harmonic function and mapping of an annulus

Question

Im working with the book "real and complex analysis" by W. Rudin. There is a part of a proof that goes like this:

Let $f:A_1 \rightarrow A_2$ be a holomorphic one-to-one mapping where $A_1 = A(1,R_1)$ and $A_2 = A(1,R_2)$ are annuli centered at the origin. We define the function

$$u(z) = 2 \log|f(z)|-2\dfrac{\log R_2}{\log R_1}\cdot \log |z|$$

with $z \in A_1$. Now let $\partial$ be one of the Cauchy-Riemann operators. Since $\partial \overline f = 0$ and $\partial f = f'$, the chain rule gives

$$\partial(2\log|f|) = \partial(\log \overline f f) = \dfrac{f'}{f},$$

so that

$$(\partial u)(z) = \dfrac{f'(z)}{f(z)}-\dfrac{1}{z}\cdot \dfrac{\log R_2}{\log R_1} \quad (z\in A_1)$$.

Thus $u$ is a harmonic function in $A_1$. But why?

We dont evaluate the laplace operator. I know a result stating that the real- and imaginary-part of a holomorphic function is also harmonic and harmonic functions on simply-connected domains are in fact the real-part of some holomorphic function. Is something from that being used here?

Answer 1

$\dfrac{f'(z)}{f(z)}-\dfrac{1}{z}\cdot \dfrac{\log R_2}{\log R_1} \quad (z\in A_1)$ is obviously analytic so applying $\bar \partial$ to it gives zero and the Laplacian is just $4\bar \partial \partial$

(as an aside one can prove this directly noting that $\log f$ is locally analytic - it may not exist globally in the annulus - so $\log |f| = \Re \log f$ is locally harmonic, while obviously $\log |z|$ is harmonic and locally harmonic implies globally harmonic by the Laplacian condition)