom} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{generaliz}\:\mathrm{it} \\ $$$$\int_{\mathrm{C}} \frac{\mathrm{e}^{\mathrm{iz}} }{\mathrm{z}^{\mathrm{n}} }\mathrm{dz}=\frac{\mathrm{2i}\pi}{\left(\mathrm{n}−\mathrm{1}\right)!}.\left(\mathrm{i}\right)^{\mathrm{n}−\mathrm{1}} \\ $$$$ \\ $$$$=\frac{\mathrm{2}\pi.\left(\mathrm{i}\right)^{\mathrm{n}} }{\left(\mathrm{n}−\mathrm{1}\right)!} \\ $$
Commented by arkanmath7@gmail.com last updated on 06/Dec/19
$${good}\:{work}\:{thanks}\:{again} \\ $$