Can-someone-solve-this-question-plz-solve-the-contour-integral-C-e-iz-z-3-dz-where-C-is-the-circle-z-2-

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

pols of f(z)=(e^(iz) /z^3 ) has pol at z=0 order 3 ∫(e^(iz) /z^3 )dz=2iπ Res(f,0)=2iπ.(1/(2!))(d^2 /dz^2 ).z^3 f(z)∣_(z=0) =(1/2)2iπ(−e^(iz) )∣_(z=0) =−iπ

$$\mathrm{pols}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{z}\right)=\frac{\mathrm{e}^{\mathrm{iz}} }{\mathrm{z}^{\mathrm{3}} }\:\:\mathrm{has}\:\mathrm{pol}\:\mathrm{at}\:\mathrm{z}=\mathrm{0}\:\mathrm{order}\:\mathrm{3} \\ $$$$\int\frac{\mathrm{e}^{\mathrm{iz}} }{\mathrm{z}^{\mathrm{3}} }\mathrm{dz}=\mathrm{2i}\pi\:\mathrm{Res}\left(\mathrm{f},\mathrm{0}\right)=\mathrm{2i}\pi.\frac{\mathrm{1}}{\mathrm{2}!}\frac{\mathrm{d}^{\mathrm{2}} }{\mathrm{dz}^{\mathrm{2}} }.\mathrm{z}^{\mathrm{3}} \mathrm{f}\left(\mathrm{z}\right)\mid_{\mathrm{z}=\mathrm{0}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{2i}\pi\left(−\mathrm{e}^{\mathrm{iz}} \right)\mid_{\mathrm{z}=\mathrm{0}} =−\mathrm{i}\pi \\ $$

Commented by arkanmath7@gmail.com last updated on 04/Dec/19

this qust. isn′t solved from Schoum′s and the answer is −πi

$${this}\:{qust}.\:{isn}'{t}\:{solved}\:{from}\:{Schoum}'{s}\:{and}\:{the} \\ $$$${answer}\:{is}\:−\pi{i} \\ $$

Commented by mind is power last updated on 05/Dec/19

yeah Res(f,0)=(1/(2!))(d^2 /dz^2 )z^3 f(z)∣_(z=0)

$$\mathrm{yeah}\:\mathrm{Res}\left(\mathrm{f},\mathrm{0}\right)=\frac{\mathrm{1}}{\mathrm{2}!}\frac{\mathrm{d}^{\mathrm{2}} }{\mathrm{dz}^{\mathrm{2}} }\mathrm{z}^{\mathrm{3}} \mathrm{f}\left(\mathrm{z}\right)\mid_{\mathrm{z}=\mathrm{0}} \\ $$

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

y′re welcom we can generaliz it ∫_C (e^(iz) /z^n )dz=((2iπ)/((n−1)!)).(i)^(n−1) =((2π.(i)^n )/((n−1)!))

$$\mathrm{y}'\mathrm{re}\:\mathrm{welcom} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{generaliz}\:\mathrm{it} \\ $$$$\int_{\mathrm{C}} \frac{\mathrm{e}^{\mathrm{iz}} }{\mathrm{z}^{\mathrm{n}} }\mathrm{dz}=\frac{\mathrm{2i}\pi}{\left(\mathrm{n}−\mathrm{1}\right)!}.\left(\mathrm{i}\right)^{\mathrm{n}−\mathrm{1}} \\ $$$$ \\ $$$$=\frac{\mathrm{2}\pi.\left(\mathrm{i}\right)^{\mathrm{n}} }{\left(\mathrm{n}−\mathrm{1}\right)!} \\ $$

Commented by arkanmath7@gmail.com last updated on 06/Dec/19

$${good}\:{work}\:{thanks}\:{again} \\ $$

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