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How would I prove the following?

If $\Omega$ is simply connected, in $\mathbb C$, and if $u$ is a harmonic function in $\Omega$ , prove that there exists a holomorphic function $f$ in $\Omega$ such that $Re(f) = u$.

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closed as off-topic by user296602, Claude Leibovici, kimchi lover, The Phenotype, Davide Giraudo Feb 6 '18 at 14:41

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  • $\begingroup$ Hint: simple-connectedness is sufficient... so using extreme details about rectangles or disks is not to the point... $\endgroup$ – paul garrett Jan 28 '18 at 0:44
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We can find such a holomorphic function $f(z)$ as follows: since we know $u(z) = u(x, y)$, and it is harmonic, we may take its derivatives

$u_x(x, y) = \dfrac{\partial u(x, y)}{\partial x}, \; u_y(x, y) = \dfrac{\partial u(x, y)}{\partial y}; \tag 1$

consider the vector field

$V(x, y) = (V_x(x, y), V_y(x, y)) = (-u_y(x, y), u_x(x, y)); \tag 2$

we evaluate

$\nabla \times V(x, y) = \dfrac{\partial V_y(x, y)}{\partial x} - \dfrac{\partial V_x(x, y)}{\partial y} = \dfrac{\partial^2 u(x, y)}{\partial x^2} + \dfrac{\partial^2 u(x, y)}{\partial y^2} = 0, \tag 4$

since $u(x, y)$ is harmonic; since $\nabla \times V(x, y)$ vanishes, $V(x, y)$ is the gradient of some function $v(x, y)$ on $\Omega$:

$V(x, y) = \nabla v(x, y) = (v_x(x, y), v_y(x,y)); \tag 5$

we let $v(z) = v(x, y)$ be the imaginary part of $f(z)$:

$f(z) = f(x, y) = u(x, y) + i v(x, y); \tag 6$

by (2) and (5) we have

$(-u_y(x, y), u_x(x, y)) = V(x, y) = (v_x(x, y), v_y(x,y)); \tag 7$

that is,

$u_x(x, y) = v_y(x, y); \; u_y(x, y) = -v_x(x, y), \tag 8$

which are the Cauchy-Riemann equations for the function $f(z)$; thus $f(z)$ is holomorphic in $\Omega$, and it is evident that

$u(x, y) = \Re(f(x, y)); \; v(x, y) = \Im(f(x, y)), \tag 9$

as per request.

Note Added in Edit, Saturday 27 January 2018 6:12 PM PST: In fact, $v(x, y)$ may be explicitly presented as follows: let $z_0 = x_0 + i y_0 \in \Omega$ be fixed and $z = x + iy \in \Omega$ be arbitrary. Let $\gamma(t):[0, 1] \to \Omega$ be any differentiable path joining $(x_0, y_0)$ and $(x, y)$, i.e.,

$\gamma(0) = (x_0, y_0),\; \gamma(1) = (x, y); \tag{10}$

then we have

$v(x, y) - v(x_0, y_0) = \displaystyle \int_0^1 \dfrac{dv(\gamma(t))}{dt} = \int_0^1 \nabla v \cdot \gamma'(t) dt = \int_0^1 (-u_y, u_x) \cdot \gamma'(t) dt, \tag{11}$

which in principle allows us to compute $v(z) = v(x, y)$ for any $z \in \Omega$; $v(x_0, y_0)$ may be arbitrarily specified. End of Note.

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