(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

for [itex]n=1,2,3,...[/itex] , evaluate the integral,

[itex]I= \int_C\frac{e^{iz}}{z^n}dz[/itex]

where [itex]C [/itex] is a curve like [itex]z(t)=e^{it}[/itex] and [itex]0 \leq t \leq 2\Pi[/itex]

2. Relevant equations

3. The attempt at a solution

I tried to use Cauchy integral formula; that

[itex]f^{(n)}(z)=\frac{n!}{2 \Pi i}\int_C\frac{f(\zeta)}{(\zeta-z)^{n+1}}d\zeta[/itex]

then we can obtain,

[itex]f^{(n-1)}(z)=\frac{(n-1)!}{2 \Pi i}\int_C\frac{f(\zeta)}{(\zeta-z)^{n}}d\zeta[/itex]

[itex]f^{(n-1)}(z)=\frac{(n-1)!}{2 \Pi i}\int_C\frac{f(\zeta)}{(\zeta-z)^{n}}d\zeta[/itex]

[itex](e^{iz})^{(n-1)}(z) \Big\vert_{z=0}=\frac{(n-1)!}{2 \Pi i}\int_C\frac{e^{iz}}{z^{n}}d\zeta[/itex]

[itex]i^{n-1}=\frac{(n-1)!}{2 \Pi i}\int_C\frac{e^{iz}}{z^{n}}d\zeta [/itex]

[itex]\int_C\frac{e^{iz}}{z^{n}}d\zeta=\frac{i^{n-1} 2 \Pi i} {(n-1)!} [/itex]

[itex]~~~~~~~~=\frac{i^n 2 \Pi} {(n-1)!} [/itex] .

can you check, is it right??????

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# Complex line integrallll

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