# Complex Analysis

1. Mar 20, 2009

### latentcorpse

Let $w_1,w_2 \in \mathbb{C}$ and $\gamma$ be some smooth curve from $w_1$ to $w_2$.

Find $\int_{\gamma} e^{\sin{z}} \cos{z} dz$

this is holomorphic on the entire copmlex plan so we can't use a residue theorem. furthermore, we can't assume $\gamma$ is a closed contour as we aren't told $w_1=w_2$ so it looks as if we're going to need to parameterise $\gamma$.

but we don't know what $\gamma$ looks like. however we do know that any two point in the copmlex plane can be joined by a finite number of horizontal and vertical lines so if we use instead of $\gamma$ a contour $\gamma_1 \cup \gamma_2$

where $\gamma_1$ is horizontal and $\gamma_2$ is vertical. this is my thoughts so far but parameterising these was still going to be pretty difficult so i decided to check if im on the right lines or not. any advice?

2. Mar 20, 2009

### gabbagabbahey

Re: Copmlex Analysis

Hint: is this integral path independent? How can you tell?

3. Mar 21, 2009

### latentcorpse

Re: Copmlex Analysis

well the fundamental theorem of calculus gives that
for
$\gamma:[a,b] \rightarrow U$ with $\gamma(a)=\alpha,\gamma(b)=\beta$
and $f: U \rightarrow \mathbb{C}$ holomorphic on the open set U then, if $\exists F'=f$,

$\int_{\gamma} f(z) dz = \int_{\gamma} F'(z) dz = \int_a^b F'(\gamma(t)) \gamma'(t) dt = \int_a^b (F \cdot \gamma(t))' dt = F \cdot \gamma(b) - F \cdot \gamma(a) = F(\beta)- F(\alpha)$

i.e. if we can find such an F, the integral will depend only on the end point values and will therefore be path independent

for our case $F=e^{\sin{z}}$ is such an F and so the answer is

$\int_{\gamma} e^{\sin{z}} \cos{z} dz = e^{\sin{w_2}} - e^{\sin{w_1}}=e^{\sin{w_2}-\sin{w_1}}$

is that ok?

is there an easier way to tell its path independent?

also, can you recommend a good book that has plenty of worked exapmles on integrals like this (and stuff using Cauchy residue theorem etc)???

4. Mar 21, 2009

### latentcorpse

Re: Copmlex Analysis

also, how did you realise to argue about path independence?

5. Mar 21, 2009

### yyat

A holomorphic function defined on $$\mathbb{C}$$ always has an antiderivative (for example compute the Taylor series and integrate each term).

As you pointed out, existence of an antiderivate implies independence of path for the integral.

More generally, suppose $$f:D\to\mathbb{C}$$ is a holomorphic function and $$\alpha, \beta$$ are paths with the same endpoints. If $$\alpha$$ can be continuously deformed in D into $$\beta$$ while keeping the endpoints fixed, then the integral of f over the two paths is the same. This follows easily from Cauchy's integral theorem.

6. Mar 21, 2009

### gabbagabbahey

Re: Copmlex Analysis

Like yyat said, whenever the integrand is holomorphic, the integral is path independent.

The question wouldn't have made much sense if the integral depended on the path now would it?

7. Mar 21, 2009