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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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    \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 \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, $z'(t)=x'(t)+iy'(t)$.2012-11-04

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What's wrong is that $f = u + i v$ may have a derivative without $u$ and $v$ having derivatives.

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    \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, $z'(t)=x'(t)+iy'(t)$.2012-10-25
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$f$ 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 $0$.