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# good morning everyone!

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

Jan 23, 2018

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

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

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

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Jan 23, 2018

#1
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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 $$a^2 + b^2 = \frac{( p^2 + 1 ) }{4q^2 + 1}$$

.
Jan 23, 2018
#2
+97497
+2

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

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

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

Melody Jan 23, 2018
#5
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What happend to the i?

PollyFAnna  Jan 23, 2018
edited by Guest  Jan 24, 2018
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Very nice, Melody  !!!

Jan 23, 2018
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Thanks Chris :)

Melody  Jan 23, 2018