# How to find the equation of a line in complex analysis?

1. Jan 21, 2009

### Bacat

*This is not homework, though a class was the origin of my curiosity.

In real analysis we could find the equation of a line that passes through two points by finding the slope and then plugging in one set of points to calculate the value of b. ie

$$y = mx + b$$

m = $$\frac{y_2-y_1}{x_2-x_1}$$

In complex analysis, we know that the equation for a line is $$Re[((m+i)z+b)]=0$$. Sitting down to derive m, I find the following:

$$m = \frac{Im[z_1] - Im[z_2]}{Re[z_1]-Re[z_2]}$$

But if I try to plug in the points (say $$z_{1}$$ and $$z_{2}$$), it doesn't give me a value for b that makes sense. what is the correct way to find the equation of a line?

Last edited: Jan 21, 2009
2. Jan 21, 2009

### slider142

The expression you gave is not an equation. What is the complete expression?
The complex plane is simply an imaginary axis and a real axis at right angles to each other. If z = (x, y) = x + iy is an arbitrary complex variable, it pretty much replaces y in our equations with Im[z] and x with Re[z]. so if you have a graph with points (x, y) that satisfy the equation y = mx + b, you will get the same graph in the complex plane with the equation Im[z] = m*Re[z] + b. This is not very geometric, however, and your equation for the slope of a line passing through the points z1 and z2 is more descriptive:
I'm not sure what equation you're plugging it into but geometrically, a line can be defined as the set of points equidistant from two distinct points in the plane, say z1 and z2. This gives us the equation |z - z1| = |z - z2| for the set of points z on the line.
If you go ahead and translate this into the point-slope form, you get your m as above and the equation b = [[Re(z2)]22 + [Im(z2)]2 - ([Re(z1)]22 + [Im(z1)]2)]/2(Re(z2) - Re(z1)) = [|z2| - |z1|]/2(Re(z2) - Re(z1)) .

3. Jan 21, 2009

### Bacat

Yes, that makes sense now. I think I was just crunching wrong. I amended my previous equation. $$Re((m+i)z + b))=0$$

If I let $$z_1=1+0i$$ and $$z_2=-1-i$$

I calculate:

$$m=\frac{0-(-1)}{1-(-1)}=\frac{1}{2}$$

$$Re((\frac{1}{2}+i)*(1+0i)+b))=0$$
$$Re(\frac{1}{2}+i+b)=0$$
$$b=-\frac{1}{2}$$

And the equation of the line is: $$Re((\frac{1}{2}+i)z - \frac{1}{2})=0$$

This works. Thank you!