Prove Complex Integral is Purely Imaginary

[owlpride]

Homework Statement

Assume that f(z) is analytic and that f'(z) is continuous in a region that contains a closed curve \gamma. Show that
\int_\gamma \overline{f(z)} f'(z) dz

is purely imaginary.

Homework Equations

If f(z) is holomorphic on the region containing a closed curve \gamma or if f(z) has a primitive (we have not yet established a relationship between these two properties), then

\int_\gamma f(z)dz = 0.

And while we did not prove that analytic functions are holomorphic, that's not hard to verify if necessary.

The Attempt at a Solution

If we let f(z) = u(z) + i v(z), where u and v are functions from the complex plane to the real line, then

\int_\gamma \overline{f(z)} f'(z) dz = \int f(z) f'(z) dz - 2 i \int v(z) f'(z) dz = 0 - 2 i \int_\gamma v(z) f'(z)dz

So I have to verify that the latter integral is real valued, and this is where I am stuck. I could parametrize z = \gamma(t) where t \in [0,1] but I am not sure that helps me in any way.

- 2 i \int_\gamma v(z) f'(z)dz = -2 i \int_0^1 v(\gamma(t)) f'(\gamma(t)) \gamma'(t) dt = 2 i \int_0^1 v'(\gamma(t))f'(\gamma(t))\gamma'(t) \gamma'(t) dt

(The last step comes from integration by parts, with one term = 0 because gamma is a closed curve.)

Is there any reason to believe that this last integral is real-valued?

 

[jbunniii]

If we let f(z) = u(z) + i v(z), where u and v are functions from the complex plane to the real line, then

\int_\gamma \overline{f(z)} f'(z) dz = \int f(z) f'(z) dz - 2 i \int v(z) f'(z) dz = 0 - 2 i \int_\gamma v(z) f'(z)dz

I am not sure what you did here.

If f(z) = u(z) + iv(z), then \overline{f(z)} = u(z) - i v(z) and f&#039;(z) = u&#039;(z) + i v&#039;(z)[/tex], so<br /> <br /> \begin{align*}\int_{\gamma} \overline{f(z)} f&amp;#039;(z) dz &amp;amp;= \int_{\gamma} (u(z) - i v(z)) (u&amp;#039;(z) + i v&amp;#039;(z)) dz \\&lt;br /&gt; &amp;amp;= \int_{\gamma} (u(z) u&amp;#039;(z) + v(z) v&amp;#039;(z)) dz +i \int_{\gamma} (u(z) v&amp;#039;(z) - v(z) u&amp;#039;(z)) dz \end{align*}<br /> <br /> All of u(z), u&amp;#039;(z), v(z), and v&amp;#039;(z) are real, so if you can show that the first integral is zero then you&#039;re done.

 

[owlpride]

\begin{align*}\int_{\gamma} \overline{f(z)} f&#039;(z) dz <br /> &amp;= \int_{\gamma} (u(z) u&#039;(z) + v(z) v&#039;(z)) dz +i \int_{\gamma} (u(z) v&#039;(z) - v(z) u&#039;(z)) dz \end{align*}

All of u(z), u&#039;(z), v(z), and v&#039;(z) are real, so if you can show that the first integral is zero then you're done.

But dz is not real. Does that mess with the value of the last integral?

Thanks for your thoughts!

 

[Dick]

You seem to be having a hard time finding a good expression for f'(z)*dz. You can do it several ways. (df/dz)*dz=df. If f(z)=u(z)+iv(z) and z=x+iy, that makes it (u+iv)_x*dx+(u+iv)_y*dy. (Underscores are partial derivatives). Or you could use f'(z)=(u+iv)_x or f'(z)=(1/i)*(u+iv)_y and dz=dx+idy. You can show they are all equal with Cauchy-Riemann. Now just collect the real and imaginary parts and figure out whether they are exact differentials. Take it from there.