# Function of a complex variable; must the real and imaginary parts be functions of two variables?

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?

• It's not that it should, it's that it can. Oct 6, 2013 at 14:23
• 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. Oct 6, 2013 at 14:24
• In your example you would have $u(x,y)=h(x)$, $v(x,y)=g(y)$, perfectly legal functions of $x$ and $y$. Oct 6, 2013 at 14:26

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.
• 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 $f(a)=b\in B$.