Question 476992
There's a little problem with your statement:


Let z = a + bi. 

Then z - 3 = (a- 3) + bi.
==> |z - 3| = |(a- 3) + bi| = {{{sqrt((a-3)^2 + b^2)}}} = 3, by the given.

==> {{{(a-3)^2 + b^2 = 3^2}}}
==> {{{a^2 - 6a + 9 + b^2 = 9}}} ==> {{{a^2 + b^2 = 6a}}}.

Now {{{(z-6)/z = ((a-6) + bi)/(a+bi) = (((a-6) + bi)/(a+bi))*((a-bi)/(a-bi))}}}

={{{(a(a-6)-(a-6)bi + abi - b^2*i^2)/(a^2 + b^2) = (a(a-6) + 6abi + b^2)/(6a)}}}

={{{(a^2 - 6a + 6abi + b^2)/(6a) = (6a -6a +6abi)/(6a) = (6abi)/(6a) = bi = Im(z)*i}}}

Now {{{arg(z) = tan^(-1)(b/a)}}}, which gives {{{tan(arg(z)) = b/a}}}, or 
{{{b = a*tan(arg(z))}}}, hence 
{{{(z-6)/z = a*i*tan(arg(z)) = Re(z)*i*tan(arg(z))}}}.
Check again the statement of your result.