# Every harmonic function is the real part of a holomorphic function

Is there a way to show that every harmonic function is the real part of a holomorphic function without using integration equations if later theorems are allowed also?

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This amounts to solving the Cauchy–Riemann equations as a system of PDEs. Integration is probably the only method that works generically... – Zhen Lin Feb 11 '12 at 12:38

Let $$h:\quad \Omega\to{\mathbb R}\ ,\qquad (x,y)\mapsto h(x,y)$$ be the given harmonic function which we assume to be $C^2$. Using $h$ we define the function $$f(x,y):= h_x(x,y)- i\ h_y(x,y)$$ with real part $u(x,y)=h_x(x,y)$ and imaginary part $v(x,y)=- h_y(x,y)$. As $h$ is harmonic one has $u_x\equiv v_y$, furthermore $u_y\equiv -v_x$ by equality of the mixed derivatives. So $f\colon \Omega\to{\mathbb C}$ is $C^1$ and satisfies the CR equations; therefore it is an analytic function of $z=x+iy\in\Omega$.
Assume that $\Omega$ is simply connected and chose a point $z_0\in\Omega$. Then by a standard theorem of complex analysis the function $$F(z)\ :=\ h(z_0)+\int_\gamma f(z)\ dz\ , \qquad \hbox{\gamma\  a path from z_0 to z}\ ,$$ is an analytic primitive of $f$ in $\Omega$. Let $(x,y)\mapsto U(x,y)$ be the real part of $F$. Then by the CR equations, this time applied to $F$, we have $$U_x(z) -i U_y(z)=F'(z)=f(z)= h_x(z)-i h_y(z)\qquad(z\in\Omega)\ .$$ It follows that $$\nabla U(x,y)\equiv\nabla h(x,y)\qquad \bigl( (x,y)\in\Omega\bigr)\ ,$$ and as $U(x_0,y_0)=h(x_0,y_0)$ we conclude that in fact $U(x,y)\equiv h(x,y)$ in $\Omega$.
Note that we had to assume $\Omega$ simply connected. The function $h(x,y):=\log\sqrt{x^2+y^2}$ is harmonic in the punctured plane but is not the real part of an analytic function in this domain.
How we know that $log \sqrt{x^2+y^2}$ is not the real part of an analytic function? – Idonknow Oct 6 '13 at 17:26
@Idonknow: This is one of the most basic facts of complex analysis. The analytic function you are looking for would be $f(z)=\log|z|+i{\rm Arg}(z)$, but this function is not well defined in all of ${\mathbb C}\setminus\{0\}$. – Christian Blatter Oct 6 '13 at 19:01