Let #alpha=r(costheta+isintheta)# and #beta=cosdelta+isindelta#. Observe that #|beta|=1#
Now as #|z_1/z_2|=|z_1|/|z_2|#, we have
#|(beta-alpha)/(1-alphabeta)|=|beta-alpha|/|1-alphabeta|#
Now #|beta-alpha|#
= #|cosdelta-rcostheta+i(sindelta-rsintheta)|#
= #sqrt{(cosdelta-rcostheta)^2+(sindelta-rsintheta)^2)#
= #sqrt(cos^2delta+r^2cos^2theta-2rcosdeltacostheta+sin^2delta+r^2sin^2theta-2rsindeltasintheta)#
= #sqrt(1+r^2-2rcos(delta-theta))#
and #|1-alphabeta|#
= #|(1-r(cosdelta+isindelta)(costheta+isintheta))|#
= #|1-r(cos(delta+theta)+isin(delta+theta)))|#
= #|(1-rcos(delta+theta))-irsin(delta+theta))|#
= #sqrt((1-rcos(delta+theta))^2+r^2sin^2(delta+theta))#
= #sqrt(1+r^2cos^2(delta+theta)-2rcos(delta+theta)+r^2sin^2(delta+theta))#
= #sqrt(1+r^2-2rcos(delta+theta))#
Hence #|(beta-alpha)/(1-alphabeta)|=sqrt((1+r^2-2rcos(delta-theta))/(1+r^2-2rcos(delta+theta)))#,
where #delta=arg(beta)# and #theta=arg(alpha)#