Let f(z)= z bar. This is an example of a function which is continuous everywhere but does NOT possess an antiderivative. To demonstrate this fact it suffices to show that the integral z bar dz is not path independent. Consider two paths: C1: the straight path from z=-1 towards z=i. Let C2 be the sequence of straight line paths along the Real axis followed by the straight line path along the Imaginary axis. Parameterize these paths and show the values do not agree. Explain how this demonstrates that there is no antiderivative for f(z)=z bar.
I'm lost..need help with this problem. Thanks!
Re: Complex analysis
What do you understand then? Do you understand that a complex number, z, can be written as z= x+ iy where x is the "real part" and y the "imaginary part? Do you understand that the "complex conjugate", "z bar", is x- iy? Do you understand that in the "real plane" the point z= -1= -1+ 0i is represented by (-1, 0) and z= i by (0, 1)?
Originally Posted by nm321
So you are asked, first, to integrate f(x, y)= x- iy along the line given by x= -1+ t, y= t, for t= 0 to 1. That is the same as the vector integration .
The second part is asking you to integrate f(x,y)= x- iy along two line. First along the x-axis from (-1, 0) to (0, 0) then along the y-axis from (0, 0) to (0, 1). Along the x-axis, we can take x= t- 1, y= 0 so the integral becomes . Along the y-axis, we can take x= 0, y= t so the integral becomes . The integral from z= -1 to z= i is the sum of those two integrals.