+0

# good morning everyone!

0
199
5
+11847

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

rosala  Jan 23, 2018

#2
+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
Sort:

#1
+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
+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
#5
-15
0

What happend to the i?

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

Very nice, Melody  !!!

CPhill  Jan 23, 2018
#4
+92164
+1

Thanks Chris :)

Melody  Jan 23, 2018

### 27 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details