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QJEEM12Ma3690: JEE Main 2012 Mathematics - Class 11 JEEM12Ma3690 Solution: If and is real, then the point represented by the complex…

If z≠1z \ne 1 and  z2z−1 \,{{{z^2}} \over {z - 1}}\, is real, then the point represented by the complex number z lies :

Multiple Choice Options

  • A either on the real axis or a circle passing through the origin.
  • B on a circle with centre at the origin
  • C either on real axis or on a circle not passing through the origin.
  • D on the imaginary axis.

Step-by-Step Verified Solution

Let z=x+iyz = x + iy

∴\therefore     z2=x2−y2+2ixy\,\,\,\,{z^2} = {x^2} - {y^2} + 2ixy

Now z2z−1{{{z^2}} \over {z - 1}} is real

⇒Im(z2z−1)=0\Rightarrow {\mathop{\rm Im}\nolimits} \left( {{{{z^2}} \over {z - 1}}} \right) = 0

⇒Im(x2−y2+2ixy(x−1)+iy)=0\Rightarrow {\mathop{\rm Im}\nolimits} \left( {{{{x^2} - {y^2} + 2ixy} \over {\left( {x - 1} \right) + iy}}} \right) = 0

⇒Im[(x2−y2+2ixy)(x−1)−iy)]=0\Rightarrow {\mathop{\rm Im}\nolimits} \left[ {\left( {{x^2} - {y^2} + 2ixy} \right)\left. {\left( {x - 1} \right) - iy} \right)} \right] = 0

⇒2xy(x−1)−y(x2−y2)=0\Rightarrow 2xy\left( {x - 1} \right) - y\left( {{x^2} - {y^2}} \right) = 0

⇒y(x2+y2−2x)=0\Rightarrow y\left( {{x^2} + {y^2} - 2x} \right) = 0

⇒y=0; x2+y2−2x=0\Rightarrow y = 0;\,{x^2} + {y^2} - 2x = 0

∴\therefore     \,\,\,\, zz lies either on real axis or on a circle through origin.

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