Proof of $5.11$. Let
$$D=] a, b[\times] c, d[\quad \text { with } \quad a<b \text { and } c<d .$$
We choose two fixed points
$$\left.x_{0} \in\right] a, b\left[\quad \text { and } \quad y_{0} \in\right] c, d[\text {. }$$
From the equation $\partial_{1} u=\partial_{2} v$ it follows that, for each $\left.x \in\right] a, b[$,
$$v(x, y)=\int_{y_{0}}^{y} \partial_{1} u(x, t) d t+h(x)$$
From LEIBNIZ’s rule the integral is differentiable as a function of $x$ and we have

I Differential Calculus in the Complex Plane $\mathbb{C}$
\begin{aligned} \partial_{1} v(x, y) &=\int_{y 0}^{y} \partial_{1}^{2} u(x, t) d t+h^{\prime}(x)=-\int_{y_{0}}^{y} \partial_{2}^{2} u(x, t) d t+h^{\prime}(x) \ &=\partial_{2} u\left(x, y_{0}\right)-\partial_{2} u(x, y)+h^{\prime}(x) \end{aligned}
and therefore
$$h^{\prime}(x)=-\partial_{2} u\left(x, y_{0}\right) .$$
This suggests the following Ansatz:
$$v(x, y):=\int_{y_{0}}^{y} \partial_{1} u(x, t) d t-\int_{x_{0}}^{x} \partial_{2} u\left(t, y_{0}\right) d t$$

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