Evaluating Complex Integral: Over |z|=4 Region

[fudgenstuff]

I've been given the problem of evaluating the integral

$$\int$$(exp^z)/Sinh(z) dz

Over the region C which is the circle |z|=4

I can't figure out how to do this,I tried parameterizing with z(t)=4e^i$$\theta$$ but the integrand just seems far too complicated. Any suggestions?

(Apologies for the terrible formatting)

 

[Hootenanny]

Welcome to PF fudgenstuff,

Have you come across the Residue Theorem before?

I'm assuming that this is homework, for future reference we have Homework & Coursework forums where such questions can be posted.

 

[fudgenstuff]

Ah sorry, if one of the mods could move this thread then that'd be great.

The question is from an exam paper,I'm not entirely sure they want us to use the residue theorem here as there's another question on it later on in the paper.
This question follows on from deducing that Sinh(x+iy) = sinh(x)cos(y) + icosh(x)sin(y), and then showing that Sinh(z)=0 only if z= i*pi*n (where n is an integer), so I'm not sure if I'm supposed to use that result somehow?

 

[Hootenanny]

The question is from an exam paper,I'm not entirely sure they want us to use the residue theorem here as there's another question on it later on in the paper.

If it doesn't explicitly say that you can't use the Residue Theorem then you'd be crazy not to!

 

[zhentil]

If they tell you how to find the zeros of sinh(z), they want you to use the residue theorem.

 

[zhentil]

Here's an alternative way, using the argument principle. We have
$$\frac{e^z}{sinh(z)} = \frac{cosh(z) + sinh(z)}{sinh(z)} = 1 + \frac{cosh(z)}{sinh(z)}$$

Now the contour integral of the first is of course zero, and the contour integral of the second is of the form $$\frac{f'(z)}{f(z)}$$ which by the argument principle is $$2 \pi i[Z-P]$$ where Z and P are the numbers of zeros and poles of sinh(z) enclosed by the contour, respectively. Since you already know the number of zeroes, and since sinh(z) has no poles, you're done.

If you haven't seen the residue theorem, you probably haven't seen this, but it's always nice to avoid computations if possible.