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Is it true that if a complex number $z_2$ times $z_1$ is the square of norm of $z_1$, then $z_2$ is the conjugate of $z_1$?

$z_2 = \bar{z_1} \Leftrightarrow z_1z_2 = \|z_1\|^2?$

It occurs to me to be true: Let $z_1 = r_1e^{i\theta_1}, z_2 = r_2e^{i\theta_2}$, then

\begin{align*} z_1z_2 = \|z_1\|^2 &\Leftrightarrow r_1r_2e^{i(\theta_1+\theta_2)} = r_1^2\\ &\Leftrightarrow r_2e^{i(\theta_1+\theta_2)} = r_1\\ & \Leftrightarrow r_2 = r_1,\theta_1 + \theta_2 = (2k+1)\pi \end{align*}

Is this correct?

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  • $\begingroup$ Shouldn't it be $\theta_1 + \theta_2 = 2 k \pi, \ k \in \mathbb{Z}$? so it covers the case $\theta_1 = -\theta_2$, i.e., $z_1$ and $z_2$ are complex conjugates. Cheers! $\endgroup$ – Dmoreno Jan 27 '14 at 9:51
  • $\begingroup$ Oh, yes, the equation should be. But what confuses me is, consider the complex conjugate numbers $e^{\pi/4}$ and $e^{3\pi/4}$ - aren't their angle sum up to be $\pi?$ Thanks cheerful mathematician @Dmoreno.. $\endgroup$ – 1LiterTears Jan 27 '14 at 9:58
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    $\begingroup$ No, this two are not complex conjugates! The sum of their arguments must be equal to $0$ (or an integer multiple of $2 \pi$). When this numbers are represented in the cartesian $(x,y)$ plane, one is the mirror reflection of the other with respect to the real ($x$) axis. $\endgroup$ – Dmoreno Jan 27 '14 at 10:01
  • $\begingroup$ Oh that's right and very helpful. Thanks @Dmoreno a lot! $\endgroup$ – 1LiterTears Jan 27 '14 at 10:02
  • $\begingroup$ You're welcome! Cheers! $\endgroup$ – Dmoreno Jan 27 '14 at 10:02
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The forward implication is true: $z_2 = \bar z_1 \Rightarrow z_1 z_2 = \| z_1 \|^2$. But the reverse is not; e.g. if $z_1 = 0$, then $z_2$ can be anything, so one must have the provision that $z_1 \ne 0$. In such a case, then $e^{i(\theta_1 + \theta_2)} = 1$, or $\theta_1 + \theta_2 = 2\pi k$ for some integer $k$, from which it follows that $\theta_2 \equiv -\theta_1$ up to an integer multiple of $2\pi$, and $z_2 = \bar z_1$, as claimed.

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It's not true for $z_1=0$..

However, if $z_1 \neq 0$, if given $z_1z_2=|z_1|^2$, then a nicer way to show it is: $$z_2=z_1^{-1}|z_1|^2=\overline{z_1}$$ But in the other direction, if given $z_2=\overline{z_1}$ then: $$z_1z_2=z_1\overline{z_1}=|z_1|^2$$ For any complex $z_1$.

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