complex numbers
I'm asked to plot the following complex number inequalities, I managed to plot the second one(PS see this attached img)
PS help me plot the first inequality and calculate the $|z|$ for the point in this region for which $arg~z$ is least.
Any help is greatly appreciated. (Also any book/web reference to learn more about these type of questions and concepts will be of great help)
Thanks,
Arif
$|z-2-i|\le 1$ is the disk centre $2+i$ radius 1. $|z-i|\le |z-2|$ is the half-plane to the left of the perpendicular bisector of the line joining the points 2 and $i$. We want the point in the intersection of these two regions which has the largest arg. That is obviously $X$ where $OX$ is tangent to the circle.
We have $\arg z=2\arg(2+i)=2\tan^{-1}0.5$.
By taking a look at your plot, it is clear that:
Hence, the difference you seek is, in radians, \begin{align} \arg z_2-\arg z_1&=\left(\frac{\pi}{2}+\arctan\frac{2}{3}\right)-\frac{\pi}{2}\\ &=\arctan\frac{2}{3}\\ &\approx0.588 \end{align}
Best Answer
Let $z=a+ib$.
First, the two absolute values are
$$|z-1|=\sqrt{(a-1)^2+b^2} \text{ and } |z-i|=\sqrt{a^2+(b-1)^2}.$$
We have the following inequality then
$$a^2-2a+1+b^2\le a^2+b^2-2b+1$$
or
$$-2a\le -2b$$
or $$a\ge b.$$
That is, the first inequality holds for those complex numbers whose real part is greater or equal than their imaginary par.
As far as the second inequality we have
$$|z-(2+2i)|=\sqrt{(a-2)^2+(b-2)^2}.$$
So we have the following inequality
$$(a-2)^2+(b-2)^2\le 1$$
which holds inside the unit circle centered at $2+i2$.
The following two figure depicts the two regions
The third figure depicts the common region in red. The red line is tangent to the circle.
The angle between the tangent line and the Real axis is the smallest argument belonging to complex numbers satisfying both of the inequalities.
Further edit:
The length of the segment (the absolute value of the complex number sought) joining the origin and the tangent point is $\sqrt7$ as explained by the figure below