# Complex Variables questions

1. May 28, 2007

### blumfeld0

Hi. I am studying on complex variables on my own using Brown and Churchill, 6th edition.
so far i am doing pretty good, but i have run into trouble doing a few contour integrals.

1. First off, the authors state that unlike in calculus, where the integral can mean area under the curve, in complex analysis, it does not have any physical meaning (except in rare special cases). this is what they claim. really? no physical meaning? i was very surprised to learn this. why isn't it just the area under the curve?

ok here are the problems i just can't get:

2. f(z) = Pi *exp(Pi z_) and C is the boundary of the square with vertices at the points 0,1, 1+i and i, the orientation of C is counterclockwise direction.
where z_ = x - i y.
i know I have to break up the integral into two. my limits of integration for both are going to be from 0 to 1 i believe.
the first one will be integral from 0 to 1 of (-i) * Pi *exp(Pi z_)
the second integral will be from 0 to 1 of (x+i) * Pi *exp(Pi z_)
then i add up the result. is that right?

3.

f(z) is the branch z^(-1+i) = exp((-1+i)*log[z]) ( |z|>0, 0<arg z <2 Pi)
and C is the positively oriented unit circe |z|=1.
I really thought i had this but my answer isn't right. (I have the solution)
I know log[z] = Ln r + i*theta
r = 1 so ln r = 0 so log[z] = i*theta

Integral from 0 to 2Pi of exp((-1+i)*i*theta) * (-1+i)*i
but it ain't workin'.

3. lastly,
f(z) is defined by the equations

f(z) = { 1 when y<0 and 4y when y>0}
and C is the arc from z = -1-i to 1+i along the curve y =x^3.

this i don't even know where to start.

thank you!

2. May 28, 2007

### mathwonk

in complex analysis the integrals are all path integrals, and they absically are similar to the path integral of dtheta, the angle form, so their interpretation is something like winding number.

i.e. integrals of differentiable functions give zero around a path, but integrands which are not defiend everywhere give an answer that depends on how many times your path winds around those places where the integrand is not defined.

(The graph of a complex function is a surface in 4 space so "area under the curve" makes no sense.)