# Complex integration (Using Cauchy Integral formula)

## 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.

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

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

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

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