On the third edition of Ahlfors' Complex Analysis, page 113 Theorem 4(Cauchy's Theorem in a Disk) it states: Hence $F(z)$ is analytic in $\Delta$ with the derivative $f(z).$
In Ahlfors' Complex Analysis, this proof is written very briefly and a lot of details are omitted. I want to check all the details about why $F(z)$ is analytic. But I have trouble in understanding it.
In what follows we assume that $\Omega$ is an open disk $|z-z_{0}|<\rho$ to be denoted by $\Delta$.
Theorem 4(Cauchy's Theorem in a Disk) If $f(z)$ is analytic in an open disk $\Delta$, then
\begin{equation*} \int_{\gamma}f(z)dz=0 \end{equation*} for every closed curve $\gamma$ in $\Delta$.
Proof:
$(x_{0},y_{0})$ is the center of $\Delta$. For every $z=x+iy\in\Delta$, we define a function:
\begin{equation*} F(z) =\int_{(x_{0},y_{0})\rightarrow(x,y_{0})}f(z)dz+\int_{(x,y_{0})\rightarrow(x,y)}f(z)dz \end{equation*}
$f(z)=u(x,y)+iv(x,y)$ is analytic and continuous.
$F(z)=U(x,y)+iV(x,y)$
$$\begin{align*} F(z) &=\int_{(x_{0},y_{0})\rightarrow(x,y_{0})}f(z)dz+\int_{(x,y_{0})\rightarrow(x,y)}f(z)dz\\ &=\int_{x_{0}}^{x}\left(u(t,y_{0})+iv(t,y_{0})\right)dt+\int_{y_{0}}^{y}\left(u(x,t)+iv(x,t)\right)dt\\ &=\left(\int_{x_{0}}^{x}u(t,y_{0})dt+\int_{y_{0}}^{y}u(x,t)dt\right)+i\left(\int_{x_{0}}^{x}v(t,y_{0})dt+\int_{y_{0}}^{y}v(x,t)dt\right) \end{align*}$$ $U(x,y) =\int_{x_{0}}^{x}u(t,y_{0})dt+\int_{y_{0}}^{y}u(x,t)dt$
$V(x,y) =\int_{x_{0}}^{x}v(t,y_{0})dt+\int_{y_{0}}^{y}v(x,t)dt$
$\frac{\partial U}{\partial y} =u(x,y)$
$\frac{\partial V}{\partial y} =v(x,y)$
According to "Theorem 2(Cauchy's Theorem for a Rectangle)", we also have: $$\begin{align*} F(z) &=\int_{(x_{0},y_{0})\rightarrow(x_{0},y)}f(z)dz+\int_{(x_{0},y)\rightarrow(x,y)}f(z)dz\\ &=\int_{y_{0}}^{y}\left(u(x_{0},t)+iv(x_{0},t)\right)dt+\int_{x_{0}}^{x}\left(u(t,y)+iv(t,y)\right)dt\\ &=\left(\int_{y_{0}}^{y}u(x_{0},t)dt+\int_{x_{0}}^{x}u(t,y)dt\right)+i\left(\int_{y_{0}}^{y}v(x_{0},t)dt+\int_{x_{0}}^{x}v(t,y)dt\right) \end{align*}$$
$U(x,y) =\int_{y_{0}}^{y}u(x_{0},t)dt+\int_{x_{0}}^{x}u(t,y)dt$
$V(x,y) =\int_{y_{0}}^{y}v(x_{0},t)dt+\int_{x_{0}}^{x}v(t,y)dt$
$\frac{\partial U}{\partial x} =u(x,y)$
$\frac{\partial V}{\partial x} =v(x,y)$
If $F(z)$ is analytic, then Cauchy-Riemann equations are satisfied. So we have $\frac{\partial U}{\partial x}=\frac{\partial V}{\partial y}$. Since $\frac{\partial U}{\partial x} =u(x,y)$ and $\frac{\partial V}{\partial y} =v(x,y)$, we have $u(x,y)=v(x,y)$. But $u(x,y)=v(x,y)$ is not true. So I get stuck here. I don't know what is going wrong so that I get $u(x,y)=v(x,y)$.
This a question from Ahlfors' Complex Analysis. I show you the whole context of the question below. I have trouble understanding the statement with red line.
