Integral of e^(-z) dz / (z - (pi*i)/2)

[laura_a]

Homework Statement

Let C denote the positively oriented boundary of the square whos sides lie on the lines (x= + / - 2 and y = + / - 2)

Evaluate int_c (e^(-z) dz) / (z - (pi*i/2))

The answer is simply 2*pi

Homework Equations

This is a textbook question after the topic of Cauchy Integrals but before residues

The Attempt at a Solution

I am not very cluely at the Contour integration as the teacher just gave us the textbook notes and said do it... so I figure to use the Cauchy Integral

f(z_0) = 1/(2*pi*i) int_c f(z) dz / z - z_0

Where my belief is that z_0 is interior to the given contour (in this case it's the square)

so I thought that the question looks like it will just fit inside the formula, so to speak, because (pi*i)/2 is interor to the contour (well if it's not then I have no idea what I'm talking about) so then i put

f(z_0) = 1/(2*pi*i) int_c e^z dz/ z - ((Pi*i) / 2)

so I sub in f(z_0) = e^(Pi*i) and get -1

so -1 = 1/(2*pi*i) int_c e^z dz/ z - ((Pi*i) / 2)
-2*pi*i = int_c e^z dz/ z - ((Pi*i) / 2)

and that's where I stop because I don't know how to integrate e^z dz/ z - ((Pi*i) / 2)

Any suggestions?

 

[Päällikkö]

Cauchy's integral formula (I'm not going to list the assumptions it requires, you should make sure that they are satisfied in this problem, though):
$$f(\alpha) = \frac{1}{2\pi i}\int_C \frac{f(z)}{z-\alpha}dz$$Substitute f(z) = e^-z:

$$\int_C \frac{e^{-z}}{z-\alpha}dz = 2\pi i e^{-\alpha}$$

Can you do the rest?
As you can see, you don't really have to integrate anything.