Limit under Integral Sign with Complex Integrand I've been revising complex analysis and came upon the following problem:

Let $h(t)$ be a complex-valued function on the unit interval and define $$H(z) = \int_0^1\frac{h(t)}{t-z}dt, \quad z \in \mathbb{C}\setminus[0,1].$$ Show that $H$ is complex differentiable.

I was thinking something along the lines of
$$\lim_{z\rightarrow z_0}\frac{H(z)-H(z_0)}{z-z_0} = \lim_{z\rightarrow z_0}\frac{1}{z-z_0}\int_0^1h(t)\left(\frac{1}{t-z} - \frac{1}{t-z_0}\right)dt = \\ = -\lim_{z\rightarrow z_0}\int_0^1\frac{h(t)}{(t-z)(t-z_0)}dt = -\int_0^1\lim_{z\rightarrow z_0}\frac{h(t)}{(t-z)(t-z_0)}dt = \\ = \int^1_0\frac{h(t)}{(t-z_0)^2}dt$$
so the limit exists and is continuous. However, I am unsure about the step where the limit and integral are exchanged, since the integrand is complex. For real integrands, we have the dominated convergence theorem and other results. Does that also hold for complex integrands, or can this step be justified in some other way?
Any help would be appreciated!

If $h$ is such that $d\nu=h(t)dt$ defines a finite complex measure, then the answer is yes. Rudin does it like this:
$\tag1 \text{define}\quad \displaystyle F(z)=\int_0^1\frac{d\nu}{t-z}\quad \text{and let}\quad  z_0\in \mathbb C\setminus [0,1]$
We can find an $r>0$ such that the ball $B_r(z_0)\subseteq \mathbb C\setminus [0,1]$. Then, $\left|\frac{z-z_0}{t-z_0}\right|<\frac{|z-z_0|}{r}<1$ as soon as $z\in B_r(z_0).$  Fix such a $z.$
Now,
$\tag2 \displaystyle \sum_{n=0}^\infty\frac{(z-z_0)^{n+1}}{(t-z_0)^n}=(z-z_0)\sum_{n=0}^\infty\left(\frac{z-z_0}{t-z_0}\right)^n=(z-z_0)\frac{1}{1-\left(\frac{z-z_0}{t-z_0}\right)}=\frac{1}{t-z}$
And since the convergence is uniform on $[0,1]$, we may interchange summation and integration, after substituting the sum into the integral. This expresses $F$ in $B_r(z_0)$ as a power series
$\tag3 \displaystyle\sum_{n=0}^\infty a_n(z-z_0)^n\quad  \text{with} \quad a_n=\int_0^1\frac{d\nu}{(t-z_0)^{n+1}}$
