The integral of a harmonic function

[Poopsilon]

Homework Statement

Show that:

[tex]\frac{1}{2\pi}\int_0^{2\pi}log|re^{i\theta} - z_0|d\theta = \left\{\begin{matrix}
log|z_0| & if & |z_0| < r \\
log|r| & if & |z_0| > r
\end{matrix}\right.[/tex]

Homework Equations

The function log|z| is harmonic in the slit plane since it is the real part of log(z) which is analytic in the slit plane. I'm pretty sure I'm supposed to use the fact that it's harmonic.

Harmonic functions have the Mean Value Property which states that the average of a harmonic function on the boundary of a ball is equal to the value of the harmonic function at the center point of this ball.

The Attempt at a Solution

What I don't understand is that this integral is the average of log|z| along a circle of radius r centered at [itex]z_0[/itex], and thus by the Mean Value Property I feel that it should always equal [itex]log|z_0|[/itex] for [itex]log|z| \neq r[/itex]. I also feel like we should have the requirement that [itex]z_0 \neq 0[/itex] since the logarithm of zero is negative infinity. What am I missing here? Thanks.

 

[mighty2000]

Hello,

Thank you for sharing your thoughts on this problem. You are correct in your understanding that we can use the Mean Value Property to solve this integral. However, there are a few key points that we need to consider in order to fully understand the solution.

Firstly, you are correct that the function log|z| is harmonic in the slit plane, meaning that it satisfies Laplace's equation. This is because it is the real part of the analytic function log(z), and the real part of an analytic function is always harmonic. This is an important property that we will use in our solution.

Next, let's consider the integral itself. As you mentioned, it represents the average value of log|z| along a circle of radius r centered at z_0. This means that we are essentially taking the average of the function over a closed curve. We can use the Mean Value Property to rewrite this integral as the value of the function at the center point of the curve, which in this case is z_0. However, we need to be careful with the boundary points of the curve.

If we consider the case where |z_0| < r, then the curve is contained entirely within the circle of radius r centered at z_0. In this case, the boundary points of the curve are all inside the circle, and so the Mean Value Property tells us that the average value of the function is equal to the value at the center point, which is log|z_0|. This is what we see in the first part of the solution.

On the other hand, if we consider the case where |z_0| > r, then the curve extends beyond the circle of radius r centered at z_0. In this case, the boundary points of the curve include some points outside the circle. This means that the Mean Value Property no longer applies, and we cannot simply take the value of the function at the center point. Instead, we need to consider the average value of the function over the entire circle of radius r centered at z_0. This is where the second part of the solution comes in – we can use the Mean Value Property to show that the average value of log|z| over the entire circle is equal to log|r|. This is because the function is harmonic, and so the average value over the entire circle is equal to the value at the center point, which is log|r|. This is what we see in the second