Complex Slope of a Line

The complex slope μ of the line through z₁ and z₂. Two lines are parallel iff their complex slopes are equal; perpendicular iff μ₁ + μ₂ = 0. Note: complex slope is not the same as the real slope m = (y₁−y₂)/(x₁−x₂). The relation is m = i(1−μ)/(1+μ) when μ ≠ −1. Connects back to: Straight Lines chapter.

The real slope $m$ of a line carries partial geometric information — it measures the angle but loses the sign. The complex slope $\mu$ carries the full directional information of a line in one complex number.

Definition

For the line through $z_1$ and $z_2$, the direction vector is $d = z_2 - z_1$. The complex slope is:

$$\mu = \frac{z_1 - z_2}{\bar{z}_1 - \bar{z}_2} = \frac{d}{\bar{d}}$$

Writing $d = re^{i\theta}$, we get $\mu = e^{2i\theta}$. So $|\mu| = 1$ always — the complex slope lies on the unit circle, encoding twice the angle of inclination.

Condition for Parallel Lines

Two lines with direction vectors $d_1$ and $d_2$ are parallel iff $d_1$ and $d_2$ point in the same or opposite direction, i.e., $d_1/d_2 \in \mathbb{R}$.

$$\frac{d_1}{d_2} \in \mathbb{R} \iff \frac{d_1}{d_2} = \frac{\bar{d}_1}{\bar{d}_2} \iff \frac{d_1}{\bar{d}_1} = \frac{d_2}{\bar{d}_2} \iff \mu_1 = \mu_2$$

Parallel lines have equal complex slopes. $\blacksquare$

Condition for Perpendicular Lines

Two lines are perpendicular iff the angle between their directions is $\pi/2$, i.e., $d_1/d_2$ is purely imaginary.

A complex number $w$ is purely imaginary iff $w + \bar{w} = 0$.

$$\frac{d_1}{d_2} + \frac{\bar{d}_1}{\bar{d}_2} = 0 \iff \frac{d_1\bar{d}_2 + \bar{d}_1 d_2}{d_2\bar{d}_2} = 0 \iff d_1\bar{d}_2 + \bar{d}_1 d_2 = 0$$

Dividing by $\bar{d}_1\bar{d}_2$:

$$\frac{d_1}{\bar{d}_1} + \frac{d_2}{\bar{d}_2} = 0 \iff \mu_1 + \mu_2 = 0$$

Perpendicular lines have complex slopes summing to zero. $\blacksquare$

Relation to Real Slope

Writing $d = \Delta x + i\Delta y$:

$$\mu = \frac{\Delta x + i\Delta y}{\Delta x - i\Delta y}$$

Multiply top and bottom by $(\Delta x + i\Delta y)$:

$$\mu = \frac{(\Delta x + i\Delta y)^2}{(\Delta x)^2 + (\Delta y)^2} = \frac{(\Delta x)^2 - (\Delta y)^2 + 2i\Delta x\Delta y}{|\Delta x + i\Delta y|^2}$$

The real slope $m = \Delta y/\Delta x$ recovers $\mu$ via: $\mu = \dfrac{1 - m^2 + 2im}{1 + m^2}$, which equals $e^{2i\arctan m}$ — consistent with $\mu = e^{2i\theta}$ where $\tan\theta = m$.

For a vertical line ($\Delta x = 0$): $\mu = -1$. For a horizontal line ($\Delta y = 0$): $\mu = 1$. Check: $1 + (-1) = 0$ ✓ horizontal and vertical are perpendicular.