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Suppose #w=(z+1)/(z-1)# where #z=a+bi#, #a, b in RR#. If #|z|=1#, can you find #Re(w)#?

i.e. can you find the real part of #w#?

1 Answer
Nov 22, 2016

Answer:

#Re[w]=0#

Explanation:

Given a complex number #w# then

#Re[w]=(w+bar(w))/2# so

#w+barw = (z+1)/(z-1)+(bar z+1)/(barz-1)=#
# = ((z+1)(barz-1)+(z-1)(barz+1))/((z-1)(barz-1))=#
#2(z barz-1)/(z barz-z-barz+1)=2(a^2+b^2-1)/(a^2+b^2-2a+1)#

and

#Re[w]=(a^2+b^2-1)/(a^2+b^2-2a+1)#

but #a^2+b^2=1# so

#Re[w]=0#

NOTE:

If #w = (a+ib+1)/(a+ib-1) = ((a+ib+1)(a-ib-1))/((a+ib-1)(a-ib-1)) = (a^2+b^2-1-2ib)/((a-1)^2+b^2)#

analogously

#bar w = (a-ib+1)/(a-ib-1) = (a^2+b^2-1+2ib)/((a-1)^2+b^2)#