Complex number, p^2 + q^2

[Immortal]

Hai, I'm a student of +1 computer science.
In the extra questions section in my maths guide, I found this question and I cant get to the answer.
The question is

(i) Prove that (|(a+i)^2|)/(|2a-i|) =(a^2+1)/(sqrt(4a^2+1))
I proved this but the second section is what I cant prove,
(ii) If ((a+i)^2)/2a-i=p+iq, prove that p^2 +q^2=((a+1)^2)/4a^2+1

I cant prove this.
The modulus of the given expression was found in the first question
modulus of a complex number \(\displaystyle a+ib=sqrt(a^2+b^2) \)
We proved it is \(\displaystyle (a^2+1)/(sqrt(4a^2+1)) in the first section \)
So naturally,\(\displaystyle a^2 +b^2 \) means just remove the square root, \(\displaystyle → P^2 +Q^2 =(a^2+1)^2/(4a^2+1) \)
which is not equal to \(\displaystyle ((a+1)^2)/(4a^2+1) \)
How do I solve this?

 

[Ishuda]

What if a were minus 1?

 

[Immortal]

What if a were minus 1?

I didnt understand what you mean, but if a is -1, then
\(\displaystyle P^2+q^2= ((-1)^2+1^2)^2/(-4^2+1) \)
\(\displaystyle =4/15 \)

I think there is something wrong with the question

 

[Ishuda]

I didnt understand what you mean, but if a is -1, then
\(\displaystyle P^2+q^2= ((-1)^2+1^2)^2/(-4^2+1) \)
\(\displaystyle =4/15 \)

I think there is something wrong with the question

Yes, something is wrong. That's the answer according to what you got, what about using the formulas in (ii). That is substitute a = -1 in
\(\displaystyle \frac{(a+i)^2}{2a-i}\)
Obviously the magnitude of that number is not zero. Yet, for a = -1,
\(\displaystyle \frac{(a+1)^2}{4a^2+1} = 0\)