Complex Variables: Area Enclosed by Contour Formula

[eaglesmath15]

Homework Statement

Show that if C is a positively oriented simple closed contour, then the area of the region enclosed by C can be written (1/2i)/∫_C\bar{}zdz.
Note that expression 4 Sec. 46 can be used here even though the function f(z)=\bar{}z is not analytic anywhere.
FORMATTING NOTE: SHOULD BE Z BAR, NOT NEGATIVE Z.

Homework Equations

exspression 4 sec. 46: ∫_Cf(z)dz=∫∫_R(-v_x-u_y)dA+i∫∫_R(u_x-v_y)dA.

 

[Simon Bridge]

Cool - what have you tried so far?
Do you now what all the terms in the problem statement mean at least?

Formatting note: here - let me help...

You want to show:

If $C$ is a, positively oriented, simple closed contour, then the area of the region enclosed by $C$ can be written $$A_C=\frac{1}{2i}\int_C\bar{z}\;dz$$

You can use: $$\int_C f(z)dz = \iint_R (-v_x-u_y)dA + i\iint_R (u_x-v_y)dA$$... even though $f(x)=\bar{z}$ is not analytic anywhere.

That better?
Use the "quote" button under this post to see how I did that ;)

 

[eaglesmath15]

Cool - what have you tried so far?
Do you now what all the terms in the problem statement mean at least?

Formatting note: here - let me help...

You want to show:

If $C$ is a, positively oriented, simple closed contour, then the area of the region enclosed by $C$ can be written $$A_C=\frac{1}{2i}\int_C\bar{z}\;dz$$

You can use: $$\int_C f(z)dz = \iint_R (-v_x-u_y)dA + i\iint_R (u_x-v_y)dA$$... even though $f(x)=\bar{z}$ is not analytic anywhere.

That better?
Use the "quote" button under this post to see how I did that ;)

Thanks! It looks like what I had tried before, but it hadn't worked, so I probably just missed something.

I haven't really tried anything yet, I'm not entirely sure where to begin.

 

[Dick]

Thanks! It looks like what I had tried before, but it hadn't worked, so I probably just missed something.

I haven't really tried anything yet, I'm not entirely sure where to begin.

What are u and v for the function $f(z)=\bar z$? Start there.

 

[Simon Bridge]

Thanks! It looks like what I had tried before, but it hadn't worked, so I probably just missed something.

If PF just gave you a funny box with your latex in it, then you probably left of a brace somewhere. No worries.

I haven't really tried anything yet, I'm not entirely sure where to begin.

Like Dick says ... look at the expression you are allowed to use: the LHS of it contains part of what you have to prove - which means you have to end up with something that has the RHS in it. The RHS has loads of u's and v's ... so you want to express f(z) in terms of u and v first - then try to work out the area enclosed.

 

[jackmell]

I'm not entirely sure where to begin.

(1) Start with the expression:

\oint_C f(z)dz=\oint_C (u+iv)dz=\oint_C (u+iv)(dx+idy)

and does f(z) even have to be analytic for that to hold?

(2) Now review Green's Theorem in the plane. Does that theorem require f(z) to be analytic?

(3) What happens when I combine (1) and (2) for the function f(z)=\overline{z}?

(4) And last and foremost, try it with a non-trivial example (circles won't do). Do it for a triangle contour.