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[tex]f(z)=\frac{e^z}{1+e^{4z}}[/tex].

and the loop

[tex]\gamma_{r_1,r_2}=I_{r_1,r_2}+II_{r_2}+III_{r_1,r_2}+IV_{r_1},\quad r_1,r_2>0,[/tex]

which bounds the domain

[tex]A_{r_1,r_2}=\{z\in\mathbb{C}\mid -r_1<\Re(z)< r_2\wedge0<\Im(z)<\pi\}.[/tex]

Now I have to show that

[tex]\int_{\gamma_{r_1,r_2}}f(z)\,dz=\tfrac{\sqrt2}{2}\pi.[/tex]

I know that the integral along [itex]\gamma_{r_1,r_2}[/itex] is the sum of the integrals along the four countours [itex]I_{r1,r_2},\dots,IV_{r_1}[/itex]. I also know that if [itex]f[/itex] is continuous in a domain D and has an antiderivative [itex]F[/itex] throughout D, then the integral along a contour lying in D is [itex]F(z_T)-F(z_I)[/itex], where [itex]z_T[/itex] is the terminal point, and [itex]z_I[/itex] the initial point, of the countour.

However, using this, I am not able to show the aforementioned result, i.e, I get [itex]\int_{\gamma_{r_1,r_2}}f(z)\,dz=0[/itex].

Could anyone try to calculate the integral, and then report back what they've found.

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# Homework Help: Complex integration

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