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  • <img alt="Let\,\, \mathrm{A}=\left\{z \in \mathbf{C}:\left|\frac{z+1}{z-1}\right|<1\right\}$ and $\mathrm{B}=\left\{z \in \mathbf{C}: \arg \left(\frac{z-1}{z+1}\right)=\frac{2 \pi}{3}\right\}."

Let\,\, \mathrm{A}=\left\{z \in \mathbf{C}:\left|\frac{z+1}{z-1}\right|<1\right\}$ and $\mathrm{B}=\left\{z \in \mathbf{C}: \arg \left(\frac{z-1}{z+1}\right)=\frac{2 \pi}{3}\right\}.

Then \: A \cap \: B\: is :

Option: 1

a portion of a circle centred at \left(0,-\frac{1}{\sqrt{3}}\right) that lies in the second and third quadrants only


Option: 2

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


Option: 3

an empty set


Option: 4

a portion of a circle of radius \frac{2}{\sqrt{3}} that lies in the third quadrant only


Answers (1)

best_answer

\mathrm{\text { Set } A} \\

\Rightarrow \mathrm{\left|\frac{z+1}{z-1}\right|<1} \\

\Rightarrow \mathrm{|z+1|<|z-1|} \\

\Rightarrow \mathrm{(x+1)^{2}+y^{2}<(x-1)^{2}+y^{2}}

\Rightarrow \mathrm{x< 0}

\mathrm{Set \: B}

Using Rotation theorem and circular arc equation

\mathrm{Centre:\left ( 0,-\frac{1}{\sqrt{3}} \right ) }

Hence, \mathrm{A \cap B}

\mathrm{\Rightarrow Centre \left ( 0,\frac{-1}{\sqrt{3}} \right )}

Hence the correct answer is option 2.

Posted by

Suraj Bhandari

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