Functions which satisfy $\mathrm{f}(wz) =w\,\mathrm{f}(z)+z\,\mathrm{f}(w)$ Let $\mathrm{f}$ be a complex-valued function with the following property:
$$\mathrm{f}(wz) =w\,\mathrm{f}(z)+z\,\mathrm{f}(w) $$
for all $w,z \in \mathbb C$. Necessary conditions are that $\mathrm{f}(0)=\mathrm{f}(\pm 1) = \mathrm{f}(\pm\mathrm{i})=0$.
One obvious example is the zero function: $\mathrm{f}(z)=0$ for all $z \in \mathbb C$. 


*

*Are there any other examples of functions which satisfy the above condition?

*If so, what are the "nicest" examples, e.g. continuous, holomorphic, biholomorphic?

 A: I finally found the answer. Perhaps there is an easier way?
First, assuming that $z \neq 0$ we can use an alernative definition for the derivative:
$$\mathrm{f}'(z) = \lim_{h \to 1} \frac{\mathrm{f}(hz)-\mathrm{f}(z)}{hz-z}$$
The function has the property that $\mathrm{f}(hz) = h\,\mathrm{f}(z)+z\,\mathrm{f}(h)$. Hence:
$$\mathrm{f}'(z) = \lim_{h \to 1}\frac{(h-1)\,\mathrm{f}(z)+z\,\mathrm{f}(h)}{(h-1)z}
=\frac{\mathrm{f}(z)}{z}+\lim_{h\to1}\frac{\mathrm{f}(h)}{h-1}$$
By definition, $\mathrm{f}(1\cdot 1) = 1\cdot \mathrm{f}(1) + 1\cdot \mathrm{f}(1)$ and so $\mathrm{f}(1)=2\mathrm{f}(1)$, i.e. $\mathrm{f}(1)=0$. Since $\mathrm{f}(1)=0$
$$\ell := \lim_{h \to 1}\frac{\mathrm{f}(h)}{h-1} < \infty$$
It follows that $\mathrm{f}$ must satisfy the differential equation 
$$\mathrm{f}'(z) = \frac{\mathrm{f}(z)}{z} + \ell$$
Solving this subject to the condition $\mathrm{f}(1)=0$ gives $\mathrm{f}(z) = \ell z\ln z$.
This seems fine until we notice that the definition of $\mathrm{f}$ tells us that 
$$0=\mathrm{f}(1)=\mathrm{f}[(-1)\cdot(-1)]=(-1)\cdot \mathrm{f}(-1)+(-1)\cdot\mathrm{f}(-1) = -2\mathrm{f}(-1)$$
and hence $\mathrm{f}(-1)=0$. Since $-1 = \mathrm{e}^{\mathrm{i}\pi(1 + 2n)}$, where $n \in \mathbb{Z}$, we have 
$$\lim_{z \to -1}z\ln z = -\ln(-1) = -\mathrm{i}\pi(1 + 2n) \neq 0$$
Finally, we conclude $ \mathrm{f}(z) = \ell z \ln|z|$ are all of the functions that satisfy the original condition. Importantly, $\mathrm{f}(z) = 0$ for all $z \in \mathbb{C}$ with $|z|=1$, and $\mathrm{f}(z) \to 0$ as $z \to 0$.
