Verified Solution • 2025–26 Session moderate

QJEEM14Mad6d0: JEE Main 2014 Mathematics - Class 11 JEEM14Mad6d0 Solution: If z is a complex number such that , then the minimum value of :

If z is a complex number such that  ∣z∣≥2 \,\left| z \right| \ge 2\, , then the minimum value of   ∣z+12∣\,\,\left| {z + {1 \over 2}} \right| :

Multiple Choice Options

  • A is strictly greater that 52{{5 \over 2}}
  • B is strictly greater that 32{{3 \over 2}} but less than 52{{5 \over 2}}
  • C is equal to 52{{5 \over 2}}
  • D lie in the interval (1, 2)

Step-by-Step Verified Solution

We know minimum value of

   ∣Z1+Z2∣   \,\,\,\left| {{Z_1} + {Z_2}} \right|\,\,\, is    ∣∣Z1∣−∣Z2∣∣\,\,\,\left| {\left| {{Z_1}} \right| - \left| {{Z_2}} \right|} \right|

Thus minimum value of

   ∣Z+12∣   \,\,\,\left| {Z + {1 \over 2}} \right|\,\,\, is    ∣∣Z∣−12∣\,\,\,\left| {\left| Z \right| - {1 \over 2}} \right|

                   ≤∣Z+12∣≤∣Z∣+12\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \le \left| {Z + {1 \over 2}} \right| \le \left| Z \right| + {1 \over 2}

Since,    ∣Z∣≥2\,\,\,\left| Z \right| \ge 2

∴\therefore \,\,\,2 - {1 \over 2} < \left| {Z + {1 \over 2}} \right| < 2 + {1 \over 2}

\Rightarrow {3 \over 2} < \left| {Z + {1 \over 2}} \right| < {5 \over 2}

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