Question 976172
COMPLEX NUMBER
If {{{z = a+bi}}} and {{{abs((z-i)/(z+2+i))=3}}}, show that {{{8a^2 + 8b^2 + 36a +20b +44 =0}}}

<pre>
{{{abs((z-i)/(z+2+i))}}}{{{""=""}}}{{{3}}}

{{{abs( ((a+bi)-i)/((a+bi)+2+i))}}}{{{""=""}}}{{{3}}}

{{{abs((a+bi-i)/(a+bi+2+i))}}}{{{""=""}}}{{{3}}}

{{{abs((a+(bi-i))/(a+2+(bi+i)))}}}{{{""=""}}}{{{3}}}

{{{abs((a+(b-1)i)/((a+2)+(b+1)i))}}}{{{""=""}}}{{{3}}}

{{{sqrt(a^2+(b-1)^2)/sqrt((a+2)^2+(b+1)^2)}}}{{{""=""}}}{{{3}}}

Square both sides:

{{{(a^2+(b-1)^2)/((a+2)^2+(b+1)^2)}}}{{{""=""}}}{{{9}}}

{{{a^2+(b-1)^2}}}{{{""=""}}}{{{9((a+2)^2+(b+1)^2)}}}

{{{a^2+(b-1)^2}}}{{{""=""}}}{{{9(a+2)^2+9(b+1)^2)}}}

{{{a^2-9(a+2)^2}}}{{{""=""}}}{{{9(b+1)^2-(b-1)^2}}}

Both sides are differences of squares, so factor:

{{{(a-3(a+2)^"")(a+3(a+2)^"")}}}{{{""=""}}}{{{(3(b+1)^""-(b-1)^"")(3(b+1)^""+(b-1)^"")}}}

{{{(a-3(a+2))(a+3(a+2))}}}{{{""=""}}}{{{(3(b+1)-(b-1))(3(b+1)+(b-1))}}}

{{{(a-3a-6)(a+3a+6)}}}{{{""=""}}}{{{(3b+3-b+1)(3b+3+b-1)}}}

{{{(-2a-6)(4a+6)}}}{{{""=""}}}{{{(2b+4)(4b+2)}}}

{{{(-2(a+3)^"")(2(2a+3)^"")}}}{{{""=""}}}{{{(2(b+2)^"")(2(2b+1)^"")}}}

{{{-4(a+3)(2a+3)}}}{{{""=""}}}{{{4(b+2)(2b+1)}}}

{{{-(a+3)(2a+3)}}}{{{""=""}}}{{{(b+2)(2b+1)}}}

{{{-(2a^2+9a+9)}}}{{{""=""}}}{{{2b^2+5b+2}}}

{{{-2a^2-9a-9}}}{{{""=""}}}{{{2b^2+5b+2}}}

{{{-2a^2-9a-2b^2-5b-11}}}{{{""=""}}}{{{0}}}

{{{2a^2+9a+2b^2+5b+11}}}{{{""=""}}}{{{0}}}

If you swap 2nd and 3rd terms and multiply through by 4, you get 

{{{8a^2 + 8b^2 + 36a +20b +44}}}{{{""=""}}}{{{0}}}

Edwin</pre>