Q. 22

# 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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