Let A = { z in mathbb{C} : |frac{z+1}{z-1}|

Question

(B) For set $A$: $|z+1| < |z-1|$. Squaring both sides,$(x+1)^2 + y^2 < (x-1)^2 + y^2$,which simplifies to $x < 0$. This represents the region to the l

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a portion of a circle centred at $(0, -\frac{1}{\sqrt{3}})$ that lies in the second quadrant only

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(B) For set $A$: $|z+1| For set $B$: $\arg(\frac{z-1}{z+1}) = \frac{2\pi}{3}$. This represents an arc of a circle passing through $(-1, 0)$ and $(1, 0)$. The equation of the circle is $x^2 + y^2 + \frac{2y}{\sqrt{3}} - 1 = 0$,with centre $(0, -\frac{1}{\sqrt{3}})$.The condition $\arg(\frac{z-1}{z+1}) = \frac{2\pi}{3}$ corresponds to the arc of this circle where $x Since $A$ requires $x < 0$ and $B$ is the arc of the circle where $x < 0$,the intersection $A \cap B$ is the entire arc of the circle lying in the second quadrant.

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

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