Complex analysis, integral independent of path

[player1_1_1]

Homework Statement

when complex integral is independent of path? i heard that its for every function $$f(z)$$ but when i have function $$f(z)=\left(x^2+y\right)+i\left(xy\right)$$ its not independent, why?

 

[player1_1_1]

well, i came to that this function is not holomorphic, and integral is inependent of path only when function is holomorphic, is that true?

 

[HallsofIvy]

If a function is analytic at every point on or between two different paths between the same points, then the integral will be the same for the two paths.

In order to have an integral "independent of the path"- that is, the same for every possible path between two points, then the function would have to be analytic at every point- "holomorphic".

You might remember from Calculus of two variable that $\int f(x,y)dx+ g(x,y)dy$ is "independent of the path" if and only if $\partial f/\partial y= \partial g/\partial x$. Here, the integral would be $\int (x^2+ y)dx+ xy dy$. $\partial (x^2+ y)/\partial y= 1$ but $\partial xy/\partial x= y$.

 

[snipez90]

Halls, I think you're right except that two-variable theory would have to confirm the cauchy riemann equations, so it should be $\partial u/\partial y= -\partial v/\partial x,$
where $u = x^2 + y$ and $v = xy.$ There is also $\partial u/\partial x= \partial v/\partial y$ to check to ensure your complex function is analytic (well you also need the partials to be C^1), so if either if these two equations fails, you don't have an analytic function, so the integral can't be path independent.