complex numbers
I'm asked to plot $|z+1-3i|\leq1$ and $\text{Im}(z) \geq3$, I've plotted both the inequalities, PS see the attachment.
Now, I'm unable to determine the diļ¬erence between the greatest and least values of $\arg z$ for points lying in this region.
PS assist,
Also any resources that will be me get better to solve such questions will be of great help.
Thanks
Arif
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
$|z-2-i|\le 1$ 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 $\arg z=2\arg(2+i)=2\tan^{-1}0.5$.
Best Answer
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}