If we have two points $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$ in the complex plane and define the relative coordinate $z=z_2-z_1$, we have that the length of $z$ is the Euclidian distance between the points:

$r=|z|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\ .$

But what about the phase of $z$? The best I have been able to come up with is

$\theta=-i\log(z/r)=-i\log\Big(\frac{x_2-x_1+i(y_2-y_1)}{\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}\Big)\ .$

I kind of expect the answer to be the angle between the points

$\theta_{12}=\cos^{-1}\Big(\frac{\frac{1}{2}(z_1^*z_2+z_1z^*_2)}{|z_1||z_2|}\Big)\ ,$

but I haven't been able to show this.


The phase is $$\theta = \arctan \frac{y_2-y_1}{x_2-x_1}$$ in the respective quadrant.

  • $\begingroup$ Thanks! This seems to indeed be related to the angle between the points; for example $\theta(x_2-x_1=y_2-y_1)=\pi/4$ and $\theta(y_2=y_1)=0$ and $\theta(x_2=x_1)=\pi/2$.. I guess one just has to be a bit careful thinking about the quadrant? $\endgroup$ – jorgen Nov 23 '14 at 22:39
  • $\begingroup$ @jorgen In general, you get the phase of some $z = x+i\cdot y = r\cdot e^{i\theta}$ from $\theta = \arctan y/x$. You'll find that in any introductory book on the topic. Yes, you always have to take care about the quadrant. In your case you have $z = z_2-z_1 = (x_2-x_1) + i(y_2-y_1)$, so $x = x_2-x_1$ and $y = y_2-y_1$, which you have to insert in the arctan-formula. $\endgroup$ – GDumphart Nov 23 '14 at 22:54

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