Dominated convergence theorem and Cauchy's integral formula

I think the only possible issue with the proposed solution in the OP is a careful proof that the proposed dominating function is a dominating function and a reference to the appropriate version of the DCT (or bounded convergence theorem, see below) that the OP would like to use.

There are at least two ways to approach this problem with the DCT. In each approach, we must identify which parameter serves as the parameter that we take a limit in, and identify an appropriate dominating function, prove it is a dominating function, and then quote the appropriate version of the DCT.

Note that the DCT is most commonly stated in terms of sequences of functions, so in any application of the "sequential" DCT to problems involving limits with a continuous parameter, we must use a characterization of limits in terms of sequences—see the second approach below. (Also see this old answer of mine regarding DCT with respect to continuous and discrete parameters for more on this.)

Now, we want to justify the equation: \begin{align*} \frac{\partial}{\partial w} \oint_{\partial B(a,r)}\frac{f(z)}{z-w}\,dz= \oint_{\partial B(a,r)}\frac{\partial }{\partial w}\frac{f(z)}{z-w}\,dz. \end{align*}

The first approach using real parameters:

We will use that $\frac{\partial}{\partial w} = \frac12\left(\frac\partial{\partial w_1}-i\frac\partial{\partial w_2}\right)$, and can show that \begin{align*} \frac{\partial}{\partial w_j} \oint_{\partial B(a,r)}\frac{f(z)}{z-w}\,dz= \oint_{\partial B(a,r)}\frac{\partial }{\partial w_j}\frac{f(z)}{z-w}\,dz\qquad(j=1,2). \end{align*} Because then by linearity and the definition of $\partial/\partial w$, we will have the equality we are after. With this approach, the parameters $w_j$ are the parameters we take limits in, and they are real parameters, which has the advantage that we can use the version of differentiating under the integral sign quoted here:

Differentating under the integral sign. Suppose that $F(x,t)$ is integrable as a function of $x \in \mathbb{R}^d$ for each value of $t \in \mathbb{R}$ and differentiable as a function of $t$ for each value of $x$. Assume also that $$\bigg| \frac{\partial}{\partial t} F(x,t) \bigg| \le G(x),$$ for all $x,t$, where $G(x)$ is an integrable function of $x$. Then $\frac{\partial}{\partial t} F(x,t)$ is integrable as a function of $x$ for each $t$ and $$\frac{d}{dt} \int F(x,t)\, dx = \int \frac{\partial}{\partial t} F(x,t)\,dx.$$

To prove this, you can mimic the second approach we will use to the problem in the involving the characterization of limits I mentioned (to prove this theorem I quoted above, the mean value theorem is also useful). To apply this, write out $\frac{f(z)}{z-w}$ as a function $F_j = F(t,w_j)$ where $t$ can be the parameter for $\partial B(a,r)$ for each $j = 1,2$ and apply this result to each of $F_1$ and $F_2$ separately.

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A second approach from first principles:

Without separating the integral into real and imaginary parts and quoting the theorem on differentiating under the integral that we are familiar with from real variables, we can choose to write the integral in a form that lets us apply the DCT for sequences of functions $([0,2\pi],\mathrm{Borel},dt)\to(\mathbb C,\mathrm{Borel})$ from first principles. We still would like to show \begin{align*} \frac{\partial}{\partial w} \oint_{\partial B(a,r)}\frac{f(z)}{z-w}\,dz= \oint_{\partial B(a,r)}\frac{\partial }{\partial w}\frac{f(z)}{z-w}\,dz. \end{align*} The DCT is stated for sequences of functions, so recall the following characterization of limits in a metric space: \begin{align*} \lim_{h\to a}g(h) = L \iff \text{for all sequences $h_j\to a$,}\ \lim_{j\to\infty}g(h_j) = L. \end{align*} (Cf. Rudin's Principles of Mathematical Analysis p. 84.) Thus, let $h_j\to 0$ be an arbitrary sequence of complex numbers and write the difference quotient corresponding to the left-hand side as (after skipping some algebra): \begin{align*} \int_0^{2\pi}\frac{f(re^{it})}{(re^{it}-w)^2-h_j(re^{it}-w)}ire^{it}\,dt. \end{align*} By continuity, $f$ is bounded by $M$ say on $\partial B(a,r)$. To bound the expression in the denominator, we use the reverse triangle inequality, \begin{align*} |(re^{it}-w)^2-h_j(re^{it}-w)| &\ge |re^{it}-w|\big(|re^{it}-w|-|h_j|\big). \end{align*} Because the distance $\delta$ from $w$ to the boundary of the disk is positive, we have $|re^{it}-w|\ge \delta > 0$ for all $t$, so if $j$ is so large that $|h_j|<\frac\delta2$, then the right-hand side of the last inequality is bounded below by $$ |re^{it}-w|\big(|re^{it}-w|-\frac\delta2\big)\ge \delta\big(\frac\delta2\big). $$ Hence we see that for $j\gg1$, $$ \bigg|\frac{f(re^{it})}{(re^{it}-w)^2-h_j(re^{it}-w)}ire^{it}\bigg| \le \frac{2M}{\delta^2}r, $$ which is bounded and hence belongs to $L^1([0,2\pi],dt)$. By the DCT (in fact, merely bounded convergence theorem will do here), \begin{align*} \lim_{j\to\infty}\int_0^{2\pi}\frac{f(re^{it})}{(re^{it}-w)^2-h_j(re^{it}-w)}ire^{it}\,dt &= \int_{0}^{2\pi}\lim_{j\to\infty}\frac{f(re^{it})}{(re^{it}-w)^2-h_j(re^{it}-w)}ire^{it}\,dt \\ &= \int_{0}^{2\pi}\frac{f(re^{it})}{(re^{it}-w)^2}ire^{it}\,dt\\ &= \oint_{\partial B(a,r)}\frac{f(z)}{(z-w)^2}\,dz\\ &= \oint_{\partial B(a,r)}\frac{\partial }{\partial w}\frac{f(z)}{z-w}\,dz. \end{align*} As the sequence $h_j\to 0$ we chose was arbitrary, we have the desired conclusion by the characterization of limits we stated.