Why is $
\frac{\partial F}{\partial x} = -i \frac{\partial F}{\partial y}
$ called the Cauchy–Riemann equation for $F(z)$?
A function of the complex variable $z = x + iy$ is also sometimes considered as a function of the pair of real variables $(x, y)$. Thus, if $F(z) = u(x, y) + i v(x, y)$, then
$$
\frac{\partial F}{\partial x} = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x}\quad \text{and} \quad
\frac{\partial F}{\partial y} = \frac{\partial u}{\partial y} + i \frac{\partial v}{\partial y}.
$$
Now, the Cauchy–Riemann equations for $F(z)$ are
$$
\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}\quad \text{and} \quad
\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}.
$$
On the other hand,
$$
\frac{\partial F}{\partial x} = -i \frac{\partial F}{\partial y}\ \iff \ \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} = \frac{\partial v}{\partial y} -i\frac{\partial u}{\partial y}.
$$
Comparing the real and imaginary parts, we see that the two Cauchy–Riemann equations for $F(z)$ can also be concisely stated as
$$
\bbox[5px,border:2px solid black]
{
\frac{\partial F}{\partial x} = -i \frac{\partial F}{\partial y}.
}
$$
Does it make sense to write $\int_\gamma f(z)\, dz = \int_\gamma f(z)\, dx + i \int_\gamma f(z)\, dy$?
Ahlfors defines the complex line integral of the continuous function $f(z)$ over a piecewise differentiable arc $\gamma$ as
$$
\int_\gamma f(z)\, dz = \int_a^b f(z(t)) z'(t)\, dt\label{defn}\tag{1}
$$
where $z = z(t)$, $a \leq t \leq b$, is a parametrization of the arc $\gamma$. This is given in $\S$4.1.1 on page 102.
On the next page, Ahlfors defines line integrals with respect to $\bar{z}$ as
$$
\int_\gamma f\, \overline{dz} = \overline{\int_\gamma \bar{f}\, dz}.
$$
Finally, line integrals with respect to $x$ and $y$ are defined as
\begin{align}
\int_\gamma f\, dx &= \frac{1}{2} \left( \int_\gamma f\, dz + \int_\gamma f\, \overline{dz} \right), \\
\int_\gamma f\, dy &= \frac{1}{2i} \left( \int_\gamma f\, dz - \int_\gamma f\, \overline{dz} \right).
\end{align}
Then, one sees that we indeed have
$$
\bbox[5px,border:2px solid black]
{
\int_\gamma f\, dz = \int_\gamma f\, dx + i \int_\gamma f\, dy.
}\label{zxy}\tag{2}
$$
Furthermore, using \eqref{defn} one can show the analogous formulas
\begin{align}
\int_\gamma f\, dx &= \int_a^b f(z(t)) x'(t)\, dt, \label{xt}\tag{3} \\
\int_\gamma f\, dy &= \int_a^b f(z(t)) y'(t)\, dt, \label{yt}\tag{4}
\end{align}
where $z(t) = (x(t),y(t))$, $a \leq t \leq b$, is a parametrization of the arc $\gamma$.
How is it immediate that $\partial F / \partial y = i f(z)$?
Here, $F(z)$ is defined as
$$
F(z) = \int_\sigma f\, dz
$$
where $\sigma$ is the horizontal line segment from $(x_0, y_0)$ to $(x, y_0)$ followed by the vertical line segment from $(x, y_0)$ to $(x, y)$.
Now, write $F(z)$ as
$$
F(z) = \int_{\sigma_1} f\, dz + \int_{\sigma_2} f\, dz,
$$
where $\sigma_1$ is the horizontal line segment from $(x_0, y_0)$ to $(x, y_0)$, and $\sigma_2$ is the vertical line segment from $(x,y_0)$ to $(x,y)$.
Then, further expanding each integral on the right using \eqref{zxy}, we get
$$
F(z) = \int_{\sigma_1} f\, dx + i\int_{\sigma_2} f\, dy,
$$
since $\int_{\sigma_1} f\, dy = 0 = \int_{\sigma_2} f\, dx$ (use \eqref{xt} and \eqref{yt} to convince yourself that this is indeed so).
