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Quoting Saff & Snider's "Fundamentals of Complex Analysis" (3rd. edition):

If w denotes the value of the function $f$ at the point $z$, we then write $w=f(z)$.

Just as $z$ decomposes into real and imaginary parts as $z=a+ib$, the real and imaginary parts of $w$ are each (real-valued) functions of $z$ or, equivalently, of $x$ and $y$, and so we customarily write

$w=u(x,y)+iv(x,y)$

with $u$ and $v$ denoting the real and imaginary parts, respectively, of $w$. Thus a complex valued function of a complex variable is, in essence, a pair of real functions of two real variables.

My question is, does $u$ and $v$ have do be functions from $\mathbb{R}^2$? If $z=x+iy$ and $w=a+ib$, can you have just $h(x)=a$ and $g(y)=b$ so that $w=h(x)+ig(y)$?

What about stuff like $w=u(x,y,0)+iv(x,y,0)$?

Basically, why is it obvious that $f(z)$ should decompose into functions of two variables? Is there a nice way to think about this?

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    $\begingroup$ It's not that it should, it's that it can. $\endgroup$
    – Git Gud
    Oct 6, 2013 at 14:23
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    $\begingroup$ You can identify $\mathbb{C}$ with $\mathbb{R}^2$, so you can interpret every function of one complex variable as a function of two real variables and vice versa. $\endgroup$ Oct 6, 2013 at 14:24
  • $\begingroup$ In your example you would have $u(x,y)=h(x)$, $v(x,y)=g(y)$, perfectly legal functions of $x$ and $y$. $\endgroup$
    – Carsten S
    Oct 6, 2013 at 14:26

1 Answer 1

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It is common to misunderstand "$u$ is a function of $x$ and $y$" as "$u$ depends on $x$ and $y$". Which leads to errors such as thinking that $u(x,y)=x^2$ is not a function of $x $ and $y$.

Saying that "$u$ is a function of $x$ and $y$" means that the domain of $u$ is the set of ordered pairs $(x,y)$. Informally, $u$ takes two real numbers as inputs, and produces a number as output. To qualify as a function, $u$ must be consistent: if given the same pair $(x,y)$ today and tomorrow, it must give the same output. Other than that, it's free to use the inputs in any way it wants; and that includes not using them at all. For example, $f(x,y)=\sqrt{2}$ is a function which always returns $\sqrt{2}$ as output. It's still a function of $x$ and $y$ -- meaning it takes $x$ and $y$ as inputs -- it just does not use those inputs.

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    $\begingroup$ There is no reason to bring up time! A function is $f:A\to B$ is a subset $f\subset A\times B$ such that for every $a\in A$ there exists one, and only one $b\in B$ such that $(a,b)\in f$; from where we write $f(a)=b$. Slightly informally it is a correspondence that assigns to every $a\in A$ one, and only one $f(a)=b\in B$. $\endgroup$
    – Pedro
    Oct 6, 2013 at 16:06
  • $\begingroup$ I just think the time analogy is a strange one. $\endgroup$
    – Pedro
    Oct 6, 2013 at 16:55

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