# Complex integration (Using Cauchy Integral formula)

1. Oct 19, 2013

### Crake

1. The problem statement, all variables and given/known data

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

with $\gamma$ defined as:

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

I need to solve this using Cauchy integral formula.

2. Relevant equations

Cauchy Integral Formula

3. 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.

Last edited: Oct 19, 2013
2. Oct 19, 2013

### Crake

3. Oct 19, 2013

### dirk_mec1

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

4. Oct 20, 2013

### Crake

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

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

Last edited: Oct 20, 2013