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I'm currently working a formula about iterated functions. One of its restrictions in certain cases, is that for a certain function $f$ and for any complex number $z$ situated inside a certain zone in the complex plane (the zone is defined), there isn't another complex number $z_1$ in that zone such that $f(z)=f(z_1)$.

It's completely analogous to real functions which are monotonic in a certain interval. Where inside that interval, there isn't 2 real numbers $x_1$ and $x_2$ such that $f(x_1)=f(x_2)$

I just wonder if there is any analogue of that for complex functions, in a certain zone of the complex plane. Like for real functions, where it's a «monotonic function on an interval»

Is there any analogue, and if yes is there any ways to mathematically check if a function satisfies that? (for the real example, that would be a strictly positive or negative derivative of this function on that same interval, that implies the function to be strictly monotonic on that interval)

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  • $\begingroup$ Note that while strictly positive or negative derivative implies one-to-one, there are one-to-one functions for which that is not true. In fact, there are continuous, strictly increasing functions that have zero derivative at all points where they are differentiable. $\endgroup$ Commented Apr 16, 2023 at 15:38

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Locally one-to-one = locally injective. The nonvanishing of the derivative at $z_0$ ensures that this property is locally true on some open ball centered at $z_0$.

Addenda.

  1. If the function is analytic, the derivative referenced above is the complex derivative. If the function is only real-variable differentiable, the derivative should be interpreted as the real-variable derivative (differential) as a multivariable map from the plane into the plane. If the function is not differentiable, then locally 1-1 still makes sense but there won't be any easy calculus test for it.

  2. There is no standard agreement on how to order complex numbers, so there is no simple way to answer your original question if you interpret monotonicity as some form of "order-preserving".

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  • $\begingroup$ Thank you very much! $\endgroup$ Commented Apr 16, 2023 at 14:31
  • $\begingroup$ There are also everywhere continuous but non-differentiable complex functions that have this property, with $\bar{z}$ being the most notable. So this won't capture all such functions. This is not the case in $\mathbb R$, where all continuous monotonic functions are differentiable almost everywhere. $\endgroup$ Commented Apr 16, 2023 at 15:21
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A complex-valued function f(z) of a complex variable z is said to be:

monotonically increasing on a domain D in the complex plane if for any two points z1 and z2 in D with z1 < z2, we have f(z1) < f(z2). monotonically decreasing on a domain D in the complex plane if for any two points z1 and z2 in D with z1 < z2, we have f(z1) > f(z2). Here, the inequality symbols < and > are interpreted in the same sense as in the real numbers.

Note that in the complex plane, the concept of "left" and "right" is not well-defined as it is in the real line. Therefore, it is more appropriate to use the concept of "less than" and "greater than" for defining monotonicity in the complex plane.

For example, the function f(z) = z^2 is not monotonically increasing or decreasing on the entire complex plane, but it is monotonically increasing or decreasing on some subsets of the plane. For instance, it is monotonically increasing on the set of points with positive real parts and monotonically decreasing on the set of points with negative real parts.

In general, the behavior of a complex function with respect to monotonicity can be quite complex, and there are many interesting examples of functions with intricate monotonicity properties in the complex plane.

The inequality f(z1) < f(z2) means that the value of the function f at the point z1 is strictly less than the value of the function f at the point z2. In other words, f(z1) is located "below" f(z2) on the complex plane.

To give a concrete example, suppose we have a function f(z) = z^2 and we consider the two points z1 = 1 + i and z2 = 2 + i on the complex plane. Then, we can calculate:

f(z1) = (1 + i)^2 = 1 + 2i - 1 = 2i

f(z2) = (2 + i)^2 = 4 + 4i - 1 = 3 + 4i

Since the imaginary part of f(z2) is greater than the imaginary part of f(z1), we have:

f(z1) < f(z2)

This means that the value of the function f at the point z1 is located "below" the value of the function f at the point z2 on the complex plane. Geometrically, this corresponds to the fact that the point f(z1) is closer to the origin than the point f(z2) on the complex plane.

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    $\begingroup$ What does $f(z_1)<f(z_2)$ mean? $\endgroup$ Commented Apr 16, 2023 at 15:06
  • $\begingroup$ See my edit above $\endgroup$ Commented Apr 16, 2023 at 15:11
  • $\begingroup$ So if, say, $f(z_1)=1$ and $f(z_2)=2$, you would say that the statement $f(z_1)<f(z_2)$ is false since $1$ is not below $2$. Am I right? $\endgroup$ Commented Apr 16, 2023 at 15:14
  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$ Commented Apr 16, 2023 at 16:03

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