SOLUTION: Prove the proposition |z|=1 if {(1+z)/(1-z)} is purely imaginary. Please help.

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Question 1026729: Prove the proposition |z|=1 if {(1+z)/(1-z)} is purely imaginary. Please help.
Found 2 solutions by robertb, richard1234:
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
%281%2Bz%29%2F%281-z%29 purely imaginary implies that %281%2Bz%29%2F%281-z%29+=+i%2Aalpha for some real number alpha.
Now let z = a+ib for real numbers a, b.
==> %281%2B+a+%2B+ib%29%2F%281-a+-+ib%29+=+i%2Aalpha
<==> 1%2B+a+%2B+ib=%281-a+-+ib%29%2Ai%2Aalpha
==> 1%2B+a+%2B+ib=i%2Aalpha-a%2Ai%2Aalpha+-+ib%2Ai%2Aalpha
<==> 1%2B+a+%2B+ib=i%2Aalpha-a%2Ai%2Aalpha+%2B+b%2Aalpha
<==> %281%2Ba%29%2Bib+=+b%2Aalpha%2B+alpha%2A%281-a%29i
==> 1%2Ba=b%2Aalpha and b+=+alpha%2A%281-a%29.
These equations become a system of linear equations in a and b.
Solving for a and b gives
a+=+%28-1%2Balpha%5E2%29%2F%281%2Balpha%5E2%29 and b+=+%282alpha%29%2F%281%2Balpha%5E2%29.
Now .
=, and the statement is proved.





Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
Using robertb's solution above:

We have and . Instead of solving for a and b explicitly and checking that |z| = 1, all we need to do is show that the system implies .

Rearranging gives and . Squaring both equations and adding them gives

(note terms cancel out)