Let $x:\Omega \subset
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\mathbb{R}^2
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\rightarrow
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\mathbb{R}
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$ harmonic (I do not know if you need) and $\frac{\partial x}{\partial u_{1}},\frac{\partial x}{%
\partial u_{2}}:\Omega \subset
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\mathbb{R}^2
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\rightarrow
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\mathbb{R}
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$ their partial derivatives, with $\Omega $ simply connected open. Show that
$$
x\left( z\right) =\Re \int_{0}^{z}\left( \frac{\partial x}{%
\partial u_{1}}\left( z\right) -i\frac{\partial x}{\partial u_{2}}\left(
z\right) \right) dz. $$
I have looked in books of complex analysis, but I can not find the property that allows me this equality.
Best Answer
I don't think this has much to do with harmonic functions or complex analysis. Below I use different notation, otherwise I'll get confused.
Assume $\Omega$ is open and connected (I don't think we need simple connectivity.) I'll be writing $w=x+iy$ for $w\in \mathbb C.$ Let $u:\Omega \to \mathbb R$ be continuously differentiable. Suppose $0,z\in \Omega$ and $\gamma:[a,b]\to \Omega$ is a contour with $\gamma(a)=0,\gamma (b)=z.$ Then
$$\tag 1\text { Re }\int_\gamma (u_x(w)-iu_y(w))\, dw = u(z)-u(0).$$
Proof: Write $\gamma(t)= x(t)+iy(t).$ Then the left side of $(1)$ equals
$$\text { Re } \left [\int_a^b (u_x(\gamma(t))-iu_y(\gamma(t)))(x'(t)+iy'(t))\,dt \right ]$$ $$ = \int_a^b [u_x(\gamma(t))x'(t) + u_y(\gamma(t))y'(t)]\,dt $$ $$ = \int_a^b [u(\gamma(t))]'\, dt = u(\gamma(b)) - u(\gamma(a)) = u(z)-u(0).$$
The only things we have used are the chain rule and the FTC.