Let A=z∈ C:| z+1 z-1|<1 and B=z∈ C: arg( z-1 z+1)= 2π 3. Then A∩ B is

Set $A$A is ${z\in C:|\frac{z+1}{z-1}|<1}$z∈C:∣z−1z+1​∣<1

$$\Rightarrow |\frac{\bar{z}+1}{\bar{z}-1}|<1$$⇒∣zˉ−1zˉ+1​∣<1

$$\Rightarrow |\bar{z}+1|<|\bar{z}-1|$$⇒∣zˉ+1∣<∣zˉ−1∣

$$\Rightarrow {(x+1)}^{2}+{y}^{2}<{(x-1)}^{2}+{y}^{2}$$⇒(x+1)2+y2<(x−1)2+y2

$$\Rightarrow x<0$$⇒x<0

Set $B$B is ${z\in C:\mathrm{arg}(\frac{z-1}{z+1})=\frac{2\pi }{3}}$z∈C:arg(z+1z−1​)=32π​

$$\Rightarrow \mathrm{arg}(\frac{z-1}{z+1})=\frac{2\pi }{3}$$⇒arg(z+1z−1​)=32π​

$$\Rightarrow {\mathrm{tan}}^{-1}(\frac{y}{x-1})-{\mathrm{tan}}^{-1}(\frac{y}{x+1})=\frac{2\pi }{3}$$⇒tan−1(x−1y​)−tan−1(x+1y​)=32π​

$$\Rightarrow {x}^{2}+{y}^{2}+\frac{2y}{\sqrt{3}}-1=0$$⇒x2+y2+3​2y​−1=0

$\Rightarrow$⇒ Centre $(0,-\frac{1}{\sqrt{3}})$(0,−3​1​)

Hence, $A\cap B$A∩B will represent a portion of a circle centred at $(0,-\frac{1}{\sqrt{3}})$(0,−3​1​) that lies in the second quadrant only