Rigorous Proof of Leibniz's Rule for Complex

The "real case" was maybe proven using the MVT, and we don't have that in the complex environment. We therefore have to start afresh.

Fix a point $z_0\in D$. We have to prove that $$\lim_{h\to0}{g(z_0+h)-g(z_0)\over h}=\int_a^b f_z(t,z_0)\>dt\ .\tag{1}$$ By assumption there is a closed disc $B$ of radius $r_0>0$ around $z_0$ such that $f_z$ is continuous, hence uniformly continuous, on the set $R:=[a,b]\times B$. For complex $h$ with $0<|h|<r_0$ we then have $$f(t,z_0+h)-f(t,z_0)=\int_{z_0}^{z_0+h}f_z(t,z)\>dz=h\int_0^1 f_z(t,z_0+\tau h)\>d\tau$$ and therefore $$\Phi(t,h):={f(t,z_0+h)-f(t,z_0)\over h}-f_z(t,z_0)=\int_0^1 \bigl(f_z(t,z_0+\tau h)-f_z(t,z_0)\bigr)\>d\tau\ .$$ The uniform continuity of $f_z$ on $R$ now allows to conclude the following: Given an $\epsilon>0$ there is a $\delta>0$ such that $$\bigl|\Phi(t,h)\bigr|<\epsilon\qquad\bigl(a\leq t\leq b, \ 0< |h|<\delta)\ .$$ This means that $\lim_{h\to0}\Phi(t,h)=0$ uniformly in $t$. Since $${g(z_0+h)-g(z_0)\over h}-\int_a^b f_z(t,z_0)\>dt=\int_a^b \Phi(t,h)\>dt$$ we may infer $(1)$.