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  • Stuck here, help me understand: - Complex numbers and quadratic equations - JEE Main-2

If z\neq 0 and z moves such that \left |z-1 \right |=\left |z+1 \right |  then \left | arg\left ( z \right ) \right | equals

  • Option 1)

    0

  • Option 2)

    \frac{\pi }{4}

  • Option 3)

    \frac{\pi }{2}

  • Option 4)

    \frac{3\pi }{4}

 

Answers (1)

best_answer

\because \left | Z-1 \right |=\left | Z+1 \right |\Rightarrow \left | Z-1 \right |=\left | Z-\left ( -1 \right ) \right |

\Rightarrow Z lies on perpendicular bisector of line joining complex numbers 

Z_{1}=1\; \; and\; \; Z_{2}=-1  , so locus of Z will be imaginary axis.

\Rightarrow \arg \left ( Z \right )=\frac{\pi }{2},\frac{-\pi }{2}

 

Perpendicular bisector -

Locus of point equidistant from two given points.

\left |z-z_{1} \right |=\left |z-z_{2} \right |

z will lie on perpendicular bisector of line joining z_{1} and z_{2} .

- wherein

z_{1} and z_{2} are any two fixed points . z is a moving point in the plain which is equidistant from z_{1} and z_{2} .so z will lie on perpendicular bisector

 

  


Option 1)

0

This is incorrect

Option 2)

\frac{\pi }{4}

This is incorrect

Option 3)

\frac{\pi }{2}

This is correct

Option 4)

\frac{3\pi }{4}

This is incorrect

Posted by

divya.saini

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