good morning everyone!

if a + bi =  p + i / 2q - i , prove that a^2 + b^2 = ( p^2 + 1 ) /4q^2 + 1.

$a + bi = \frac{ p + i }{2q - i}\\ a + bi = \frac{ p + i }{2q - i}\times \frac{2q+i}{2q+i}\\ a + bi = \frac{2pq +(p+2q)i-1}{4q^2 +1}\\ a + bi = \frac{2pq-1 }{4q^2 +1}+\frac{(p+2q)i}{4q^2 +1}\\ so\\ a = \frac{2pq-1 }{4q^2+1}\qquad and \qquad b = \frac{(p+2q)}{4q^2 +1}\\~\\ a^2+b^2=\frac{(2pq-1)^2+ (p+2q)^2}{(4q^2+1)^2}$

$a^2+b^2=\frac{(2pq-1)^2+ (p+2q)^2}{(4q^2+1)^2}\\ a^2+b^2=\frac{(4p^2q^2-4pq+1)+ (p^2+4pq+4q^2)}{(4q^2+1)^2}\\ a^2+b^2=\frac{4p^2q^2+p^2+4q^2+1}{(4q^2+1)^2}\\ a^2+b^2=\frac{(4q^2+1)(p^2+1)}{(4q^2+1)^2}\\ a^2+b^2=\frac{p^2+1}{4q^2+1}\\~\\ QED$

Hi Rosala,

a + bi =  p + i / 2q - i , prove that a^2 + b^2 = ( p^2 + 1 ) /4q^2 + 1.

The accurate interpretation of this is

$a + bi =  p +\frac{ i }{ 2q} - i$ , prove that    $a^2 + b^2 =\frac{ ( p^2 + 1 ) }{4q^2} + 1$

BUT

Do you mean this one underneath, or something differnent again?

if $a + bi =  \frac{p + i }{ 2q - i }$, prove that $a^2 + b^2 = \frac{( p^2 + 1 ) }{4q^2 + 1}$

Very nice, Melody  !!!