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## Homework Statement

Using the Cauchy Integral Formula compute the following integrals,where C is a circle of radius 2a centered at z=o, where 2a<pi

## Homework Equations

[tex]\oint\frac{(z-a)e^{z}}{(z+a)sinz}[/tex]

- Thread starter hancock.yang@
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Using the Cauchy Integral Formula compute the following integrals,where C is a circle of radius 2a centered at z=o, where 2a<pi

[tex]\oint\frac{(z-a)e^{z}}{(z+a)sinz}[/tex]

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Dick

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yes.

I have done the form like this:

[tex]\oint\frac{(z-a)e^{z}}{(z+a)}\frac{dz}{sinz}[/tex] + [tex]\oint\frac{(z-a)e^{z}}{sinz}\frac{dz}{(z+a)}[/tex]

however the first one is not the standard Cauchy Integral Formula

Last edited:

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Dick

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I really don't understand what you are saying there. Did you alter the original problem statement? If so, what was the original?yes.

I have sperate the form like this:

[tex]\oint\frac{(z-a)e^{z}}{(z+a)}\frac{dz}{sinz}[/tex]+[tex]\oint\frac{(z-a)e^{z}}{sinz}\frac{dz}{(z-a)}[/tex]

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The original problem is this:I really don't understand what you are saying there. Did you alter the original problem statement? If so, what was the original?

[tex]\oint\frac{(z-a)e^{z}}{(z+a)sinz}[/tex]dz c=2a centered at z=0 2a<pi

we can express the integral around the contour as the sum of the integral around z1 and z2 where the contour is a small circle around each pole. Call these contours C1 around z1 and C2 around z2.

So I am tying to express this original problem like:

[tex]\oint\frac{(z-a)e^{z}}{(z+a)}\frac{dz}{sinz}[/tex] + [tex]\oint\frac{(z-a)e^{z}}{sinz}\frac{dz}{(z+a)}[/tex]

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Dick

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Well,thanks. Could you please give a quick explanation of Cauchy Residue Theorem?

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Dick

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Thanks, I have worked it out

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