# Lines in the Complex Plane

What is the equation of a straight line in the complex plane? There are many different forms, but I want to look at some of the simplest ones.

If you know the slope $$m \in \mathbb{R}$$ and intercept $$b \in \mathbb{R}$$ of the line, you can write an equation in parametric form $z = x + i (m x + b) ,$ where $$x \in \mathbb{R}$$ and $$z \in \mathbb{C}$$.

Another simple form is the equation $|z - z_1| = |z - z_2|,$ where $$z_1 \neq z_2$$. This gives you the locus of points in the complex plane that are equidistant from $$z_1$$ and $$z_2$$, which is a straight line. This form is less practically useful, since we don’t usually describe lines in this way.

What if you don’t know the slope and intercept of the line, but you do know two points on the line? You could use the two points to calculate the slope and intercept, and plug into the above parametric equation, but there is a nicer form. To see this, first note that if a point $$z$$ is collinear with two other points $$z_1$$ and $$z_2$$, then $z - z_1 = \lambda (z - z_2), \quad \lambda \in \mathbb{R} .$ One way to see this is to view the complex numbers as vectors. Then $$z-z_1$$ needs to point in the same direction as $$z-z_2$$, modulo 180 degrees. Note that $$\lambda$$ depends on $$z$$.

Rearranging, we get $\frac{z - z_1}{z-z_2} = \lambda,$ or $\text{Im}\left(\frac{z-z_1}{z-z_2}\right) = 0 ,$ since $$\lambda$$ is real. This can also be written as $\text{Im} \left((z - z_1)\cdot \overline{(z - z_2)} \right) = 0 .$

We can get another form by writing $$z_2 = z_1 + z_3$$, where $$z_3 = z_2 - z_1$$. Then $\frac{z - z_1}{z - z_1 - z_3} = \lambda.$ Cross-multiplying and gathering everything on one side allows us to write this as $(z - z_1)(\lambda - 1) - \lambda z_3 = 0,$ or $\frac{z - z_1}{z_3} = \frac{\lambda}{\lambda - 1} .$ This gives another equation for a line $\text{Im}\left( \frac{z - z_1}{z_3} \right) = 0, \quad \text{or} \quad \text{Im} \left( (z-z_1) \cdot \overline{z}_3 \right) = 0.$ This is the form given in this document by William Kahan. Also, the method of arriving at this form allows us to see that the previous form can be written as $\text{Im}\left( \frac{z}{z_2 - z_1}\right) = 0$ or equivalently $\text{Im}\left( (z_2 - z_1) \cdot \overline{z}\right) = 0.$

This final form can be further massaged. Since $$z_2 - z_1 = z_3$$, we can write $\text{Im}(z_3\cdot \overline{z}) = \frac{1}{2i} \left( z_3\cdot \overline{z} - \overline{z}_3 \cdot z\right) = 0,$ or $z_3 \cdot \overline{z} = \overline{z}_3 \cdot z.$ or $\frac{z^2}{|z|^2} = \frac{z_3^2}{|z_3|^2}.$ Hopefully by now you can see that there are many many equivalent ways to write these expressions.

Another form can be obtained by starting with the fact that $$z + \overline{z} = d \in \mathbb{R}$$ defines a vertical line at $$x = d/2$$. Then since complex multiplication is equivalent to rotation (along with dilation), we can rotate all points in the line by $$\theta$$ by taking $$z \rightarrow z_0 z$$, where $$z_0 = r e^{i\theta}$$. The equation of the line becomes $z z_0 + \overline{z z}_0 = r d = D .$ This form is referenced in this discussion on Mathematics Stack Exchange. An example might be useful. Assume you are given the two points $$z_1 = 0 + 0i$$ and $$z_2 = 1 + i$$. Then $$z_3 = z_2 - z_1$$ can be written $$z_3 = \sqrt{2}e^{i\pi/4}$$. The initial equation for a vertical line at $$x = 0$$ is $$z + \overline{z} = 0$$. We want to rotate the plane by $$\pi/4$$ so that the line ends up with the same angle as $$z_3$$. Since $$r$$ is arbitrary, choose $$r = \sqrt{2}$$, so $$z_0 = 1 +i$$, which gives $(1+i) z + (1-i)\overline{z} = 0,$ which reduces to $2(x-y) = 0, \quad \text{or} \quad y = x.$ Note that in general you want to rotate by $$\pi/2 - \text{arg}(z_3)$$.

What interesting equations for lines in the complex plane do you know? What interpretations (geometric or algebraic) do they have?

##### Landon Lehman
###### Data Scientist

My research interests include data science, statistics, physics, and applied math.