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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?

flag	asked Apr 4 2012 at 4:05 Grant Rotskoff 353●3●9

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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 2012 at 4:45

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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 2012 at 4:57

Amit · Accepted Answer · 2012-04-04 04:21:11Z UTC

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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.

answered Apr 4 2012 at 4:21

Amit
50●5

When are conformal maps holomorphic?

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