I suppose it natural to write $f(x,y) = u(x,y) + iv(x,y)$ but it make me wonder are we losing anything in that process? When we talk about things component-wise, is that limiting anything?

Can we always assume that given any complex function $f(x,y)$ that it could be written $u(x,y)+iv(x,y)$? (yes, I think)

Can we always assume that we can find formulas/equations for $u$ and $v$? (no, I think)

Are there situations where even though we are given the equation for $f$, it is not possible to find equations for u and v - more substantial ones than $u=Re(f)$ and $v=Im(f)$ obviously ? (I'm not sure about this one but I suspect that it can never happen but then again...)

Anyone with anything to add is appreciated.


If you have a formula for $\mathrm{f}$ then you are guaranteed a formula for $\mathrm{u}$ and $\mathrm{v}$.

The real and imaginary parts of any complex number, and hence any complex valued function, can be found using the simple formulae

$$\Re(z) = \frac{1}{2}\left(z+\overline{z}\right)$$ $$\Im(z) = \frac{1}{2\mathrm{i}}\left(z-\overline{z}\right)$$

The formulae can be expressed in terms of $x$ any $y$ by making the substitution $z=x+y\mathrm{i}$ and doing lots of unsightly algebra.

A more interesting question is this: If I give you $\mathrm{u}(x,y)$ and $\mathrm{v}(x,y)$, can you find $\mathrm{f}(z)$?

  • $\begingroup$ This requires the knowledge of the conjugate, which is not easy to find for some functions, no? $\endgroup$ – Daniel R May 5 '14 at 11:48
  • $\begingroup$ @DanielR Not at all. If you have a formula then the conjugate is trivial to find. Give me an example of a function where you can't find the conjugate. $\endgroup$ – Fly by Night May 5 '14 at 14:56
  • $\begingroup$ I was thinking along the lines of the "complex real numbers" and related forms, where the real part cannot be expressed by radicals. (A related question of mine here). For example, what's the conjugate of $$f(z)=-\frac{1-iz\sqrt3}{2^{2/3}\sqrt[3]{-z+i\sqrt3}}-\frac12\sqrt[3]{\frac12\left(-1+i\sqrt3\right)}\left(1+\sqrt3\z^{2/5}right)$$? I haven't really thought this through, it might be trivial. I just wanted to check. $\endgroup$ – Daniel R May 5 '14 at 16:18
  • $\begingroup$ Please disregard the above comment, MathJax in comments is obviously not working very well for longer expressions (possibly due to the []s). Instead, imagine the second expression for $x$ in this question. Add a couple of random $z$ expressions here and there. What would the conjugate be? $\endgroup$ – Daniel R May 5 '14 at 16:25
  • $\begingroup$ Not the most satisfying answer but I have to agree in theory with it. But if possible, I'm not looking for something so generic. It should be as close to the function as possible without using such generic expressions. You say after doing lots of "unsightly algebra". Is that all it takes? Are you sure? How about the Riemann-Zeta function? Would that work with it? Given u and v, I could find f as follows: f(x,y)=u+iv but I think what you mean is find a function f(z) which equals u+iv so please do tell! $\endgroup$ – user147835 May 6 '14 at 2:10

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