Find the real value of a for which 3i3 – 3ai2 + (1 – a) i + 5 is real.
a = 2
Explanation
Z is a purely real, it means Im (z) = 0
Z = 3i3-3ai2+ (1-a) i+5
= -3i+3a+ (1-a) i+5
= (3a+5)+i(-3+1-a)
= (3a+5)+i(-2-a)
Re(z) = 3a+5 , Im(z) = (-2-a)
Z is a real so, Im (z) = 0
-2-a = 0
a = -2