Verified Solution • 2025–26 Session moderate

QJEEM10Ma5ffa: JEE Main 2010 Mathematics - Class 11 JEEM10Ma5ffa Solution: The number of complex numbers z such that equals :

The number of complex numbers z such that ∣z−1∣=∣z+1∣=∣z−i∣\left| {z - 1} \right| = \left| {z + 1} \right| = \left| {z - i} \right| equals :

Multiple Choice Options

  • A 1
  • B 2
  • C ∞\infty
  • D 0

Step-by-Step Verified Solution

Let z=x+iyz=x+iy

∣z−1∣=∣z+1∣(x−1)2+y2\left| {z - 1} \right| = \left| {z + 1} \right|{\left( {x - 1} \right)^2} + {y^2}

                     =(x+1)2+y2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left( {x + 1} \right)^2} + {y^2}

⇒Re z=0              ⇒x=0\Rightarrow {\mathop{\rm Re}\nolimits} \,z = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow x = 0

∣z−1∣=∣z−i∣(x−1)2+y2\left| {z - 1} \right| = \left| {z - i} \right|{\left( {x - 1} \right)^2} + {y^2}

                     =x2+(y−1)2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {x^2} + {\left( {y - 1} \right)^2}

⇒x=y\Rightarrow x = y

∣z+1∣=∣z−i∣(x+1)2+y2\left| {z + 1} \right| = \left| {z - i} \right|{\left( {x + 1} \right)^2} + {y^2}

                     =x2+(y−1)2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {x^2} + {\left( {y - 1} \right)^2}

Only (0,0)(0,0) will satisfy all conditions.

⇒\Rightarrow Number of complex number z=1z=1

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