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

 

[owlpride]

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

All of $u(z)$, $u'(z)$, $v(z)$, and $v'(z)$ 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.