Existence of a holomorphic function on unit disc such that Re(f) = u, for u a twice continuously differentiable and harmonic real-valued function? Let $u$ be a real-valued function defined on the unit disc $\mathbb{D}$. Suppose that $u$ is twice continuously differentiable and harmonic. That is,
$$\Delta u(x, y) = 0$$
for all $(x, y) \in \mathbb{D}$.

Precisely, this means that
$$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$$
right?

Prove that there exists a holomorphic function $f$ on the unit disc such that $Re(f) = u$. Also show that the imaginary part of $f$ is uniquely defined up to an additive (real) constant.
My questions:
The problem gives a hint that states we have 
$$f'(z) = \frac{2\partial u}{\partial z}$$

Why are this and the below equation true?

Therefore, we can let
$$g(z) = \frac{2 \partial u}{\partial z}$$
and prove that $g$ is holomorphic. Why can we find an $F$ with $F' = g$? Prove that $Re(F)$ differs from $u$ by a real constant.

I'm a little confused at how to approach this problem, as well as the notation used... (my questions are in italics, and are separated by page divisions). What exactly is $\frac{\partial u}{\partial z}$ if $u$ is a real-valued function? How do I approach the proof?
 A: For any function $h\colon \Bbb C \to \Bbb C$ it is defined that
$$\frac{\partial h}{\partial z} = \frac12 \left( \frac{\partial h}{\partial x} - i \frac{\partial h}{\partial y} \right).$$
In particular, $u$ is such a function, so the above definition shows what $\frac{\partial u}{\partial z}$ means.
Now, a holomorphic function $f\colon U \subset \Bbb C \to \Bbb C$ has the property that, if $f = u + iv$, then
$$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \qquad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}.$$
These are called the Cauchy-Riemann equations. Furthermore, it is not hard to see that also
$$f' = \frac{\partial f}{\partial x} \ \left( = \frac{\partial f}{\partial z} \right),$$
the second equation being an immediate consequence of the Cauchy-Riemann equations.
Using them, it is easy to see that if $f = u + iv$ is holomorphic, then $f'(z) = \frac{2 \partial u}{\partial z}$.
Now what you are to do is to take this function $g = \frac{2 \partial u}{\partial z}$ and show that it is holomorphic. To do this, the easiest way would be to show that $g$ satisfies the Cauchy-Riemann equations (they work in both ways!). This should not be hard, since this is basically just your assumption on $u$: We have $g = s + it$ where $s = \frac{\partial u}{\partial x}$ and $t = -\frac{\partial u}{\partial y}$, so, for example
$$\frac{\partial s}{\partial x} = \frac{\partial^2 u}{\partial x^2} = -\frac{\partial^2 u}{\partial y^2} = \frac{\partial t}{\partial y}.$$
The last step is to show that $g$ has an antiderivative $F$. This is true because $g$ is a holomorphic function on a simply connected region. To construct $F$, fix a point $z_0$ in the domain (i. e., the unit disc). For every $z$ in the domain define
$$F(z) = \int_{\gamma} g(w) dw, \quad \text{where $\gamma$ is a path from $z_0$ to $z$}.$$
That this actually works is not too hard to see if you know some theory behind it. Basically, we use the fact that the path integral of $g$ is equal for homotopic paths (in particular, it is $0$ on closed loops). Some basic calculations with the integral show that indeed $F' = g$.
Finally $F - f$ has zero derivative, so is constant.
