The chain rule for a function to $\mathbf{C}$

Question

Let $f:U\longrightarrow \mathbf{C}$ be a holomorphic function, where $U$ is a Riemann surface, e.g., $U=\mathbf{C}$, $U=B(0,1)$ or $U$ is the complex upper half plane, etc.

For $a$ in $\mathbf{C}$, let $t_a:\mathbf{C} \longrightarrow \mathbf{C}$ be the translation by $a$, i.e., $t_a(z) = z-a$.

What is the difference between $df$ and $d(t_a\circ f)$ as differential forms on $U$?

My feeling is that $df = d(t_a\circ f)$, but why?

Answer 1

The forms will be different if $a\not=0$, namely if $\mathrm{d} f = w(z) \mathrm{d}z$ locally, then $\mathrm{d}\left( t_a \circ f\right) = w(z-a) \mathrm{d} z$.

Added: Above, I was using the following, unconventional definition for the composition, $(t_a \circ f)(z) = f(t_a(z)) = f(z-a)$.

The conventional definition, though, is $(t_a \circ f)(z) = t_a(f(z)) = f(z)-a$. With this definition $\mathrm{d} (t_a \circ f) = \mathrm{d}(f-a) = \mathrm{d} f$.