As complex numbers can be represented in the complex plane $\mathbb{C} \cong \mathbb{R}^2$ there are two related notions of derivatives.

For a real 2d function with continuous partial derivatives $ f:\left( \begin{array}{c} x \\ y \\ \end{array} \right) \mapsto \left( \begin{array}{c} u \\ v \\ \end{array} \right) $

the derivative at a point $z_0 = \left( \begin{array}{c} x_0 \\ y_0 \\ \end{array} \right) $ is defined as

$$ \lim_{h \to 0}\frac{\|f(z_0+h)-f(z_0)-Df(z_0)h \|}{\|h\|} = 0 $$

where $Df$ is the Jacobian $\left( \begin{array}{cc} u_x & u_y \\ v_x & v_y \\ \end{array}\right)$

and has due to the Cauchy-Riemann equations the structure $\left( \begin{array}{cc} a & b \\ -b & a \\ \end{array}\right)$

For a complex differentiable function $f(x,y) = u(x,y) + iv(x,y)$ the complex derivative is defined as

$$ \lim_{h \to 0}\frac{|f(z_0+h)-f(z_0)-f'(z_0)h |}{|h|} = \lim_{h \to 0}\left|\frac{f(z_0+h)-f(z_0)}{h} -f'(z_0)\right| = 0 $$

where now $f'(z_0)\in \mathbb{C}$.

Comparing the two we have

$$ Df(z_0)h \cong f'(z_0)h = f'(z_0)(h_1 + ih_2) $$

On the right hand side, the complex derivative $f'(z_0)$ is just a complex number multiplying $h=h_1 + ih_2$.

How can $f'(z_0)$ explicitely be calculated from the partial derivatives that appear in the Jacobian of the corresponding 2d function?

From the structure of the Jacobian and interpreting it as the matrix representation of a complex number, my first guess would be that

$$ f'(x_0) = a + ib = u_x(x_0) + iu_x(x_0) = v_y(x_0) - iv_x(x_0) $$

  • $\begingroup$ $a+ib = \left( \begin{array}{cc} a & b \\ -b & a \\ \end{array}\right)$ is the usual representation of $\mathbb{C}$ in $M_2(\mathbb{R})$ $\endgroup$
    – reuns
    Dec 10, 2016 at 15:01
  • 2
    $\begingroup$ If the Jacobian of $f(z)$ has the form $D f(z) = \left( \begin{array}{cc} a(z) & b(z) \\ -b(z) & a(z) \\ \end{array}\right)$ for every $z$, then $f(z)$ is complex-differentiable (this is trivial) but also $f(z) $ is complex analytic, that is $f'(z) = a(z)+ib(z)$ and all its derivatives exist (highly non-trivial, this is a consequence of the en.wikipedia.org/wiki/Cauchy's_integral_theorem ) $\endgroup$
    – reuns
    Dec 10, 2016 at 15:04

1 Answer 1


Your guess is correct.

From $z=x+iy$ and $f(z)=u(z)+iv(z)$, we form the vector function $$(x,y) = {u(z) \choose v(z)} = {\Re f(x+iy) \choose \Im f(x+iy)}$$ from $R^2$ to $R^2$. Now use the chain rule.


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