Holomorphic functions at stationary points

  • Thread starter ayae
  • Start date
  • #1
ayae
20
0
Recently I have been self teaching myself complex analysis.

I am interested in the conformal mapping property of holomorphic functions and why and how it breaks down at stationary points.

Could anyone suggest further reading for this or shed some light on the subject.

Many thanks,
Ayae
 

Answers and Replies

  • #2
henry_m
160
2
An intuitive explanation for why holomorphic implies conformal can be seen from the power series expansion of a function.

A complex function, differentiable at 0 can be expressed as a power series in some finite circle around the origin: [itex]f(z)=f(0)+f'(0)z+f''(0)\frac{z^2}{2!}+\cdots[/itex]. Sufficiently near the origin, we can ignore the higher powers of z. So this is approximately just a (affine) linear function, as long as [itex]f'(0)\neq0[/itex]: a multiplication by a fixed complex number, which is just a rotation (by the argument) and dilation (by the modulus), followed by a translation. This is clearly conformal (and conversely every conformal transformation must locally look like this).

On the other hand, if f'(0) vanishes, the higher order terms are dominant so there is an additional power of z. So, for example, if the first nonvanishing derivative is the second, the function locally looks like [itex]z\mapsto z^2 [/itex] (followed by rotation, dilation & translation). You just have to look at the polar form to see that this doubles angles at the origin so it isn't conformal.

Locally, any complex differentiable function looks like a power, so to understand the local behaviour you need only understand z, z2, z3 etc. This is all just heuristic, but hopefully it will help to understand why it works.
 
  • #3
ayae
20
0
Thanks that really made sense. :approve:

You just have to look at the polar form to see that this doubles angles at the origin so it isn't conformal.
One thing; does it not half the angles?
[tex]e^{i\theta}\rightarrow e^{2i\theta}[/tex]

http://www.wolframalpha.com/input/?i={Re[%28x%2Biy%29]%3D0%2C+Im[%28x%2Biy%29]%3D0}
http://www.wolframalpha.com/input/?i={Re[%28x%2Biy%29^2]%3D0%2C+Im[%28x%2Biy%29^2]%3D0}
 
  • #4
henry_m
160
2
One thing; does it not half the angles?
[tex]e^{i\theta}\rightarrow e^{2i\theta}[/tex]
I think you've got the direction of the map the wrong way round; a curve [itex]\gamma[/itex] gets mapped to [itex]\{z^2|z\in \gamma\}[/itex], not [itex]\{z|z^2\in \gamma\}[/itex]. Think about where each of the points goes after the function is applied to it. For example, the ray [itex]iy[/itex] with [itex]y>0[/itex]; after getting squared each of the points ends up on the negative real axis, doubling the ray's angle with the positive real axis. Hope that makes sense.
 

Suggested for: Holomorphic functions at stationary points

  • Last Post
Replies
3
Views
575
  • Last Post
Replies
8
Views
856
  • Last Post
Replies
4
Views
463
  • Last Post
Replies
7
Views
371
  • Last Post
Replies
2
Views
277
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
5
Views
808
  • Last Post
Replies
8
Views
958
  • Last Post
Replies
4
Views
2K
Top