An identity involving the chain rule I have a question concerning a demonstration in Pommerenke's Univalent Functions :
Let $\Phi(t)$ be twice continuously differentiable and
$\Psi(t) = t\frac{d}{dt}[t \Phi'(t)]$, $(0\leq t < \infty)$.
Let $f(z)$ be analytic in $D$ (open unit disk). Since
$r\frac{d}{dr}|f(z)| = |f(z)|\text{Re}z \frac{f'(z)}{f(z)}$ and
$\frac{d}{d\theta}|f(z)| = -|f(z)|\text{Im}z \frac{f'(z)}{f(z)}$, a short calculation shows that
$\left(r \frac{d}{dr}\right)^2 \Phi(|f(z)|) + \left(\frac{d}{d\theta}\right)^2 \Phi(|f(z)|)  = \Psi(|f(z)|)\left|z \frac{f'(z)}{f(z)}\right|^2$.
This short calculation turns out to be rather confusing to me.
 A: Expanding the operators
\begin{align}
\left(r \frac{d}{d r}\right)^2  = \left(r \frac{d}{d r}\right) \left(r \frac{d}{d r }\right) = r \frac{d}{d r} \left(r \frac{d}{d r}\right)\\
\left(\frac{d}{d \theta}\right)^2  = \left(\frac{d}{d \theta}\right) \left(\frac{d}{d \theta}\right) = \frac{d}{d \theta} \left(\frac{d}{d \theta}\right)
\end{align}
we have
\begin{align}
\left(r \frac{d}{d r}\right)^2 \Phi\big(|f(z)|\big) &= r \frac{d}{d r} \left\{r \frac{d \Phi\big(|f(z)|\big)}{d r}\right\} = r \frac{d}{d r}\left\{r\,\Phi'\big(|f(z)|\big) \frac{d}{d r}|f(z)|\right\} \\ \\
&= r \frac{d}{d r}\left\{|f(z)|\,\Phi'\big(|f(z)|\big) \Re\left[\frac{zf'(z)}{f(z)}\right]\right\}\\ \\
&= |f(z)| \left\{|f(z)|\,\Phi'\big(|f(z)|\big)\right\}' \Re^2\left[\frac{zf'(z)}{f(z)}\right] \\
& \hskip2in + r |f(z)|\,\Phi'\big(|f(z)|\big) \Re'\left[\frac{zf'(z)}{f(z)}\right] \\ \\ \\
\left(\frac{d}{d \theta}\right)^2 \Phi\big(|f(z)|\big) &= \frac{d}{d \theta} \left\{\frac{d \Phi\big(|f(z)|\big)}{d \theta}\right\} = \frac{d}{d \theta}\left\{\Phi'\big(|f(z)|\big) \frac{d}{d \theta}|f(z)|\right\} \\ \\
&= -\frac{d}{d \theta}\left\{|f(z)|\,\Phi'\big(|f(z)|\big) \Im\left[\frac{zf'(z)}{f(z)}\right]\right\}\\ \\
&= |f(z)| \left\{|f(z)|\,\Phi'\big(|f(z)|\big)\right\}' \Im^2\left[\frac{zf'(z)}{f(z)}\right] \\
& \hskip2in - |f(z)|\,\Phi'\big(|f(z)|\big) \Im'\left[\frac{zf'(z)}{f(z)}\right]
\end{align}
Adding the two
\begin{multline}
\left(r \frac{d}{d r}\right)^2 \Phi\big(|f(z)|\big) + \left(\frac{d}{d \theta}\right)^2 \Phi\big(|f(z)|\big) = \\
\Psi\big(|f(z)|\big) \left|\frac{zf'(z)}{f(z)}\right|^2 + |f(z)|\,\Phi'\big(|f(z)|\big)\left\{\Re'\left[\frac{zf'(z)}{f(z)}\right] - \Im'\left[\frac{zf'(z)}{f(z)}\right]\right\}
\end{multline}
Finally, using that $f(z)$ is analytic in $D$, you need to prove that the second term is zero. I'm very rusty on my complex analysis theorems, but I believe is pretty straigth forward since
$$
\Re'\left[\frac{zf'(z)}{f(z)}\right] - \Im'\left[\frac{zf'(z)}{f(z)}\right] = \frac{d}{dz} \left(\frac{\bar{z}\bar{f}'(z)}{\bar{f}(z)}\right) = \frac{d}{dz} \left(\frac{\bar{z} f'(\bar{z})}{f(\bar{z})}\right) = 0
$$
Please doublecheck that the last argument is correct.
A: First of all we get by a straightforward application of the chain rule
$$
r \frac{d}{dr} \Phi(|f(z)|) = \Phi'(|f(z)|) |f(z)| \operatorname{Re} \left[z \frac{f'(z)}{f(z)}\right] = \eta(|f(z)|) u(z)
$$
and
$$
\frac{d}{d\theta} \Phi(|f(z)|) = -\Phi'(|f(z)|) |f(z)| \operatorname{Im} \left[z \frac{f'(z)}{f(z)}\right] = -\eta(|f(z)|) v(z)
$$
where we write for ease of notation
$$ \eta(t) = \Phi'(t) t \quad \text{ and } \quad g(z) = z \frac{f'(z)}{f(z)} = u(z) + i v(z). $$
Then we get
$$
\left(r \frac{d}{dr}\right)^2 \Phi(|f(z)|) = \eta'(|f(z)|) |f(z)| u(z)^2 + \eta(|f(z)|) r \frac{du}{dr}(z)
$$
and
$$
\left(\frac{d}{d\theta}\right)^2 \Phi(|f(z)|) = \eta'(|f(z)|) |f(z)| v(z)^2 - \eta(|f(z)|) \frac{dv}{d\theta}(z)
$$
Now the polar form of the Cauchy-Riemann equations for $g=u+iv$ gives $r \frac{du}{dr} = \frac{dv}{d\theta}$, so
$$
\begin{split}
\left(r \frac{d}{dr}\right)^2 \Phi(|f(z)|) + \left(\frac{d}{d\theta}\right)^2 \Phi(|f(z)|) &= \eta'(|f(z)|) |f(z)| (u(z)^2+v(z)^2) \\
&= \Psi(|f(z)|) |g(z)|^2
\end{split}
$$
which is exactly the claim from Pommerenke's book.
