Is the value of a harmonic function must be real?

Question

I am reading Stein's Complex Analysis and confused about the proof of the following lemma on Page 214 in chapter 8:

Lemma 1.3 Let $V$ and $U$ be open sets in $\mathbb{C}$ and $F:V \to U$ a holomorphic function. If $u: U \to \mathbb{C}$ is a harmonic function, then $u \circ F$ is harmonic on $V$.

Proof. The thrust of the lemma is purely local, so we may assume that $U$ is an open disc. We let $G$ be a holomorphic function in $U$ whose real part is $u$ (such a $G$ exists by Exercise 12 in Chapter 2, and is determined up to an additive constant). Let $H = G \circ F$ and note that $u \circ F$ is the real part of $H$. Hence $u \circ F$ is harmonic because $H$ is holomorphic.

How can a function $u$, whose value is probably not real, become the real part of a holomorphic function? What's wrong with my understanding?

Answer 1

Suppose $u = g + ih$ for some harmonic funtion $g$ and $h$, then $u \circ F = g \circ F + ih \circ F$, then the same argument applies to function $g \circ F$ and $h \circ F$, and the conclusion is the same.

Answer 2

It follows from the following fact:

For any real harmonic function $u$ on $\mathbb R^2$, there is another real harmonic function $v$ on $\mathbb R^2$, called its harmonic conjugate, such that the two satisfy the Cauchy-Riemann equations: $$ \frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y} \;\;\;\;\;,\;\;\;\;\;\frac{\partial v}{\partial y} = \frac{\partial u}{\partial x}. $$

Proof: Because $u$ is harmonic, the vector field $(-\partial_y u, \partial_x u)$ is irrotational, meaning its line integrals are path independent. Therefore there exists a function $v$ with $\nabla v =(\partial_x v, \partial_y v) = (-\partial_y u, \partial_x u)$, satisfying the Cauchy-Riemann equations. Additionally, $\nabla^2v = -\partial_x\partial_y u +\partial_y\partial_x u = 0$, so $v$ is also harmonic.

Since $u$ and $v$ satisfy the Cauchy-Riemann equations by definition, the function $f(x + i y) = u(x,y) + i v(x,y)$ is holomorphic. Thus for any harmonic $u$, there is a holomorphic function $f$ such that $\mathrm{Re}[f(x + i y )] = u(x, y)$.