Finding the angle between two lines in the complex plane without converting to real and imaginary components

Question

How do you find the angle between two lines in the complex plane, without going through the real route (breaking $z$ into $x+yi$ and solving by finding the tangent from the slopes)?

For example, if my lines were in the form

$$\begin{align} az+\overline{az} + b &=0 \\ dz+\overline{dz} + c &=0 \\ \end{align}$$

for complex $a$, $d$ and real $b$, $c$.

I know we can determine if they are parallel or perpendicular by multiplying $ad$ and seeing if it is zero, or if $d$ is a scalar of $a$, but does this help? Would this in any way involve finding the arccos() between two vectors?

Any help is appreciated. Thanks~

Answer 1

Switch your notation to polar and express your lines as vectors with angles $\theta_1$ and $\theta_2$. Angle in between is simply their difference, so $\Delta \theta$.

Answer 2

Let us denote $$ L_1 = \{ z \in \Bbb C \mid az+\overline{az} + b=0 \} \, ,\\ L_2 = \{ z \in \Bbb C \mid dz+\overline{dz} + c=0 \} \, . $$ with $a, d \in \Bbb C\setminus \{ 0 \}$ and $b, c\in \Bbb R$. Since $$ az+\overline{az} + b = d\left( \frac ad z+\frac{b-c}{2d}\right) + \overline {d\left( \frac ad z+\frac{b-c}{2d}\right)} + c $$ we have $$ z \in L_1 \iff \frac ad z+\frac{b-c}{2d} \in L_2 \, . $$

If $a/d$ is not a real number then the two lines intersect at a single point and the oriented angle from $L_1$ to $L_2$ is $\arg(a/d)$.

Otherwise the lines are identical ($b=c$) or parallel ($b \ne c$).

Answer 3

It's simpler, WLOG, to work at the origin, bringing by translation the equations of the lines to be:

$$az+\overline{az}=0, \ \ dz+\overline{dz}=0 $$

otherwise said, as @dxiv has recalled it:

$$\Re(az)=0, \ \ \Re(dz')=0 \tag{1}$$

where we can assume, WLOG again, that $|a|=|d|=|z|=|z'|=1$ (where $z$ and $z'$ are now considered as representative points of their resp. lines at distance $1$ from the origin).

With evident notations, (1) becomes:

$$\Re(e^{i \alpha}e^{i\theta})=0, \ \ \Re(e^{i \delta}e^{i\theta'})=0$$

$$\cos(\alpha+\theta)=0, \ \ \ \ \cos(\delta+\theta')=0$$

$$\alpha+\theta=\pi/2 + k \pi, \ \ \ \ \delta+\theta'=\pi/2 + k' \pi \tag{2}$$

Therefore, by subtracting the relationships in (2), we obtain the angular "gap" between the two lines:

$$\theta'-\theta = \underbrace{ \alpha - \delta}_{arg(\tfrac{a}{d})} + k'' \pi$$