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

 

thanks in advance!smiley

rosala  Jan 23, 2018

Best Answer 

 #2
avatar+92164 
+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
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 #1
avatar+92164 
0

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

Melody  Jan 23, 2018
 #2
avatar+92164 
+2
Best Answer

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
avatar-15 
0

What happend to the i?

PollyFAnna  Jan 23, 2018
edited by Guest  Jan 24, 2018
 #3
avatar+85596 
+1

Very nice, Melody  !!!

 

 

cool cool cool

CPhill  Jan 23, 2018
 #4
avatar+92164 
+1

Thanks Chris :)

Melody  Jan 23, 2018

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