Prove that $4w^2 + 5z^2 = 4z^2 + 5w^2$ if $|z| = |w|$ and $4z^2+5w^2=azw$ 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?
 A: 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.


A: 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 ), 
$4\cos2(A-B)+a\cos(A-B)+5=0--->(1)$
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, $z=pcis A=p(-1)^n cis B=(-1)^nw\implies z^2=(-1)^{2n}w^2=w^2$
