Why does int_c cosh(z)/z^4 dz = 0 ?

[laura_a]

Homework Statement

Why does Int_c cosh(z)/z^4 dz = 0 ?
I have been working on the Cauchy Integral formula and this is one of the textbook questions. This is my working out, the answer in the book is zero but not sure why

Homework Equations

So I am using Cauchy Integral

that is

f(z_0) = 1/(2*pi*i) ⌠_c f(x) / (x-x_0)

The Attempt at a Solution

My guess is that the function is not inside the domain so there is no integral? But if that is even true is this how you show it.

The question is

Let C denote the positively oriented boundary of the square whose sides lie along the curve lines x = +/- 2 and y = +/- 2. Evaluate

⌠_c cosh(z)/z^4 dz



so I have f(x) as cosh(z)/z^3

and z_0 = 0

so when I try to calculate f(z_0) I get a division by zero, now to ME that means it is UNDEFINED, but the answer in the book says zero. Am I right in my working out and if the integral doesn't exist does that mean it is zero?

Any info would be appreciated.

THanks
Laura

 

[HallsofIvy]

Cauchy integral formula:
$$\frac{d^nf(z)}{dz^n}= \frac{n!}{2\pi i}\int \frac{f(z)dz}{(z-z_0)^{n+1}}$$
Since you have z⁴ in the denominator, look at the third derivative of cosh(z) at z= 0.