Quesiton regarding the mean value prroperty

Question

Let $f$ be analytic in and inside a disk of radius R around $z_o$

Then we can write the Cauchy integral formula, and parametrize the circle to get the following:

$f(z_0)= (1/2\pi)\int_{0}^{2\pi}f(z_o+Re^{it})dt$

This is known as the mean value property.

Here is my question: In my textbook, it says that this formula shows that $f(z_0)$ is just the average of the value of the function on the circle.

I don't see how this is the case though; say that $f(z)=M$ for any $z$ on the circle as above. Then the integral just becomes $2\pi R M$ and we get:

$f(z_0)=MR$

But this makes no sense, since changing the radius would imply that $f(z_0)$ would take on different values. Where is my mistake?

Answer 1

I'm confused as to where the $R$ is coming from. If $f(z) = M$, then $$\frac{1}{2\pi} \int_0^{2\pi} f(z_0 + Re^{it}) dt = \frac{1}{2\pi} \int_0^{2\pi} M dt = \frac{1}{2\pi} \cdot 2\pi M=M$$