Question Number 215417 by MATHEMATICSAM last updated on 06/Jan/25
$${a}\:\mathrm{and}\:{b}\:\mathrm{are}\:\mathrm{complex}\:\mathrm{numbers}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mid{b}\mid\:=\:\mathrm{1}.\:\mathrm{Find}\:\mid\frac{{b}\:−\:{a}}{\mathrm{1}\:−\:\overline {{a}b}}\mid \\ $$
Answered by MrGaster last updated on 06/Jan/25
$$\mathrm{Let}\:{a}={x}+{yi},{b}=\mathrm{cos}\theta+{i}\mathrm{sin}\theta \\ $$$${b}−{a}=\left(\mathrm{cos}\theta−{x}\right)+{i}\left(\mathrm{sin}\theta−{y}\right) \\ $$$$\mathrm{1}=\bar {{a}b}=\mathrm{1}−\left({x}+{iy}\right)\left(\mathrm{cos}\:\theta−{i}\:\mathrm{sin}\:\theta\right) \\ $$$$=\mathrm{1}−\left({x}\:\mathrm{cos}\theta+{y}\:\mathrm{sin}\:\theta\right)−{i}\left({x}\:\mathrm{sin}\theta−{y}\:\mathrm{cos}\:\theta\right) \\ $$$$\mid{b}−{a}\mid=\sqrt{\left(\mathrm{cos}\theta−{x}\right)^{\mathrm{2}} +\left(\mathrm{sin}\theta−{y}\right)^{\mathrm{2}} } \\ $$$$\mid\mathrm{1}−\bar {{a}b}\mid=\sqrt{\left.\mathrm{1}−{x}\:\mathrm{cos}\theta−{y}\:\mathrm{sin}\theta\right)^{\mathrm{2}} +\left({x}\:\mathrm{sin}\:\theta−{y}\:\mathrm{cos}\:\theta\right)^{\mathrm{2}} } \\ $$$$\mid\frac{{b}−{a}}{\mathrm{1}−\bar {{a}b}}\mid=\frac{\sqrt{\left(\mathrm{cos}\theta−{x}\right)^{\mathrm{2}} +\left(\mathrm{sin}\theta−{y}\right)^{\mathrm{2}} }}{\:\sqrt{\left(\mathrm{1}−{x}\:\mathrm{cos}\theta−{y}\:\mathrm{sin}\theta\right)^{\mathrm{2}} +\left({x}\:\mathrm{sin}\:\theta−{y}\:\mathrm{cos}\theta\right)^{\mathrm{2}} }} \\ $$$$=\mathrm{1} \\ $$$$\mid\frac{{b}−{a}}{\mathrm{1}−\bar {{a}b}}\mid=\mathrm{1}\:\checkmark \\ $$