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?

3$\begingroup$ Depends if you allow conformal maps to change the orientation. In any case, you get either just holomorphic or both holomorphic and antiholomorphic 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. $\endgroup$ – Misha Apr 4 '12 at 4:45

5$\begingroup$ 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 CauchyRiemann equations, otherwise the CauchyRiemann 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 $C^1$.) $\endgroup$ – Michael Greenblatt Apr 4 '12 at 4:57
$\begingroup$
$\endgroup$
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.