# Is a function $u,v$ in $f(z)=u+iv$ a real function of a complex variable?

In chapter 2 of "Complex Analysis" by Lars V. Alfors, the author concluded that "a real function of a complex variable either has the derivative 0 or else the derivative does not exist."

$\displaystyle \lim_{h \to 0} \frac{f(x+h+iy)-f(x+iy)}{h}$ is a real.

$\displaystyle \lim_{k \to 0} \frac{f(x+i(y+k))-f(x+iy)}{ik}$ is a pure imaginary number.

If both are the same, it must be 0. I understand this argument, but I came up with a question.

If we write a complex function $f$ of a complex variable $z$ as a sum $f(z)=u(z)+iv(z)$ where $u,v$ are real function of a complex variable, the only possible derivative of $f$ would seem to be only 0 according to the above conclusion. I know that this is ridiculous and there are plenty of counter examples ($f(z)=z, f(z)=z^2,...$), but I cannot find out what's wrong with my argument.

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You argument above, showing that the only possible derivative of $f$ is $0$, assumes that $f$ is real. – Stefan Smith Oct 25 '12 at 22:19
@bogus $f=u+iv$ and i means an imaginary number and $v$ doesn't necessarily identically vanish. So I don't think I assumed that $f$ is real though. – Tengu Oct 26 '12 at 22:05
how do you know your first limit above is real and your second limit above is pure imaginary unless $f$ is real? – Stefan Smith Oct 28 '12 at 17:14
@bogus Sorry I don't quite understand what you mean. – Tengu Oct 30 '12 at 20:33
@bogus Ah I understood what you meant. Of course, $f$ in the first half of my question is real function. But in the second half, $f$ means a complex function. I wrote "...a complex function $f$..." – Tengu Oct 30 '12 at 20:46

What's wrong is that $f = u + i v$ may have a derivative without $u$ and $v$ having derivatives.
If $f$ has a derivative, $u,v$ must have derivatives? – Tengu Oct 25 '12 at 0:38
@Tengu: No, it's the other way around. If $f$ is differentiable, then $u$ and $v$ are either both $0$ (and thus $f=0$) or neither of them is differentiable. – kahen Oct 25 '12 at 0:45
@kahen Oh then could you point out what's wrong with the following? $z(t)=x(t)+iy(t)$, $z$ differentiable at $t=a$, and $z'(a)=p+iq$. For each $\varepsilon >0$, there exists $\delta >0$ such that $|h|<\delta$ implies $\displaystyle \left | \frac{z(a+h)-z(a)}{h}-z'(a) \right|<\varepsilon$ $\to\displaystyle \left | \left( \frac{x(a+h)-x(a)}{h}-p \right)+i\left( \frac{y(a+h)-y(a)}{h}-q \right) \right |<\varepsilon$ – Tengu Oct 25 '12 at 0:52
$\to\displaystyle \sqrt{\left ( \frac{x(a+h)-x(a)}{h}-p \right)^2+ \left (\frac{y(a+h)-y(a)}{h}-q \right)^2}<\varepsilon$ Thus $u, v$ have the limits at $t=a$. Since this is true for every t, $z'(t)=x'(t)+iy'(t)$. – Tengu Oct 25 '12 at 0:53
$f$ is differentiable in the complex sense if and only if $u$ and $v$ satisfy the Cauchy-Riemann equations, $u_x = v_y$ and $u_y = -v_x$. If $v \equiv 0$, then $u_x \equiv u_y \equiv =0$ and $u$ is constant, hence $f$ is constant, and its derivative is $0$.