If a complex number z = 4+3 i 1-2 i ( where i= -1 ) is purely real, then the value of is

Step-by-step solution

The complex number z = 4+3i 1-2i is purely real if and only if its imaginary part is zero. Rationalizing the denominator by multiplying numerator and denominator by the conjugate 1+2i : z = (4+3i )(1+2i ) (1-2i )(1+2i ) = 4 + 8i + 3i + 6i^2 ^2 1 - 4i^2 ^2 Substituting i^2 = -1 : z = 4 + 11i - 6 ^2 1 + 4 ^2 = 4 - 6 ^2 1 + 4 ^2 + i 11 1 + 4 ^2 The imaginary part must vanish for z to be real. Since 1 + 4 ^2 > 0 for all , we require 11 = 0 , hence = 0 . The general solution is = n for n Z , which corresponds to option