Complex integration (Using Cauchy Integral formula)

[Crake]

Homework Statement

$$\int_\gamma \frac{\cosh z}{2 \ln 2-z} dz$$

with $\gamma$ defined as:

1. $|z|=1$
2. $|z|=2$

I need to solve this using Cauchy integral formula.

Homework Equations

Cauchy Integral Formula

The Attempt at a Solution

With $|z|=2$ I've solved already, as it is quite easy. All one needs to do is re-write the denominator as $z-2 \ln 2$ and since $2 \ln 2$ is inside $\gamma$, then one can apply easily the formula.

With 1 I'm not sure how to re-write the integral so I can use the Cauchy Integral formula.

 

[Crake]

This has been solved already. No need to answer.

 

[dirk_mec1]

For others who are still interested: obviously the function is holomorphci in the circle with radius 1.

 

[Crake]

For others who are still interested: obviously the function is holomorphci in the circle with radius 1.

Oh, right.

So yes, by the Cauchy theorem, it's 0.

One could always define $f(z)$ as

$$f(z)=\frac {z \cosh z}{2 \ln 2 -z}$$

and so the Integral becomes

$$\int_\gamma \frac {f(z)}{z}=0$$