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$f(z)=u(x,y)+iv(x,y), u$ and $v$ are real differentiable. Is $f$ complex differentiable?

I know that if $u$ and $v$ are continuous, then $f$ is continuous. How about the differentiability? If $f$ is differentiable, can we write $f'$ in terms of the partial derivatives of $u$ and $v$? Thanks!

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1 Answers1

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No, that's incorrect. Let me illustrate this with an example found in some cv notes in the comments under @Rebellos's answer. (First example under Theorem 1 in the notes.)

Let $f(z)=\bar{z}$ be defined on $\Bbb{C}$. Then $f$ induces $u(x,y)=x$ and $v(x,y)=-y$. Both of them are real differentiable, but $f$ is nowhere differentiable (holomorphic) since the Cauchy–Riemann equations is never satisfied. $$\because u_x=1 \quad u_y=0 \qquad v_x=0 \quad v_y=-1 \\ \therefore u_x(z) \neq v_y(z) \quad\forall\,z\in\Bbb{C}$$

To address the last prompt, $f'(z)=f'(x+iy)=u_x(x,y)+iv_x(x,y)$.

  • For a derivation of this identity, you may consult another cv notes. (Equations (3)-(5))
  • For the link between $$f'(z_0)= \frac{\partial f}{\partial x}(z_0) =\frac{\partial f}{\partial z}(z_0),$$ you may see another related Math.SE question.
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