How do you draw on an argand diagram: $\{z\in{\mathbb C}: \arg(z-1) < \arg(z-i)\}$?
I can plot both points but I don't know what to do with arguments and inequalities.


If you simply interpret what it means to say that $\arg(z-1)<\arg(z-i)$ geometrically, then it shouldn't be too hard to visualize the region. In each of the regions below, the argument of the point ($z$) minus the number (either $1$ or $i$) simply the angle from the dashed line to the arrow. It's easy to see that this angle is smaller from the complex number $1$ in exactly the blue regions.

Note that I'm assuming a branch cut at $\pi$ in the argument function so that, in the horizontal strip, $\arg(z-1)$ is definitely negative while $\arg(z-i)$ is positive.

enter image description here


Setting $\displaystyle z=x+iy,$ we need $$\arctan\frac y{x-1}<\arctan\frac{y-1}x$$

$$\iff\arctan\frac y{x-1}-\arctan\frac{y-1}x<0 $$

$$\iff\arctan\left(\frac{\dfrac y{x-1}-\dfrac{y-1}x}{1+\dfrac y{x-1}\cdot\dfrac{y-1}x}\right)<0 $$

$$\iff\arctan\left(\frac{x+y-1}{x^2+y^2-x-y}\right)<0 $$

$$\iff\frac{x+y-1}{x^2+y^2-x-y}<0 $$

So, we need the sign of $\displaystyle x+y-1$ and $\displaystyle x^2+y^2-x-y=\left(x-\frac12\right)^2+\left(y-\frac12\right)^2-\frac12$ to be opposite

We know, $\displaystyle \left(x-\frac12\right)^2+\left(y-\frac12\right)^2-\frac12=0$ is a circle with radius $\frac1{\sqrt2}$ with center at $\left(\frac12,\frac12\right)$.


As can easily be plotted on WolframAlpha, this region looks like so:

enter image description here

Unfortunately, the arguments can easily be read off of the image as the angles between the vectors and the dashed lines. It certainly appears that $\arg(z-1)>\arg(z-i)$ for certain points in the region.

  • $\begingroup$ Clearly there are some good ideas here, but possibly a mistake. $\endgroup$ – Mark McClure Mar 2 '14 at 19:02
  • $\begingroup$ @MarkMcClure, thanks for your feedback.We need to use math.stackexchange.com/questions/138310/… and $$\tan^{-1}(-x)=-\tan^{-1}x$$ $\endgroup$ – lab bhattacharjee Mar 3 '14 at 16:14
  • $\begingroup$ @Downvoter, thanks for the notification. I was searching this to rectify $\endgroup$ – lab bhattacharjee Mar 3 '14 at 16:15

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