If |z - 4| < |z - 2|, its solution is given by

- Re(z) > 3
- Re(z) > 0
- Re(z) < 0
- Re(z) > 2

**Solution**

|z – 4| < |z – 2|

By taking, z = a + ib

|(a – 4) + ib| < |(a – 2) + ib|

(a – 4)^{2} + b^{2} < (a – 2)^{2} + b^{2}

a^{2} + 16 - 8a < a^{2} + 4 - 4a

-8a + 4a < -16 + 4

- 4a < -12

4a > 12

a > 3

**The correct option is A.**