Question

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`?

Answer 1

`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)`