If #alpha=r(costheta+isintheta)#, #beta=cosdelta+isindelta# and #z_1=beta-alpha# and #z_2=1-alphabeta#, find #(z_1)/(z_2)#?

1 Answer
Feb 10, 2018

#|(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)#

Explanation:

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)#