# What is complex slope of a line in complex numbers

• Feb 7th 2009, 06:46 AM
pankaj
What is complex slope of a line in complex numbers
If equation of a line in complex plane is given by
$\displaystyle \bar az+a\bar z+b=0$
where $\displaystyle a$ is a complex constant and $\displaystyle b$ is a real number
Then complex slope of the line is defined as $\displaystyle \omega =-\frac{a}{\bar a}$

Now,what I want to know is that what is the complex slope of the line,i.e. what is its geometrical meaning.Also,what do $\displaystyle a$ and $\displaystyle b$ signify geometrically
• Feb 7th 2009, 09:14 AM
HallsofIvy
Let a= u+ iv, b= p+ iq, z= x+ iy. Then [tex]\overline{a}z+ a\overline{z}+ b= 0[tex] becomes (u- iv)(x+ iy)+ (u+ iv)(x- iy)+ p+ iq= 0

(ux+ vy)+ i(uy- vx)+ (ux+ vy)+ i(vx- uy)+ p+ iq= 0
Notice that i(uy- vx) and i(vx-uy) cancel. Separating real and imaginary parts, 2ux+ 2vy+ p= 0 and q= 0 (the latter is why b= p is a real number).
We can rewrite the first equation as 2vy= -2ux+ b or y= -(u/v)x+ b/(2v) so that is, in fact, a straight line with slope -u/v and y-intercept b/(2v).

Now, I honestly don't see what that has to do with the "complex slope", $\displaystyle -\frac{a}{\overline{a}}$! Where did you see that formula?
• Feb 7th 2009, 06:12 PM
pankaj
It was given in a textbook .

Among other things it was also given that complex slope of line joining $\displaystyle z_{1}$ and $\displaystyle z_{2}$ is given by $\displaystyle \frac{z_{1}-z_{2}}{\bar z_{1}-\bar z_{2}}$.

also if complex slopes of two lines equal then the two lines are parallel and the two lines will be perpendicular if the sum of the slopes is $\displaystyle 0$.

It is also given that complex slope of line making angle $\displaystyle \theta$ with the real axis is given by $\displaystyle \omega=e^{2i\theta}$

All this I have verified and they are true.

But I want to understand the geometrical interpretation of this concept.