Separating a Complex Valued Function

Question

Is there a formula (with mathematical reasoning) for separating a complex-valued function $f(z)=f(x+iy)$ into the form $ f(z)=u(x,y) + iv(x,y)$?

Thank You,

C.A

Answer 1

I don't believe there exists one that's quite that clean. But let us assume that over some local area you can model the function with a complex exponential series (the upper bound k need not be finite)

$$ \sum_{j = 0 }^{k} \left[ e^{w_j x} \right] $$

where each $w_j$ is a complex number. We can decompose the $w_j$ into two parts $u_j + y_ji$ (where u and y are real) Now given an argument $x = a + bi$ (a and b are real) each term decomposes into the form

$$e^{w_jx} = e^{(u_j + y_ji)(a + bi)} = e^{(u_ja - y_jb) + (u_jb + y_ja)i} = e^{(u_ja - y_jb)} e^{(u_jb + y_ja)i} $$

So here is the trick $e^{(u_jb + y_ja)i}$ is guaranteed to be a complex number of absolute value 1 and can be computer quite easily and decomposed into elements $q_j + r_ji$

$e^{(u_ja - y_jb)}$ is a real number that sort of determines how far from 0 the value is, so the first part of the product decides where on the complex field the number lies (which point on the unit circle) and the second product answers the question of how far out

That means that each number can then be expressed of the form:

$$ \sum_{j = 0 }^{k} \left[ q_je^{(u_ja - y_jb)} \right] + \sum_{j = 0 }^{k} \left[ r_je^{(u_ja - y_jb)} \right]i $$

Now you can take the each of these newly found series and see if a closed form exists

So the full form of the answer is:

$$ \sum_{j = 0 }^{k} \left[ e^{(u_j + y_ji)(a + bi)} \right] = \sum_{j = 0 }^{k} \left[ Re(e^{u_jb + y_ja})e^{(u_ja - y_jb)} \right] + \sum_{j = 0 }^{k} \left[ Im(e^{u_jb + y_ja})e^{(u_ja - y_jb)} \right]i $$

Which shows you how to do splitting if given an exponential series for an answer (or an approximation).

It's a famous theorem that ALL functions have exponential series so that shouldn't be an issue

Answer 2

$u = {1 \over 2} (f + \bar{f})$, $v = {1 \over 2i} (f - \bar{f})$.

It is straightforward to add the above and see that $u+iv = f$.