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Suppose that $z_{1}$ and $z_{2}$ are complex numbers, with $z_{1} z_{2}$ real and non-zero. Show that there exists a real number $r$ such that $z_{1}=r \bar{z}_{2}$.
in Mathematics by Gold Status (10,261 points) | 21 views

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Proof. Let $z_{1}=x_{1}+i y_{1}$ and $z_{2}=x_{2}+i y_{2}$ with $x_{1}, x_{2}, y_{1}, y_{2} \in \mathbb{R}$. Thus
$$
z_{1} z_{2}=x_{1} x_{2}-y_{1} y_{2}+\left(x_{1} y_{2}+y_{1} x_{2}\right) i
$$
Since $z_{1} z_{2}$ is real and non-zero, $z_{1} \neq 0, z_{2} \neq 0$, and
$$
x_{1} x_{2}-y_{1} y_{2} \neq 0 \quad \text { and } \quad x_{1} y_{2}+y_{1} x_{2}=0
$$
Thus, since $z_{2} \neq 0$, then
$$
\begin{aligned}
\frac{z_{1}}{z_{2}} &=\frac{x_{1}+i y_{1}}{x_{2}-i y_{2}} \cdot \frac{x_{2}+i y_{2}}{x_{2}+i y_{2}} \\
&=\frac{x_{1} x_{2}-y_{1} y_{2}+\left(x_{1} y_{2}+y_{1} x_{2}\right) i}{x_{2}^{2}+y_{2}^{2}} \\
&=\frac{x_{1} x_{2}-y_{1} y_{2}}{x_{2}^{2}+y_{2}^{2}}
\end{aligned}
$$
By setting $r=\frac{x_{1} x_{2}-y_{1} y_{2}}{x_{2}^{2}+y_{2}^{2}}$, we have the result.
by Gold Status (10,261 points)

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