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Suppose the product of two complex numbers $z_1$ and $z_2$ is real and not zero. Prove there exists a real number $p$ such that $$$$ $z_1 = p \overline z_2 $. $$$$

My solution thus far:

Using the equivalence $z \in \Bbb R \iff z = \overline z. $

Assuming $z_1 . z_2 \in \Bbb R$. Then $z_1 z_2 = \overline{ z_1z_2 }$. As $z_2 \ne 0$. Then $\frac{z_1}{\overline z_2}$ = $\frac{\overline z_1}{z_2}$ = $\overline (\frac{z_1}{\overline z_2})$, i.e. $\frac{z_1}{\overline z_2}$ equals its conjugate. It follows that $\frac{z_1}{\overline z_2} = p \in \Bbb R$.

If I then rearrange would this proof suffice? Is this the right approach? I'm still getting to grips with the proofs needed for this course and would appreciate some feedback. (The $\overline (\frac{z_1}{\overline z_2})$ part is supposed to have a bar above but \overline seems to have cut part of it off).

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  • $\begingroup$ It looks just fine to me. $\endgroup$ – DonAntonio Mar 13 '17 at 13:51
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What you did is OK.

To transform it to a proper proof, I would say, assume that $z_1 z_2 = k \in \mathbb R$. Then $$z_1 z_2 \bar{z_2} = z_1 \vert z_2 \vert^2 = k \bar{z_2}$$ Also $\vert z_2 \vert^2 \neq 0$ as by hypothesis $z_2 \neq 0$. Dividing both sides of the equality by $\vert z_2 \vert^2$ we get $$z_1= \frac{k}{\vert z_2 \vert^2 } \bar{z_2}$$ as desired with $p=\frac{k}{\vert z_2 \vert^2 }$.

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  • $\begingroup$ I feel like I might kick myself after your response to this but would you mind explaining how you went from $z_1 z_2 = k \in \Bbb R$ to your next line? EDIT: Or better perhaps, what you substituted that into? $\endgroup$ – B Taylor Mar 13 '17 at 14:16
  • $\begingroup$ @BTaylor You just have to multiply both sides of the equation $z_1 z_2 = k \in \mathbb R$ by $\bar{z_2}$ $\endgroup$ – mathcounterexamples.net Mar 13 '17 at 14:20

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