Is it always possible to split a complex function into the sum of its real + imaginary parts? I'm looking for any kind of proof or at least an intuitive explanation. I am curious because I want to know:
if I have an unknown complex function that I want to manipulate, can I assume it has the structure a(x) + ib(x).
It seems to me this should definitely be possible. I'm hoping someone can provide a lucid argument that makes the answer seem stupidly obvious
 A: Of course you can.  The real and imaginary parts of a complex number uniquely determine it.  You can identify $\mathbb{C}$ with the plane $\mathbb{R}^2$ via $a + ib \mapsto (a,b)$.  
Let $X$ be a set, and let $f$ be a function from $X$ to $\mathbb{C} = \mathbb{R}^2$.  Let $\pi_1: \mathbb{R}^2 \rightarrow \mathbb{R}$ be the function $\pi(a,b) = a$, and let $\pi_2: \mathbb{R}^2 \rightarrow \mathbb{R}$ be the function $\pi_2(a,b) = b$.
Let $u: X \rightarrow \mathbb{R}$ be the composition $\pi_1 \circ f$, and let $v: X \rightarrow \mathbb{R}$ be the composition $\pi_2 \circ f$.  Then obviously 
$$f(x) = (u(x),v(x))$$
for any $x \in X$.  Or in the usual notation of complex numbers,
$$f(x) = u(x) + i v(x)$$
for all $x \in X$.  This is written more succinctly as $f = u + iv$.
A: Yes. Any complex number $z$ is of the form $x+iy$, where
$$x=\frac{z+\overline{z}}{2} \quad\text{and}\quad
  y=\frac{z-\overline{z}}{2i}
$$ are real numbers called the real and imaginary parts of $z$, respectively. If $f:A\to\mathbb{C}$ is a function from a set into the complex numbers, then define functions $u,v:A\to\mathbb{C}$ by
$$u(a)=\frac{f(a)+\overline{f(a)}}{2} \quad\text{and}\quad
  v(a)=\frac{f(a)-\overline{f(a)}}{2i}.
$$
Then $f(a)=u(a)+iv(a)$ for every $a\in A$, and as one would expect, we call $u$ the real part of $f$ and $v$ the imaginary part of $f$.
A: Hint: Any complex number $z$ can be written in the form $re^{i\theta}$...