So, consider the values $F(z(k))$ where $z(k) = x + i(y + k)$. Note that $z(0) = z$. By the definition of partial derivative, we have
$$
\frac{\partial F}{\partial y}(z) = \lim_{k \to 0} \frac{F(z(k)) - F(z)}{k} = i \lim_{k \to 0} \frac{1}{k} \int_{z}^{z(k)} f\, dy,
$$
where the last integral is along the straight line segment from $z = (x,y)$ to $z(k) = (x, y+k)$. Using \eqref{yt}, we have
$$
\int_{z}^{z(k)} f\, dy = \int_{y}^{y+k} f(x,t) \cdot 1\, dt,
$$
so by the first fundamental theorem of calculus, the limit is precisely $f(z)$. Thus,
$$
\bbox[5px,border:2px solid black]
{
\frac{\partial F}{\partial y}(z) = i f(z).
}
$$
Best Answer
We have $\frac{d}{dz}\frac{\phi(\zeta)}{(\zeta - z)^{n}} = n\frac{\phi(\zeta)}{(\zeta - z)^{n + 1}}$. Let $\Omega$ be the region defined by $\gamma$. By differentiating under the integral sign, for any $z_0 \in \mathbb{\Omega}$, $$F_n'(z_0) = \int_{\gamma}n\frac{\phi(\zeta)}{(\zeta - z)^{n + 1}}\,d\zeta.$$ All we need to do is justify differentiating under the integral sign. As usual, this is done using the dominated convergence theorem. Fix $z_0 \in \mathbb{C}$. As usual, to use the dominated convergence theorem to justify the differentiation, it suffices to find a dominating function $g(\zeta) \in L^1(\gamma)$ such that $|n\frac{\phi(\zeta)}{(\zeta - z)^{n + 1}}| \leq g(\zeta)$ for all $z$ in some chosen neighborhood of $z_0$ and all $\zeta \in \gamma$. Here $L^1(\gamma)$ uses the arclength measure on $\gamma$. Integral with respect to this measure is $\int_{\gamma} g(\zeta) ds(\zeta) = \int_{I}g(\gamma(t))|\gamma'(t)|\,dt$. Pick a neighborhood $U$ of $z_0$ such that $d(U, \gamma) = C > 0$. Then for $z \in U$, $\zeta \in \gamma$, $$|n\frac{\phi(\zeta)}{(\zeta - z)^{n + 1}}| \leq n\frac{|\phi(\zeta)|}{C^{n + 1}}.$$ $|\phi(\zeta)|$ is in $L^1(\gamma)$ since $\phi$ is continuous on $\gamma$, and therefore bounded since $\gamma$ is compact. So the differentiation is justified.
Edit: Above is a situation when we had to compute a derivative of an integral. We have $g(z) = \int_{\gamma}f(z, \zeta)\,d\zeta$ defined for $z$ in a neighborhood of $z_0$. We want to show that $$g'(z_0) = \int_{\gamma}\frac{\partial f}{\partial z}(z_0, \zeta)\,d\zeta.$$ Here we assume that $f$ is complex differentiable in $z$. To show this equality, we use difference quotients: $$\frac{g(z_0 + h) - g(z_0)}{h} = \int_{\gamma}\frac{f(z_0 + h, \zeta) - f(z_0, \zeta)}{h}\,d\zeta = \int_{a}^{b}\frac{f(z_0 + h, \gamma(t)) - f(z_0, \gamma(t))}{h}\gamma'(t)\,dt.$$ Now we want to take limits as $h \to 0$ and bring the limit inside the integral on the right to get the desired result. We do this using dominated convergence. It suffices to get a bound of the form $$|\frac{f(z_0 + h, \gamma(t)) - f(z_0, \gamma(t))}{h}\gamma'(t)| \leq H(\gamma(t))|\gamma'(t)|\text{ with }\int_{a}^{b}H(\gamma(t))|\gamma'(t)|\,dt < \infty$$ for all $h$ in some disk $D_r(0)$ of our choice around $0$ and all $t \in (a, b)$. In other words, it suffices to find a bounding function $H \in L^1(\gamma)$ such that $$|\frac{f(z_0 + h, \zeta) - f(z_0, \zeta)}{h}| \leq H(\zeta)\text{ for all }\zeta \in \gamma \text{ and } h \in D_r(0).$$ By the fundamental theorem of calculus, for any $h \in D_r(0)$, $$|\frac{f(z_0 + h, \zeta) - f(z_0, \zeta)}{h}| \leq \sup_{z \in D_r(z_0)}|\frac{\partial f}{\partial z}(z, \zeta)|.$$ Thus it is enough to find $H \in L^1(\gamma)$ with $|\frac{\partial f}{\partial z}(z, \zeta)| \leq H(\zeta)$ for all $z \in D_r(z_0)$, $\zeta \in \gamma$.