Once you understand the basic trick of using Hahn-Banach to reduce everything to the situation where the target space is $\Bbb{C}$, I'm sure you can easily extend several results by yourself. What I did a few months back (and what I recommend you do) is pick your favorite complex analysis text (e.g I took Cartan), and I went through each of the basic theorem one-by one and extended them to the case where the target is an arbitrary Banach space, and the domain is an open subset of $\Bbb{C}^n$. Usually such extensions aren't too hard. Of course, from time to time I would refer to Mujica's text to see if I was missing out any details in the argument.
As an example of how Hahn-Banach can be used here, I shall prove Cauchy's integral formula:
Let $U\subset \Bbb{C}$ be an open set, $X$ a complex Banach space, and $f:U\to X$ a holomorphic mapping. Let $z\in U$ and let $\gamma:[a,b]\to U$ be a $C^1$ loop such that $z\notin \text{image}(\gamma)$. Then,
\begin{align}
I(\gamma,z)f(z)&=\frac{1}{2\pi i}\int_{\gamma}\frac{f(\zeta)}{\zeta-z}\,d\zeta
\end{align}
where $I(\gamma,z)$ denotes the index of $\gamma$ wrt $z$ (which is $+1$ for a circle containing $z$ "inside" of it).
Just so we're clear, by holomorphic on $U$, I mean once-complex differentiable at every point of $U$ ($X$ is a Banach space, so the difference quotient makes sense and limits make sense). Also, on the RHS, I'm referring to the Bochner integral of a continuous function $[a,b]\to X$, namely $\frac{1}{2\pi i}\int_a^b\frac{f(\gamma(t))}{\gamma(t)-z}\,\gamma'(t)\,dt$.
We prove this by letting $\lambda\in X^*$ be arbitrary (i.e a continuous linear map $X\to\Bbb{C}$). Then, note that $\lambda\circ f:U\to\Bbb{C}$ is holomorphic (chain rule), so by the one-dimensional Cauchy-integral formula,
\begin{align}
\lambda\bigg(I(\gamma,z)f(z)\bigg)&=I(\gamma,z)(\lambda\circ f)(z)\\
&=\frac{1}{2\pi i}\int_{\gamma}\frac{(\lambda\circ f)(\zeta)}{\zeta-z}\,d\zeta\tag{by 1D version}\\
&=\frac{1}{2\pi i}\int_{\gamma}\lambda\left(\frac{f(\zeta)}{\zeta-z}\right)\,d\zeta\tag{since $\lambda$ is linear}\\
&=\lambda\left(\frac{1}{2\pi i}\int_{\gamma}\frac{f(\zeta)}{\zeta-z}\,d\zeta\right)
\end{align}
In the last line, I used one of the basic properties of Bochner integrals, that for a Bochner-integrable function $\phi:S\to X$ from a measure space $S$ to a Banach space $X$, and $\lambda\in X^*$, we have $\lambda(\int_S\phi\,d\mu)=\int_S(\lambda\circ \phi)\,d\mu$ (in our case, you may wish to write out $\int_{\gamma}$ as an integral over $[a,b]$ to see that I'm using the theorem with $S=[a,b]$ and $\mu$ being Lebesgue measure).
Now, since $\lambda\in X^*$ was arbitrary, the Hahn-Banach theorem tells us the two things inside of $\lambda$ must be equal:
\begin{align}
I(\gamma,z)f(z)&=\frac{1}{2\pi i}\int_{\gamma}\frac{f(\zeta)}{\zeta-z}\,d\zeta
\end{align}
From here of course, the fact that "holomorphic on a disc $D_R(a)$ implies analytic on the disc $D_R(a)$" follows immediately by Taylor expanding the integrand as in the single variable case (I'm being slightly sloppy here, see this answer for the correct details). We fix $0<r<R$, then (by the just-established Banach-valued case of Cauchy's integral formula)
\begin{align}
f(z)&=\frac{1}{2\pi i}\int_{|\zeta-a|=r}\frac{f(\zeta)}{\zeta-z}\,d\zeta\\
&=\frac{1}{2\pi i}\int_{|\zeta-a|=r}\frac{f(\zeta)}{\zeta-a}\cdot\frac{1}{1-\frac{z-a}{\zeta-a}}\,d\zeta\\
&=\frac{1}{2\pi i}\int_{|\zeta-a|=r}\frac{f(\zeta)}{\zeta-a}\sum_{n=0}^{\infty}\left(\frac{z-a}{\zeta-a}\right)^n\,d\zeta\\
&=\sum_{n=0}^{\infty}\left(\frac{1}{2\pi i}\int_{|\zeta-a|=r}\frac{f(\zeta)}{(\zeta-a)^{n+1}}\,d\zeta\right)\cdot (z-a)^n,
\end{align}
where, as explained in the link, the exchange of series integral is possible due to uniform convergence.
This proves "holomorphic implies analytic". The fact that "analytic implies holomorphic" is much easier; simply take your favorite complex-analysis book and look at the proof there. The fact that $f$ is Banach-valued makes no difference (just replace absolute values by norms in appropriate places).
Now that you have the equivalence of holomorphic and analytic, I doubt you'll need Goursat's lemma (but you can still prove it if you want; just use Hahn-Banach to reduce to the one-dimensional case, and then invoke the already-known version of Goursat's lemma). Likewise, you can prove Morera's theorem, and establish Cauchy's inequalities, Louiville's theorem. With Morera's theorem in the Banach-case, you can once again prove that uniform limits of holomorphic functions are holomorphic.
