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. The biholomorphisms \(\Omega\to\Omega\) form a group under composition, the automorphism group \(\Aut_\CC(\Omega)\).
Let \(\Omega\subseteq\CC\) be open and let \(f\) be conformal on \(\Omega\).
\(f\) preserves signed angles: if \(\gamma_1,\gamma_2\colon(-1,1)\to\Omega\) are \(C^1\) curves with \(\gamma_1(0)=\gamma_2(0)=z_0\) and nonzero tangents \(\gamma_1'(0),\gamma_2'(0)\), then the signed angle from \((f\circ\gamma_1)'(0)\) to \((f\circ\gamma_2)'(0)\) equals the signed angle from \(\gamma_1'(0)\) to \(\gamma_2'(0)\).
If \(U,V\subseteq\CC\) are open and \(f\colon U\to V\) is a holomorphic bijection, then \(f'\) has no zeros and \(f^{-1}\colon V\to U\) is holomorphic and conformal. In particular, the inverse of a biholomorphism is a biholomorphism.
By the chain rule \((f\circ\gamma_j)'(0)=f'(z_0)\,\gamma_j'(0)\). Multiplication by \(f'(z_0)=\abs{f'(z_0)}e^{i\alpha}\neq0\) scales by \(\abs{f'(z_0)}>0\) and rotates by \(\alpha\), adding \(\alpha\) to the argument of both tangent vectors, so the difference of their arguments is unchanged.
A holomorphic bijection \(U\to V\) is univalent, so by the proposition on that card \(f'\) has no zeros and \(f^{-1}\) is holomorphic. Differentiating \(f(f^{-1}(w))=w\) gives \((f^{-1})'(w)=1/f'(f^{-1}(w))\neq0\).