# graph of the size of a complex function [closed]

Hi Here there are two graphs for two functions from $R^2\mapsto R$.

Is there similar graph for the absolute value of a complex variable function $f:C\mapsto C$ that has the same point (like saddle point or transition). I know some functions that have the point $(x,y,|f(x+iy)|)$ on that such that in one direction it is maximum, and in the other direction it is minimum.

My question here is that: is there any such point such that in one direction it is maximum (or minimum) but in the other direction it is not maximum nor minimum (similar to $(0,0)$ in $y=x^3$ in the real case).

Thanks

-

## closed as too localized by Igor Rivin, Alain Valette, George Lowther, Ryan Budney, Yemon ChoiDec 27 '11 at 20:33

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Without further restrictions on your function $f$ (analytic? anti-analytic?) the answer is that a function $C \to R$ is exactly the same as a function $R^2\to R$. – Yemon Choi Dec 27 '11 at 20:35
$f$ is analytic – asd Dec 28 '11 at 14:40

For any nonconstant analytic function $f$, if $f'(p) = 0$ but $f(p)$ and $f''(p)$ are nonzero, then the graph of $|f(z)|$ will have a saddle point at $p$.
@Robert Israel: Yes, it is. but I was meaning that in one direction similar to $x^2$ and in the other direction similar to $x^3$. For example: $f(z)=cos(z)$ in $z=0$ has that condition, $|f|$ in one direction is like $x^2$ and in the other is like $(−x^2)$ – asd Dec 28 '11 at 14:42
Is it possible $f'(z)\neq0$ and $f(z)\neq0$, but $z$ is saddle point of the above type. – asd Dec 28 '11 at 14:46
If $f'(p) \ne 0$, it's not a saddle point. The definition of a saddle point requires a stationary point. – Robert Israel Dec 28 '11 at 19:26