How to prove: When the product of two complex numbers is a real number, the complex numbers are proportional to each other's conjugates. [closed]

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Looking for a solution to the problem: Given $ z_1, z_2 \in \mathbb{C}$ and $\mathbf(z_1)\cdot(z_2) \in \mathbb{R}$, prove $ z_1 = p\cdot\overline z_2$ for any $ p\in\mathbb{R} $.

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2 Answers

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All you need to use is the equivalence $z \in \mathbb{R} \Longleftrightarrow z = \overline{z}$.

Assume $z_1 . z_2 \in \mathbb{R}$. Then $z_1z_2 = \overline{z_1z_2}$. Assume moreover ${z_2 \neq 0}$. 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 $\frac{z_1}{\overline{z_2}} = p \in \mathbb{R}$. If $z_2 = 0$, then $z_1 = p.\overline{z_2}$ if and only if $z_1=0$. If $z_1 \neq 0$, there is no solution for $p$.

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Since $|z|^2=z\bar{z}$, if $wz=r\in\mathbb{R}$, then $$ w|z|^2=wz\bar{z}=r\bar{z}\implies w=\frac{r}{|z|^2}\bar{z} $$ So $p=\dfrac{r}{|z|^2}$.

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