If $9z + 2/z$ is a real number, find the value of $zz^*$ I would like to seek some guidance with the following question.
Let $z$ be a complex number with $Im(z)$ not equal to $0$.
If $9z + 2/z$ is a real number, find the value of $zz^*$
My solutions are as follow
Suppose $z = x + iy$
I would have the following equation
$9(x+iy) + 2/z$
$9(x+iy) + 2(x-iy)/zz^*$
$9x + 9iy + 2x/zz* - 2iy/zz^*$
Since it is a real number, the Imaginary Part would equate to $0$.
Any guidance on how to proceed further after step $3$ would be greatly appreciated, broken down into simpler terms if possible.
I did attempt to remove $(9iy$ and $2iy/zz^*)$ since it equates to $0$. However i'm stuck afterwards.
 A: If $\,9z + 2/z=r \in \mathbb R\,$ then $\,z\,$ is a root of the equation $\,9z^2- r z + 2=0\,$. Since the coefficients are real and $\,z \in \mathbb C \setminus \mathbb R\,$, the other root must be its conjugate $\,z^*\,$, then the product of the roots is $\,zz^*=2/9\,$ by Vieta's formulas.
A: If imaginary part is $0$, we get the equation
$$9y- \frac{2y}{zz^*} =0$$ Now, as $y$ is non zero, we have
$$9=\frac{2}{zz^*}$$ which gives us $$zz^*=\frac{2}{9}$$
A: If you've worked with the "polar form" for complex numbers, you might also argue this way.  For $ \ z \ = \ r·e^{\ i·\theta} \ \ , \ \ \theta \neq 0 \ \ , $ we have
$$ 9·z \ + \ \frac{2}{z} \ \ = \ \ 9·(r·e^{\ i·\theta}) \ + \ \frac{2}{r·e^{\ i·\theta}} \ \ = \ \ (9· r)·e^{\ i·\theta}  \ + \ \left(\frac{2}{r} \right)·e^{\ -i·\theta}  $$ $$ = \ \ (9· r)·( \ \cos \theta \ + \ i·\sin \theta \ )  \ + \ \left(\frac{2}{r} \right)·( \ \cos [-\theta] \ + \ i·\sin [-\theta] \ )  $$
$$ = \ \ \left(9r \ + \ \frac{2}{r} \right)·    \cos \theta \ + \ i·9r·\sin \theta    \ - \ i·\left(\frac{2}{r} \right)· \sin  \theta \ \ .  $$
This is a real number for $ \ 9r      -  \left(\frac{2}{r} \right) \ = \ 0 \ \Rightarrow \ r^2 \ = \ z·z^{*} \ = \ \frac29 \ \ . $
