Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

It is a standard fact from elementary complex analysis that a holomorphic function $f:\mathbb{C}\to \mathbb{C}$ is a conformal mapping. Now, suppose I have a map $f':\mathbb{R}^2\to \mathbb{R}^2$ which is a conformal mapping of the plane onto itself. Write $$f'(x,y) = (f_1(x,y),f_2(x,y)).$$ Is $f_1 + if_2$ holomorphic?

share|improve this question
3  
Depends if you allow conformal maps to change the orientation. In any case, you get either just holomorphic or both holomorphic and anti-holomorphic maps. This is a part of any standard complex analysis course. By the way, in the "standard fact" you should also assume that $f$ is nonconstant. –  Misha Apr 4 '12 at 4:45
4  
The fact that it's conformal is equivalent to a condition on the matrix of first derivatives. If the determinant is nonneegative the condition is the same as the Cauchy-Riemann equations, otherwise the Cauchy-Riemann equations are off by a factor of -1. So if you stipulate the determinant of the Jacobian is nonnegative (orientation is preserved) then it implies holomorphicity (as long as $f$ is $C^1$.) –  Michael Greenblatt Apr 4 '12 at 4:57

1 Answer 1

up vote 1 down vote accepted

If we define $f' : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ by $f'(x,y) = (x, -y)$, then it is conformal, but the corresponding map $f_1 + i f_2$ is not holomorphic.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.