Let \(\HH_R\coloneqq\ts{z\in\CC \st \Re z>0}\) be the right half-plane and \(\DD\coloneqq\ts{w\in\CC \st \abs{w}<1}\) the unit disc. The map \[ \varphi\colon\HH_R\to\DD,\qquad \varphi(z)=\frac{1-z}{1+z}, \] is a biholomorphism, with inverse \(\varphi^{-1}\colon\DD\to\HH_R\), \(\varphi^{-1}(w)=\frac{1-w}{1+w}\).
For \(z\in\HH_R\) we have \(1+z\neq0\), and \[ \abs{1-z}^2-\abs{1+z}^2=-4\Re z<0, \] so \(z\) is closer to \(1\) than to \(-1\) and \(\abs{\varphi(z)}<1\). Hence \(\varphi\) is a holomorphic map \(\HH_R\to\DD\). The same map is the composite of the rotation \(z\mapsto iz\), which maps \(\HH_R\) onto the upper half-plane \(\HH=\ts{u\in\CC\st\Im u>0}\), with the Cayley map \(u\mapsto\frac{i-u}{i+u}\) from \(\HH\) onto \(\DD\): \[ \frac{i-iz}{i+iz}=\frac{1-z}{1+z}. \]
For \(w\in\DD\) we have \(w\neq-1\), so \(\psi(w)\coloneqq\frac{1-w}{1+w}\) is holomorphic on \(\DD\), and \[ \Re\psi(w)=\frac{\Re\big((1-w)(1+\bar w)\big)}{\abs{1+w}^2}=\frac{1-\abs{w}^2}{\abs{1+w}^2}>0, \] so \(\psi\) maps \(\DD\) into \(\HH_R\). If \(w=\varphi(z)\), then \(w+wz=1-z\), so \(z(1+w)=1-w\) and \(z=\psi(w)\); thus \(\psi\circ\varphi=\id_{\HH_R}\). The same computation with the roles of \(z\) and \(w\) exchanged gives \(\varphi\circ\psi=\id_{\DD}\).
The formula \(\frac{1-z}{1+z}\) also maps the imaginary axis into the unit circle: for \(t\in\RR\), the numbers \(1-it\) and \(1+it\) are complex conjugates, so \(\abs{\frac{1-it}{1+it}}=1\).