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I know that:

$$|z| = |w| = p$$

$$4z^2 + 5w^2 = azw ,\qquad a \in R$$

I need to prove that $4w^2 + 5z^2 = azw$

How I solved it is:

$$|z| = p \implies |z|^2 = p^2 \implies zz^* = p^2$$

then solved as $z$ and replaced it to the original equation.

Then I wanted to prove that:

$$4z^2 + 5w^2 = 4w^2 + 5z^2$$ and find that $0 = 0$.

But this covers more than 5 pages and it contains lots of math, so what is an easier approach?

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Please...!!! Use Latex to write mathematics in this site... – DonAntonio Oct 28 '12 at 14:17
I think if $4z^2+5w^2=4w^2+5z^2$, then you have $4(z^2-w^2)=5(z^2-w^2)$, which means $z^2-w^2=0$, so either $z=w$ or $z=-w$. Am I right? – user46090 Oct 28 '12 at 14:17
@DonAntonio I am not very good at computers so I apologize – Chris Oct 28 '12 at 14:18
@user46090 okay but how can i prove the equation? – Chris Oct 28 '12 at 14:21
but why is it not correct? That's what i have to prove – Chris Oct 28 '12 at 14:30

One assumes that $|z|=|w|$ and $4z^2+5w^2=azw$ with $a$ real and one wants to prove that $z^2=w^2$. The proof is as follows:

  • If $w=0$ then $z=0$ hence $z^2=w^2$. Done.
  • Otherwise $u=z/w$ solves $|u|=1$ and $4u^2-au+5=0$.
    • If $u$ is real, $|u|=1$ implies $u=\pm1$ hence $u^2=1$. Done.
    • Otherwise $4u^2-au+5$ has one non real root, hence two non real conjugate roots, hence the roots are $u$ and $\bar u$. Their product $u\cdot\bar u=|u|^2=1$ equals $5/4$. Impossible.
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Nice. One really needs the full premise including $azw$. – Hagen von Eitzen Oct 28 '12 at 14:39
what is the u variable? – Chris Oct 28 '12 at 14:50
I don't understand, did we prove that 4z^2 + 5w^2 = 4w^2 + 5z^2 ? – Chris Oct 28 '12 at 15:00
what is the u variable?... u=z/w when w is not zero. Isn't this written in the post? – Did Oct 28 '12 at 15:25
did we prove that... Indeed we did. Quote: One assumes that $|z|=|w|$ and $4z^2+5w^2=azw$ with $a$ real and one wants to prove that $z^2=w^2$. Unquote. Thus, we proved that, if $|z|=|w|$ and $4z^2+5w^2=azw$ with $a$ real, then $z^2=w^2$ (which is equivalent to $4z^2+5w^2=5z^2+4w^2$). – Did Oct 28 '12 at 15:28

This's strongly inspired by the other answer.

Let's denote $\cos C+i\sin C=cis C$

If $z=p(cis A),w=p(cis B)$ with $p\ne 0$

So, $\frac z w=cis(A-B),4cis 2(A-B)+acis(A-B)+5=0$

Equating real & imaginary parts(using the fact that $a$ is real ),


and $4\sin2(A-B)+a\sin(A-B)=--->(2)$

$(2)\implies \sin(A-B)\{8\cos(A-B)+a\}=0$

If $\cos(A-B)=-\frac a 8,$

from (1) we get, $4(2\{-\frac a 8\}^2-1)+a\{-\frac a 8\}+5=0$

or, $-4+5= 0$, which is impossible.

So, $\sin(A-B)=0\implies A=n\pi+B$ where $n$ is any integer.

$\cos A=\cos(n\pi+B)=(-1)^n\cos B $ and $\sin A=\sin(n\pi+B)=(-1)^n\sin B $

So, $z=pcis A=p(-1)^n cis B=(-1)^nw\implies z^2=(-1)^{2n}w^2=w^2$

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What happened to the first sentence of your post? :-) – Did Oct 29 '12 at 8:04
Really? If one does not use the hypothesis that $a$ is real, the result is false. – Did Oct 29 '12 at 8:16
Sorry to insist but you assume that $a$ is real as soon as you want to Equat(e) real & imaginary parts. – Did Oct 29 '12 at 8:51
@did, I've edited the answer. – lab bhattacharjee Oct 29 '12 at 8:59

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