Let \(\Omega\subseteq\CC\) be open. A map \(f\colon\Omega\to\CC\) is conformal on \(\Omega\) if \(f\) is holomorphic on \(\Omega\) and \(f'(z)\neq0\) for every \(z\in\Omega\).
Let \(U,V\subseteq\CC\) be open. A bijective conformal map \(f\colon U\to V\) is a biholomorphism, and \(U\) and \(V\) are biholomorphic, or conformally equivalent, if a biholomorphism \(U\to V\) exists.
A conformal map preserves signed angles between \(C^1\) curves with nonzero tangent vectors, because its derivative at each point is multiplication by a nonzero complex number; see the proposition on D-PCDNH.