If $f:\mathbb{R}\rightarrow\mathbb{R}$ is differentiable, then, for example, if $f'(x_0)=0$ and $f'$ is again differentiable at $x_0$ and $f''(x_0)\neq 0$, then $f$ has a maximum or minimum at $x_0$.

But if $f:\mathbb{C}\rightarrow \mathbb{C}$ is holomorphic, how can I think of the zeros of $f'$? What does it tell me if $f'(z_0)=0$ and $f''(z_0) \neq 0$? For example, is it true that $|f(z_0)|$ is a local maximum or minimum of $|f|$?


2 Answers 2


Felix Klein published a small book in 1893, called On Riemann's Theory of Algebraic Functions and Their Integrals (available for free at the the linked URL). This book contains a wonderful geometrico-physical interpretation of the zeroes and poles of meromorphic functions.

Klein interprets the level curves of the real part of a meromorphic function as the stream-lines of an electric field, and interprets the level curves of the imaginary part (which are orthogonal to those of the real part, by Cauchy-Riemann) as the lines of constant potential. The zeroes are then interpreted as sources, or point charges; the poles are interpreted as sinks (or maybe the other way around - it's been a while since I read the book. Anyways, it doesn't matter too much). The residues reflect the strength of the corresponding source or sink. Many theorems of complex function theory then take on a very beautiful and concrete meaning. For instance, the fact that the residues of a meromorphic function on a compact Riemann surface sum to $0$ becomes a manifestation of the principle of conservation of charge, or an analogue of the preservation of currents, Kirchoff's circuit law.

The book, although over a century old, has aged very well, and it's worth taking a look at if only for its beautiful illustrations:

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No. It tells you that locally (in a ball around $z_0$) the function $f$ is looking like $f(z_0) + c(z-z_0)^2$, so it wraps twice around a ball centered at $f(z_0)$. Stuff over $\Bbb R$ is very particular because of the fact that squares of real numbers are nonnegative. :)

  • $\begingroup$ Thanks! So in general, if the $1$st to the $n$th derivative are zero and the $n+1$st is not zero, then it wraps $n+1$ times around? $\endgroup$ May 17, 2014 at 21:57
  • 1
    $\begingroup$ Yup! Precisely. It looks locally like $z\mapsto z^{n+1}$. :) $\endgroup$ May 17, 2014 at 22:14

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