# Thread: Basic Complex Integral - A bit lost

1. ## Basic Complex Integral - A bit lost

Evaluate along the straight line segment lying along x + y = 1

Now i take that i must paramaterize this.

x = t
y = 1 - t

dx = dt
dy = -dt

so my integral is now of (t - (1-t)i)(dt + idt)or something

so my question is, do i evaluate this from i to 1 ?

or from 0 to 1?

I'm not sure

2. We are doing something simiar to this but with real numbers so im thinking it should work. If not im sorry.

After you reparamertize the funcion you need to differeniate with respect to t:

x = t
y = 1 - t

I dont believe you do this implicitly...
dx/dt = 1
dy/dt = -1

Now your equation is a function of time and it should be (i think) integrated from -1 to 1.

EDIT:
Also your need to up in "t" everyhwere in there was a "z" before in the equation. This is because the function should be changed from "F(z)=" to "F(z(t))=", assuming your first function was "f".

3. Ok... Update on this question.

Sorry if it isn't conventional to do things this way on these forums...

This is the entire question

Now I think I have figured it out...

Now my problem is the last part ...

How can I explain that the complex conjugate does this?

4. Originally Posted by douber
Evaluate along the straight line segment lying along x + y = 1

Now i take that i must paramaterize this.

x = t
y = 1 - t

dx = dt
dy = -dt

so my integral is now of (t - (1-t)i)(dt + idt)or something

so my question is, do i evaluate this from i to 1 ?

or from 0 to 1?

I'm not sure
z goes from i to 1 and z= x+ iy= t+ (1- t)i. z= i when x+iy= t+(1-t)i= i. That is, t= 0. z= 1 when x+ iy= t+(1-t)i= 1, t= 1. The integral is from t=0 to t= 1.

5. Originally Posted by douber
Ok... Update on this question.

Sorry if it isn't conventional to do things this way on these forums...

This is the entire question

Now I think I have figured it out...

Now my problem is the last part ...

How can I explain that the complex conjugate does this?
f(z)= z is an analytic function. f(z)= $\displaystyle \overline{z}$ is not.

6. Originally Posted by HallsofIvy
f(z)= z is an analytic function. f(z)= $\displaystyle \overline{z}$ is not.
Thanks for that.. I just came to this realization in my lecture today as well!