∣a∣=1|a| = 1∣a∣=1 から aaˉ=1a\bar{a} = 1aaˉ=1 ∣a−β1−aβˉ∣=(a−β)(a−β‾)(1−aβˉ)(1−aβˉ‾)=aaˉ−aβˉ−aˉβ+ββˉ1−aβˉ−aˉβ+aβˉaˉβ\left|\frac{a-\beta}{1-a\bar{\beta}}\right| = \sqrt{\frac{(a-\beta)(\overline{a-\beta})}{(1-a\bar{\beta})(1-\overline{a\bar{\beta}})}} = \sqrt{\frac{a\bar{a} - a\bar{\beta} - \bar{a}\beta + \beta\bar{\beta}}{1 - a\bar{\beta} - \bar{a}\beta + a\bar{\beta}\bar{a}\beta}}1−aβˉa−β=(1−aβˉ)(1−aβˉ)(a−β)(a−β)=1−aβˉ−aˉβ+aβˉaˉβaaˉ−aβˉ−aˉβ+ββˉ aaˉ=1a\bar{a} = 1aaˉ=1 を代入して =1−aβˉ−aˉβ+ββˉ1−aβˉ−aˉβ+ββˉ=1= \sqrt{\frac{1 - a\bar{\beta} - \bar{a}\beta + \beta\bar{\beta}}{1 - a\bar{\beta} - \bar{a}\beta + \beta\bar{\beta}}} = 1=1−aβˉ−aˉβ+ββˉ1−aβˉ−aˉβ+ββˉ=1