Hoolomorphic functions stretching angles

In summary, the function f given by f(z)=z^m multiplies angles at z0 by a factor of m. This mapping is given by f(z)=z^m*(z-z0)^m*(1+...), where the '...' stuff becomes small as z->z0. The derivative of f with respect to t gives the slope of a parametrized curve in R^2.
  • #1
sin123
14
0

Homework Statement



If z0 is a critical point of f(z), and m is the smallest integer such that f(m)(z0)=!= 0, then the mapping given by f multiples angles at z0 by a factor of m.

Hint: Taylor's Theorem

Homework Equations





The Attempt at a Solution



I don't even know where to start. Any hints would be appreciated.
 
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  • #2
Start by writing down the taylor expansion. What does the given condition tell you about the coefficients of the series?
 
  • #3
It tells me that the first couple of coefficients are zero. But what's the connection between the Taylor series of a function and the angles in its image?
 
  • #4
Yes, it's telling you that you can write the taylor series as f(z)=a_m*(z-z0)^m*(1+...) where the '...' stuff becomes small as z->z0. How do angles of straight lines through the origin behave under the function f(z)=z^m?
 
  • #5
I am not sure.

Suppose f(z) = z^n, and g is a line of slope m, say g(t) = t + m t i. Then

[tex]f(g(t)) = (t + m t i)^n = \Sigma \left( \begin{array}{c} n \\ 2k \end{array} \right) (-1)^k m^{2k} t^n \, + \, i \, \, \Sigma \left( \begin{array}{c} n \\ 2k+1 \end{array} \right) (-1)^{k} m^{2k+1} t^n [/tex]

Interpreting this as a parametrized curve in R^2, I would get

[tex]x(t) = \Sigma \left( \begin{array}{c} n \\ 2k \end{array} \right) (-1)^k m^{2k} t^n [/tex]
[tex]y(t) = \Sigma \left( \begin{array}{c} n \\ 2k+1 \end{array} \right) (-1)^{k} m^{2k+1} t^n[/tex]

Their derivatives with respect to t are

[tex] \frac{dx}{dt} = n \Sigma \left( \begin{array}{c} n \\ 2k \end{array} \right) (-1)^k m^{2k} t^{n-1} [/tex]
[tex]\frac{dy}{dt} = n \Sigma \left( \begin{array}{c} n \\ 2k+1 \end{array} \right) (-1)^{k} m^{2k+1} t^{n-1}[/tex]

Resulting in a slope of

[tex]\frac{dx}{dy} = \frac{\Sigma \left( \begin{array}{c} n \\ 2k \end{array} \right) (-1)^k m^{2k}}{\Sigma \left( \begin{array}{c} n \\ 2k+1 \end{array} \right) (-1)^{k} m^{2k+1} }[/tex]

Does this look right? And if it does, does this relate to m in any obvious way?
 
  • #6
Yeow. Nice try. But you are taking the long way around! Write the line in polar notation r*exp(i*theta) where r is variable and theta is the constant angle.
 
  • #7
Hahaha, yes, it should have occurred to me that I am in complex analysis and not multivariable calculus :)

Let's see. In this case g(r) = r exp(i theta), and f(g(r)) = r^n exp(i n theta). Now my angle is n*theta. Makes sense.

By pre- and post-composing with translations, I can WLOG assume that I am interested in a pair of lines through the origin, and that f(0) = 0. But why do higher-order terms not impact the angle?
 
  • #8
The angles they are talking about are the tangent angles of curves through the point z0. You really just need the angle in the limit as z->z0. You can make the higher order terms as small as you want relative to the first term. The higher order terms will make lines transform into curves away from z=z0, but they won't affect the tangent angle.
 
  • #9
It makes sense intuitively. I am still struggling to come up with a rigorous argument why higher-order terms don't matter - I guess I am not sure how tangent angles are defined in complex-analysis-land to begin with. But at least now I get the idea for the proof. I can probably figure the rest out on my own.

Thank you very much for your help!
 
  • #10
I wouldn't worry about making this overly rigorous. The result isn't THAT interesting or fundamental. I would feel free to do some hand waving, as long as you communicate the idea. That's about the only interesting part.
 

1. What are hoolomorphic functions?

Hoolomorphic functions, also known as complex analytic functions, are functions that are defined and differentiable on a complex domain. These functions have a special property that their derivatives exist at every point within the domain.

2. What is the significance of stretching angles in hoolomorphic functions?

Stretching angles in hoolomorphic functions refers to the transformation of the complex plane under the mapping of a hoolomorphic function. This is significant because it allows us to visualize and understand the behavior of these functions in a geometric way.

3. How do hoolomorphic functions behave under stretching angles?

Hoolomorphic functions behave in a predictable manner under stretching angles. They preserve angles and shapes, but can stretch or shrink them depending on the function. This property allows us to use stretching angles to analyze and manipulate hoolomorphic functions.

4. Can hoolomorphic functions have multiple stretching angles?

Yes, hoolomorphic functions can have multiple stretching angles depending on the function and the chosen domain. These angles can interact with each other to produce complex mappings of the complex plane.

5. What real-world applications do hoolomorphic functions with stretching angles have?

Hoolomorphic functions with stretching angles have various applications in physics, engineering, and mathematics. They are used to model and analyze complex systems, such as fluid dynamics, electromagnetic fields, and quantum mechanics. They also have applications in signal processing, image processing, and computer graphics.

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