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  • Solve this problem - Complex numbers and quadratic equations - JEE Main-9

if     z\neq 1\; and\, \frac{z^{2}}{z-1}     is real, then the point represented by the complex number z lies

  • Option 1)

    either on the real axis or on a circle passing through the origin

  • Option 2)

    on a circle with centre at the origin

  • Option 3)

    either on the real axis or on a circle not passing through the origin

  • Option 4)

    on the imaginary axis

 

Answers (2)

best_answer

z\neq 1,\frac{z^{2}}{z-1}\; is\; real

If\; z\; is \; a\; real\; number, then\; \frac{z^{2}}{z-1}\; is \; real

Let\;\; z=x+iy

\therefore \; \; \; \frac{(x^{2}-y^{2}+2xiy)((x-1)-iy)}{(x-1)^{2}+y^{2}}\; is\; real

\Rightarrow \; \; -y(x^{2}-y^{2})+2xy(x-1)=0

\Rightarrow \; \; y(x^{2}+y^{2}-2x)=0\Rightarrow y=0\; or\; x^{2}+y^{2}-2x=0

\therefore \; \; \; z\; \; lies \; on\; real \; axis\; or \; on\; a \; circle \; passing \; through\: origin.


Option 1)

either on the real axis or on a circle passing through the origin

Option 2)

on a circle with centre at the origin

Option 3)

either on the real axis or on a circle not passing through the origin

Option 4)

on the imaginary axis

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