Q22 of 153 Page 14

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


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