Question 1184217
If {{{ (z-1)/(z+1) }}} is purely imaginary, then {{{ (z-1)/(z+1) = Ai}}} for some {{{A<>0}}}.

Let z = a + bi.
===> z - 1 = A(z+1)i  ===> a + bi - 1 = A(a + bi + 1)i

===> a + bi - 1 = A((a + 1)i - b) = A(a+1)i - Ab


<===> (a - 1 + Ab) + (b - A(a+1))i = 0

===> a - 1 + Ab = 0  and b - A(a+1) = 0, or 

===> a + Ab = 1 and -Aa + b = A.

===> {{{a = (1-A^2)/(1+A^2)}}} and  {{{b = (2A)/(1+A^2)}}} after solving the system for a and b.

===> {{{z  = a+bi =  (1-A^2)/(1+A^2) + i*((2A)/(1+A^2))}}}

===> {{{abs(z) = sqrt( ((1-A^2)/(1+A^2))^2 + ((2A)/(1+A^2))^2)}}}


={{{sqrt( ((1-2A^2 + A^4)/(1+A^2)^2 + (4A^2)/(1+A^2)^2)) = sqrt ((1+2A^2 + A^4)/(1+A^2)^2 ) = 1}}}


Therefore, {{{highlight(abs(z) = 1)}}}