# Two complex numbers $z_1$ and $z_2$ so that $z_1=kz_2 \iff z_1 \bar{z_2} \in \Bbb R^+$

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?

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.

• without considering real, imaginary parts! Nice...Thanks! – Chinnapparaj R Apr 2 '20 at 5:37
• @ChinnapparajR: thank you for the kind words my friend, and the "acceptance"! Cheers! – Robert Lewis Apr 2 '20 at 5:38

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{aligned} x_1(kx_1) + y_1(ky_1) & \gt 0 \\ k(x_1^2 + y_1^2) & \gt 0 \end{aligned}\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 $$z_1$$.

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

• The problem specifies $k > 0$; how do you handle that here? – Robert Lewis Apr 2 '20 at 5:44
• yes. I will edit it – Mojbn Apr 2 '20 at 5:50
• @RobertLewis now I correct it – Mojbn Apr 2 '20 at 6:11
• Thank you for your speedy response! – Robert Lewis Apr 2 '20 at 6:13
• 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. – Robert Lewis Apr 2 '20 at 6:19

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 $$z = 0$$.