Sketch the set $A\cap B$ in the Argand plane, where $A=\{z:|\frac{z+1}{z-1}|\leq 1\}$ and $B=\{z:|z|-\operatorname{Re}z\leq 1\}$

Question

Sketch the set $A\cap B$ in the Argand plane, where $A=\{z:|\frac{z+1}{z-1}|\leq 1\}$ and $B=\{z:|z|-\operatorname{Re}z\leq 1\}$

In this problem, I can only make out $A=\{z:|\frac{z+1}{z-1}|\leq 1\}=\{z:|z+1|\leq |z-1|\}$ . So, if, we consider $z=a+ib$ , then, $z\in \mathbb {C}$ if $(a+1)^2+b^2\leq (a-1)^2+b^2$ which means, $(a+1)^2\leq (a-1)^2$ and hence, $a\leq 0$. Also, if $|z|-\operatorname{Re}z\leq 1$, then, $\sqrt{a^2+b^2}-a\leq 1$ which essentially means, $a^2+b^2\leq (1+a)^2$ due to which , $b^2\leq 2a+1$ . But I dont have a clue how how to proceed further. I am not quite getting this. Thanks in advance...

Answer 1

I will be replacing $a$ with $x$ and $b$ with $y$ in the following in order to match the graphs on WolframAlpha. This only changes labels, as the Argand plane is the $(a,b)$-plane and we will be graphing on the $(x,y)$-plane.

We are left with the question of how to graph the set of points in the $(x,y)$-plane such that $x \leq 0$ and $y^2 \leq 2x+1$.

The most common method is to note that all the expressions involved ($x$, $0$, $y^2$, $2x+1$) are continuous functions of $x,y$, and so we will first graph the inequalities $x=0$ and $y^2 = 2x+1$ and then check the regions these graphs split the plane into.

For instance, the graph of $x=0$ is a vertical line on the $y$-axis and then checking the region to the left and to the right of this vertical line shows that the region to the left and the vertical line itself is $A$.

Similarly, the graph of $y^2 = 2x+1$ can be recognized as equivalent to $x = \frac12y^2 - \frac12$, which is a parabola open to the right with vertex as $(-1/2, 0)$ and $y$-intercepts as $(0, \pm 1)$. Then once again testing the regions, we find $B$ is this parabola along with the region to its right

Finally, this means that the intersection $A \cap B$ is the region between $x=0$ and $y^2 = 2x+1$ along with its boundary.

Answer 2

Pay attention to the fact that, in the original definition of $A$, the numbers $z \in \mathbb{C} $ for which $z-1 = 0$ are to be excluded, so you have to keep track of this when manipulating the expressions:

$$ A = \{ x : |\frac{z+1}{z-1}| \leq 1 \} = \{ x : |z+1| \leq |z-1| \wedge z \neq 1 \} . $$

Now, having that in the end $z=a+ib \in A \ \ \text{iff} \ \ a < 0$ the additional condition $z \neq 1$ is useless, but in other circumstances this could not be the case.
Apart from that, the request is to sketch the set resulting from the intersection of $A$ and $B$ on the complex plane, so you can proceed on the Argand plane $(a, b)$ like you would in the ordinary $(x,y)$ coordinate plane, once you have the analytic expression of the two sets in terms of $a$ and $b$. So, actually, you have done almost all the preliminary work in obtaining the pair of conditions:

$$ \cases{ a \leq 0 & (from the definition of $A$) \\ b^2 < 2a + 1 \Rightarrow a > \frac{1}{2} \left( b^2 - 1 \right) & (from the definition of $B$)} $$

So you have now to proceed to:

• sketch $A$ (a half-plane)
• sketch $B$ (a curve)
• outlining the intersection of the two sets

In doing this, if you are familiar with the ordinary cartesian plane, consider that in the Argand plane the $a$- and $b$-axis are omologous to the $x$- and $y$-axis in the cartesian plane.