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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.

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    $\begingroup$ Is z* the conjugate of z? $\endgroup$
    – Ilovemath
    Sep 4, 2021 at 3:41
  • $\begingroup$ @Ilovemath yes it is $\endgroup$
    – X-men
    Sep 4, 2021 at 3:44
  • $\begingroup$ Use the fact that if $z=x+iy, zz^*=x^2+y^2$ and is real. Also the things you call equations are not that as they have no equal sign. They are expressions. You are claiming correctly that they are all equal, but do not say so. In 3 only the second and fourth terms are imaginary, so set their sum to $0$ and see where it leads. $\endgroup$ Sep 4, 2021 at 3:44
  • $\begingroup$ If $9z + \frac{2}{z}$ is real, then $9z + \frac{2}{z} = (9z + \frac{2}{z})^* = 9z^* + \frac{2}{z^*}$. A big hint is to multiply through by $(zz^*)^2$ $\endgroup$
    – Sasha
    Sep 4, 2021 at 3:50

3 Answers 3

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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.

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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 $$zz^*=\frac{2}{9}$$

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  • $\begingroup$ Hi ! I see where you are coming from. By forming an equation with the fact that Im equates to 0 since it is a real number. But for line 2, since y is non-zero, why is it that we can remove y from the equation? $\endgroup$
    – X-men
    Sep 4, 2021 at 4:00
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    $\begingroup$ If $r,s,t$ are real numbers, with $r \neq 0$, and if $r(s-t) = 0$, then either $r = 0$ or $(s - t) = 0$. With the premise that $r \neq 0$, you can conclude that $(s - t) = 0$. $\endgroup$ Sep 4, 2021 at 4:03
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    $\begingroup$ In line 2, $y$ can't be factored out from all the terms to give something of the type $y$.(something free from $y$). So, there is no benefit in removing $y$ from line 2. In line 3, as you can see, $y$ can be factored out from all the terms. $\endgroup$
    – Ilovemath
    Sep 4, 2021 at 4:05
  • $\begingroup$ thank you everyone ! $\endgroup$
    – X-men
    Sep 4, 2021 at 4:18
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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 \ \ . $

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