Given that the product of two complex numbers $z_1$ and $z_2$ is real and different from $0$, show that there exists a real number $p$ such that $z_1 = p\overline{z_2}$.
Please help!!
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Zev Chonoles
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Sophie
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Use polar coordinates: $z=r e^{i\theta}$. – J.R. Mar 02 '14 at 16:33
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No need for polar I think – user76568 Mar 02 '14 at 16:35
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Isn't z2bar the complex conjugate of z2? But yeah, i am not really sure where to start? – Sophie Mar 02 '14 at 16:44
4 Answers
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If $z_{1} z_{2} = r \neq 0$ is real then $z_1, z_2 \neq 0$, and $z_1 z_2= \overline{z_1 z_2} = \overline{z_1} \cdot \overline{z_2} $ and thus: $$z_1 |z_2|^2 = z_1 z_2 \overline{z_2} = \overline{z_1 z_2} \cdot \overline{z_2} \implies z_1 = \frac{\overline{z_1 z_2}}{|z_2|^2} \overline{z_2} = \frac{r}{|z_2|^2} \overline{z_2}$$ $\frac{r}{|z_2|^2}$ is a real number and wer'e done.
user76568
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1It is not $z_2^2$, it is $|z_2|^2 = z_2 \cdot \overline{z_2} = \Re({z_2})^2 + \Im({z_2})^2$. Take a $z = a + ib, \overline{z} = a - ib$, multiply them, and see for your self. – user76568 Mar 02 '14 at 16:59
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$$z_1 = \frac{z_1 z_2 \bar z_2}{z_2 \bar z_2} = \underbrace{\frac{z_1 z_2}{|z_2|^2}}_p \ \ \bar z_2$$
themaker
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Hint:
$z_{1}\left|z_{2}\right|^{2}=z_{1}\left(z_{2}\bar{z}_{2}\right)=\left(z_{1}z_{2}\right)\bar{z}_{2}$
(So it is enough if $z_2\neq 0$. Then you can divide by $\left|z_{2}\right|^{2}$)
drhab
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Hint $\,\ z_1z_2\in\Bbb R\,\Rightarrow\, z_1z_2 = \bar z_1\bar z_2 \,\Rightarrow \dfrac{z_1}{\bar z_2} = \dfrac{\bar z_1}{z_2} = \overline{\left(\dfrac{z_1}{\bar z_2}\right)}\,\Rightarrow\,\dfrac{z_1}{\bar z_2}\in\Bbb R$
Bill Dubuque
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