Complex Analysis - Harmonic function as real part of holomorphic function 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 $Re(f) = u$.
 A: 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.
