Let A=z C: |z+1z-1|

Question

Let $A=\left\{z\in\mathbb C: \left|\dfrac{z+1}{z-1}\right|<1\right\}$ and $B=\left\{z\in\mathbb C: \arg\left(\dfrac{z-1}{z+1}\right)=\dfrac{2\pi}3\right\}$. Then $A\cap B$ is:

Accepted Answer

$B$ is an arc of the circle through $\pm1$ subtending an angle of $\dfrac{2\pi}3$ at points on it --- this circle has centre $\left(0,-\dfrac1{\sqrt3}\right)$ and radius $\dfrac2{\sqrt3}$ (standard result for the locus $\arg\frac{z-1}{z+1}=$ const.).

$A$ is the region $\operatorname{Re}(z)<0$ (points closer to $-1$ than to $1$).

Intersecting the specific arc of $B$ (which by the orientation of the angle $\dfrac{2\pi}3$ lies in the upper half-plane) with $\operatorname{Re}(z)<0$ leaves only the portion of the arc lying in the second quadrant.

Solutions · 1