+0  
 
0
454
5
avatar+11852 

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+94105 
+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
 #1
avatar+94105 
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+94105 
+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+92674 
+1

Very nice, Melody  !!!

 

 

cool cool cool

CPhill  Jan 23, 2018
 #4
avatar+94105 
+1

Thanks Chris :)

Melody  Jan 23, 2018

7 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.