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I recently came across the following:

Two non zero complex numbers $z_1$ and $z_2$ is a positive multiple of other if and only if $z_1\bar z_2$ is real and positive

$\implies$: Suppose $z_1=kz_2$ where $k>0$ and where $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$.

Then $$x_1=kx_2\\ y_1=ky_2$$

Now, $z_1 \bar {z_2}=(x_1+iy_1)(x_2-iy_2)=x_1x_2+y_1y_2+i(y_1x_2-x_1y_2)=\color{red}{kx_2}x_2+\color{red}{ky_2}y_2+i(\color{red}{ky_2}x_2-\color{red}{kx_2}y_2)=k(x_2^2+y_2^2)>0$

How to prove other direction? Any help?

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It is not necessary to look at the real and imaginary parts of $z_1$ and/or $z_2$ to effect the solution, to wit:

If

$\Bbb R \ni k > 0 \tag 1$

and

$z_1 = kz_2, \tag 2$

then

$z_1 \bar z_2 = k z_2 \bar z_2 > 0, \tag 3$

since

$z_2 \bar z_2 > 0; \tag 4$

going the other way, if

$z_1 \bar z_2 = l > 0, \tag 5$

then

$z_1 z_2 \bar z_2 = l z_2, \tag 6$

from which

$z_1 = \dfrac{l}{z_2 \bar z_2} z_2; \tag 7$

we take

$k = \dfrac{l}{z_2 \bar z_2} > 0; \tag 8$

then (7) becomes

$z_1 = kz_2, \ k > 0, \tag 9$

as per request.

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    $\begingroup$ without considering real, imaginary parts! Nice...Thanks! $\endgroup$ – Chinnapparaj R Apr 2 '20 at 5:37
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    $\begingroup$ @ChinnapparajR: thank you for the kind words my friend, and the "acceptance"! Cheers! $\endgroup$ – Robert Lewis Apr 2 '20 at 5:38
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You've already shown that where $z_1 = x_1 + iy_1$ and $z_2 = x_2 + iy_2$ that

$$z_1 \bar {z_2} = x_1x_2+y_1y_2+i(y_1x_2-x_1y_2) \tag{1}\label{eq1A}$$

Given that $z_1 \bar {z_2}$ is real and positive, then

$$x_1x_2 + y_1y_2 \gt 0 \tag{2}\label{eq2A}$$

$$y_1x_2 - x_1y_2 = 0 \iff y_1x_2 = x_1y_2 \tag{3}\label{eq3A}$$

In \eqref{eq3A}, if $x_1 = 0$, then $y_1x_2 = 0$, so $y_1 = 0$ and/or $x_2 = 0$. If $y_1 = 0$, however, then the LHS of \eqref{eq2A} is $0$, so you must have $x_2 = 0$. Then \eqref{eq2A} simplifies to $y_1y_2 \gt 0$, so $y_1 \lt 0$ and $y_2 \lt 0$, or $y_1 \gt 0$ and $y_2 \gt 0$. In either case, you have that $y_2 = ky_1$ for some positive $k$ and, of course, $x_2 = kx_1 = 0$, so $z_2$ is a positive multiple $k$ of $z_1$.

Next, consider $x_1 \neq 0$. Then there exists a real $k$ such that

$$x_2 = kx_1 \tag{4}\label{eq4A}$$

Substitute this into \eqref{eq3A} and divide both sides by $x_1$ to get

$$y_1(kx_1) = x_1y_2 \iff y_2 = ky_1 \tag{5}\label{eq5A}$$

Substitute \eqref{eq4A} and \eqref{eq5A} into \eqref{eq2A} to get

$$\begin{equation}\begin{aligned} x_1(kx_1) + y_1(ky_1) & \gt 0 \\ k(x_1^2 + y_1^2) & \gt 0 \end{aligned}\end{equation}\tag{6}\label{eq6A}$$

Since $x_1 \neq 0$, then $x_1^2 + y_1^2 \gt 0$ and $k \gt 0$. Thus, once again, you have using \eqref{eq4A} and \eqref{eq5A} that $z_2$ is a positive multiple $k$ of $z_1$.

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let $z_1=r_1e^{i\theta_1}$ and $z_2=r_2e^{i\theta_2}$ then $\bar{z}_2=r_2e^{-i\theta_2}$$$$$ $z_1\bar{z}_2\in \Bbb R^+,z_1\bar{z}_2\gt 0 \Longrightarrow z_1\bar{z}_2=r_1r_2e^{i{(\theta_1-\theta_2)}}= r_1r_2 $ $\Longrightarrow e^{i{(\theta_1-\theta_2)}}=1\Longrightarrow \theta_1 - \theta_2 = 2n\pi , n\in \Bbb Z$ $$\Longrightarrow z_1=kz_2, k\gt 0$$

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    $\begingroup$ The problem specifies $k > 0$; how do you handle that here? $\endgroup$ – Robert Lewis Apr 2 '20 at 5:44
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    $\begingroup$ yes. I will edit it $\endgroup$ – Mojbn Apr 2 '20 at 5:50
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    $\begingroup$ @RobertLewis now I correct it $\endgroup$ – Mojbn Apr 2 '20 at 6:11
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    $\begingroup$ Thank you for your speedy response! $\endgroup$ – Robert Lewis Apr 2 '20 at 6:13
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    $\begingroup$ I upvoted your answer because shows the right idea but it could still use a little polish. For example $\theta_1 - \theta_2 = 2n\pi$, $n \in \Bbb Z$ etc. $\endgroup$ – Robert Lewis Apr 2 '20 at 6:19
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Conclusion:

If $z_1$ is a non-zero complex number and $z_2$ is any complex number, then $z_1\overline z_2$ will by positive real/zero/negative real/non-real complex if and only if $k=\frac {z_1}{z_2}$ is.

Reaching the conclusion:

I think the "gyst" of this is realizing that $z\overline z = Re(z)^2 + Im(z)^2 \ge 0$[1] and, indeed, that is why we use $\sqrt{z\overline z}$ as the definition of $|z|$.

Now as $z_1 \ne 0$ then $\frac {z_2}{z_1} = k$ is well defined. And $z_2 = kz_1$.

So $z_1\overline z_2 = kz_1\overline z_1 = k|z_1|^2$.

Now $|z_1|^2$ is a non-negative real number and as $z_1\ne 0$ it is a positive real number. So $k|z_1|^2=z_1\overline z_2$ will be .... whatever $k$ is.

If $k$ is a positive real number then $z_1\overline z_2$ will be a positive real number. If $k$ is zero then $z_1\overline z_2$ (but it's not as we presumed $z_2\ne 0$ --- which actually wasn't required). $k$ is a negative real or a complex non-real, then so is $z_1\overline z_2$.

So....

If $z_1$ is a non-zero complex number and $z_2$ is any complex number, then $z_1\overline z_2$ will by positive real/zero/negative real/non-real complex if and only if $k=\frac {z_1}{z_2}$ is.

.......

[1] Review (hopefully unnecessary): a) $z\overline z = (Re(z) + Im(z)i)(Re(z)-Im(z)i) = Re(z)^2 - (Im(z)i)^2 = Re(z)^2 + Im(z)^2\ge 0$ with equality holding if and only if $Re(z)=Im(z) = 0$ and therefore $z =0$.

b) Let $z = re^{\theta i}$ so $\overline z = r^2e^{-\theta i}$ and $z\overline z = r^2e^{\theta i - \theta i} = r^2\ge 0$ with equality holding if and only if $r = 0$ and therefore $z = 0$.

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