What is the mutual position of points z_1 and z_2 if

What is the mutual position of points $z_1$ and $z_2$ if:

$1^{\circ}\ \text{Re}(z_1z_2)=0$; $2^{\circ}\ \text{Im}(z_1z_2)=0$; $3^{\circ}\ \text{Re}(z_1\bar z_2)=0$; $4^{\circ}\ \text{Im}(z_1\bar z_2)=0$.

Explanation of Method

Write $z_1=r_1e^{i\theta_1}$, $z_2=r_2e^{i\theta_2}$ (both nonzero). Then $z_1z_2=r_1r_2e^{i(\theta_1+\theta_2)}$ encodes the sum of the arguments, while $z_1\bar z_2 = r_1r_2e^{i(\theta_1-\theta_2)}$ encodes their difference. Setting the real or imaginary part of each to zero forces that angle to be an odd or even multiple of $\pi/2$, which translates directly into a geometric relationship between the two points.

Case 1° · $\text{Re}(z_1z_2)=0$

In Cartesian form, $z_1z_2=(x_1x_2-y_1y_2)+i(x_1y_2+y_1x_2)$, so the condition is $x_1x_2=y_1y_2$. In polar form:

$$ \text{Re}(z_1z_2) = r_1r_2\cos(\theta_1+\theta_2) = 0 \quad\Longrightarrow\quad \cos(\theta_1+\theta_2)=0 $$

$$ \theta_1+\theta_2 = \frac\pi2+n\pi, \quad n\in\mathbb Z $$

Meaning: the product $z_1z_2$ is purely imaginary — the sum of the two arguments is an odd multiple of $\dfrac\pi2$. Equivalently, since $z_1z_2 = z_1\overline{(\bar z_2)}$, this says $z_1$ is perpendicular to the mirror image of $z_2$ across the real axis.

Case 2° · $\text{Im}(z_1z_2)=0$

$$ \text{Im}(z_1z_2) = r_1r_2\sin(\theta_1+\theta_2) = 0 \quad\Longrightarrow\quad \theta_1+\theta_2 = n\pi,\quad n\in\mathbb Z $$

Meaning: the product $z_1z_2$ is real. Equivalently, $z_1$ and the reflection $\bar z_2$ lie on the same line through the origin.

Case 3° · $\text{Re}(z_1\bar z_2)=0$

In Cartesian form, $z_1\bar z_2 = (x_1x_2+y_1y_2)+i(y_1x_2-x_1y_2)$, so the condition is $x_1x_2+y_1y_2=0$ — exactly the dot product of the position vectors $(x_1,y_1)$ and $(x_2,y_2)$. In polar form:

$$ \text{Re}(z_1\bar z_2) = r_1r_2\cos(\theta_1-\theta_2) = 0 \quad\Longrightarrow\quad \theta_1-\theta_2 = \frac\pi2+n\pi $$

Meaning: the vectors from the origin to $z_1$ and $z_2$ are perpendicular — the angle between them is exactly $90°$.

Case 4° · $\text{Im}(z_1\bar z_2)=0$

$$ \text{Im}(z_1\bar z_2) = r_1r_2\sin(\theta_1-\theta_2) = 0 \quad\Longrightarrow\quad \theta_1-\theta_2 = n\pi, \quad n\in\mathbb Z $$

Equivalently, $y_1x_2=x_1y_2$, i.e. $\dfrac{y_1}{x_1}=\dfrac{y_2}{x_2}$ (equal slopes from the origin).

Meaning: $z_1$ and $z_2$ lie on the same line through the origin — either $\theta_1=\theta_2$ (same ray) or $\theta_1=\theta_2+\pi$ (opposite rays).

Left: $\text{Re}(z_1\bar z_2)=0$ — $z_1\perp z_2$. Right: $\text{Im}(z_1\bar z_2)=0$ — $z_1$, $z_2$, and the origin lie on one line.

Result

$$ \begin{aligned} 1^{\circ}\ \text{Re}(z_1z_2)=0: &\quad \theta_1+\theta_2 = \tfrac\pi2 \ (\text{mod }\pi) \ \ (z_1z_2 \text{ purely imaginary}) \\ 2^{\circ}\ \text{Im}(z_1z_2)=0: &\quad \theta_1+\theta_2 = 0 \ (\text{mod }\pi) \ \ (z_1z_2 \text{ real}) \\ 3^{\circ}\ \text{Re}(z_1\bar z_2)=0: &\quad z_1 \perp z_2 \ \ (\text{perpendicular through the origin}) \\ 4^{\circ}\ \text{Im}(z_1\bar z_2)=0: &\quad z_1,\, z_2,\, 0 \ \text{collinear} \end{aligned} $$

The key distinction: $z_1z_2$ encodes the sum $\theta_1+\theta_2$ of the arguments, while $z_1\bar z_2$ encodes their difference $\theta_1-\theta_2$ — and $\text{Re}(z_1\bar z_2)=x_1x_2+y_1y_2$ is exactly the Euclidean dot product, making case $3^{\circ}$ the direct complex-number test for perpendicularity.