Melody

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

The digits have to add up to a multiple of 3

The biggest one will be 9999 

Lets see

There are loads of patterns happening here - can you find them all ??      

1221             2

1551             5

1881             8

2112             1

2442             4

2772             7

3003             0

3333             3

3663             6

3993             9

4224             2

4554

4884

5115             1

5445

5775

6006              0

6336             

6666

6996 

7227             2

7557

7887 

8118              1

8448

8778

9009              0

9339

9669

9999

6*3 + 3*4 = 18+12 = 30

There are 30 four digit  palindromes  

Proper numbers only, I didn't include any begining with 0 but if i had there would be 4 or those.

Let \( a_1,a_2,a_3,\dots \)   be an arithmetic sequence. If  \(\dfrac{a_4}{a_2} = 3\), what is  \(\dfrac{a_5}{a_3}\) ?

\(d=(a_2-a_1)\\ so\\ a_1=a_1\\\ a_2=a_1+ (a_2-a_1)\\ a_3=a_1+ 2(a_2-a_1)\\ a_4=a_1+ 3(a_2-a_1)\\ a_5=a_1+ 4(a_2-a_1)\\~\\ \)

\(\dfrac{a_1+ 3(a_2-a_1)}{a_1+ (a_2-a_1)}=3\qquad find \qquad \dfrac{a_1+ 4(a_2-a_1)}{a_1+ 2(a_2-a_1)}\\ \dfrac{a_1+ 3(a_2-a_1)}{a_1+ (a_2-a_1)}=3\\ a_1+ 3(a_2-a_1)=3(a_1+ (a_2-a_1))\\ a_1+ 3(a_2-a_1)=3a_1+ 3(a_2-a_1)\\ a_1=3a_1\\ a_1=0\\ a_2=d\)

so

\(\dfrac{a_5}{a_3}=\dfrac{a+4d}{a+2d}=\dfrac{4a_2}{2a_2}=2\)

Am I allowed to answer??

I just realized that you posted this question twice in quick succession.

PLEASE DO NOT DO THAT!

If your question has been there a long time then you can make a knew post that directs people to the original post.

Just copy and post the address of the original post in with a short explanation.

A long time is subjective but there should be either a day or around 20 questions between the two . 

the latus rec tum of parabola with equation (x-1)^2 = -16(y-4) coincides with the diameter of a circle. find equation of circle?

Consider the parabola

it is of the from     \((x-h)^2=4a(y-k)\)

Vertex  \((h,k) = (1,4)\)

Focal length = a = -4

So it is a concave down parabola with focus point  x=1 and y=4-4=0     focus (1,0)

The latus r****m is 16 units across and that is the diameter of the circle.

So the radius of the circle is 8 units and the centre of the circle is (1,0)

Circle:

\((x-1)^2+y^2=64\)

Hi    TheXSquaredFactor,

That is really cool.  I have never seen that trick before!