Evaluation of contour integration help involving exponential and cosh$z$
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Let the contour $gamma$ be a positively oriented cifrcle of radius 2 centered at zero, traversed once.
Evaluate I = $int_{gamma}$ $frac{dz}{(1-e^{iz})cosh(z)}$.
This is what Ive done so far
let $z = 2e^{itheta}$ for 0 < $theta$ < 2$pi$
$dz$ = $2ie^{itheta}$$dtheta$
So I = $int_0^{2pi}$ $frac{2ie^{itheta}dtheta}{(1-e^{2ie^{itheta}}))cosh(2e^{itheta})}$.
Is this the best way to do this question and where do i go from here?
Edit Just realised I can probably incorporate cosh($z$) = ($e^z$ + $e^{-z}$)/2
contour-integration
asked Nov 18 at 5:58
sam
498
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up vote
1
down vote
favorite
Let the contour $gamma$ be a positively oriented cifrcle of radius 2 centered at zero, traversed once.
Evaluate I = $int_{gamma}$ $frac{dz}{(1-e^{iz})cosh(z)}$.
This is what Ive done so far
let $z = 2e^{itheta}$ for 0 < $theta$ < 2$pi$
$dz$ = $2ie^{itheta}$$dtheta$
So I = $int_0^{2pi}$ $frac{2ie^{itheta}dtheta}{(1-e^{2ie^{itheta}}))cosh(2e^{itheta})}$.
Is this the best way to do this question and where do i go from here?
Edit Just realised I can probably incorporate cosh($z$) = ($e^z$ + $e^{-z}$)/2
contour-integration
asked Nov 18 at 5:58
sam
498
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let the contour $gamma$ be a positively oriented cifrcle of radius 2 centered at zero, traversed once.
Evaluate I = $int_{gamma}$ $frac{dz}{(1-e^{iz})cosh(z)}$.
This is what Ive done so far
let $z = 2e^{itheta}$ for 0 < $theta$ < 2$pi$
$dz$ = $2ie^{itheta}$$dtheta$
So I = $int_0^{2pi}$ $frac{2ie^{itheta}dtheta}{(1-e^{2ie^{itheta}}))cosh(2e^{itheta})}$.
Is this the best way to do this question and where do i go from here?
Edit Just realised I can probably incorporate cosh($z$) = ($e^z$ + $e^{-z}$)/2
contour-integration
asked Nov 18 at 5:58
sam
498
Let the contour $gamma$ be a positively oriented cifrcle of radius 2 centered at zero, traversed once.
Evaluate I = $int_{gamma}$ $frac{dz}{(1-e^{iz})cosh(z)}$.
This is what Ive done so far
let $z = 2e^{itheta}$ for 0 < $theta$ < 2$pi$
$dz$ = $2ie^{itheta}$$dtheta$
So I = $int_0^{2pi}$ $frac{2ie^{itheta}dtheta}{(1-e^{2ie^{itheta}}))cosh(2e^{itheta})}$.
Is this the best way to do this question and where do i go from here?
Edit Just realised I can probably incorporate cosh($z$) = ($e^z$ + $e^{-z}$)/2
contour-integration
contour-integration
asked Nov 18 at 5:58
sam
498
asked Nov 18 at 5:58
sam
498
edited Nov 18 at 7:34
asked Nov 18 at 5:58
sam
498
asked Nov 18 at 5:58
sam
498
asked Nov 18 at 5:58
sam
498
498
add a comment |
add a comment |
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