Question 1116382
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        Surely, the condition MUST be rewritten in this form

<pre>
          If {{{(z-1)/(z+1)}}} = ki,   <U>where k is a real number</U>,  then show that |z| = 1.
</pre>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;to be correct  &nbsp;&nbsp;(adding that k is a real number).



<pre>
Let  z = a + bi.

We are given  {{{(z-1)/(z+1)}}} = ki,  which means that

{{{(a+bi-1)/(a+bi+1)}}} = ki.


Left side is

{{{(a+bi-1)/(a+bi+1)}}} = {{{(a+bi-1)/(a+bi+1)}}}.{{{(a-bi+1)/(a-bi+1)}}} = {{{(((a-1)+bi)*((a+1)-bi))/((a+1)^2+b^2)}}}.


The denominator is now a real number.


The numerator is  (a-1)*(a+1) + bi*(a+1) - bi*(a-1) + b^2.


Since the ratio  {{{Num/Den}}} is purely imaginary number ki,  it means that the real part of the numerator is zero:


    (a-1)*(a+1) + b^2 = 0,   or

    a^2 - 1 + b^2 = 0,  which is equivalent to

    a^2 + b^2 = 1.


    The last equality precisely means that  |z| = a^2 + b^2 = 1,   QED.
</pre>

Solved.