Question:
The real value of 'a' for which 3i3 – 2ai2 + (1 – a) i + 5 is real is ____________.
Solution:
$3 \hat{}^{3}-2 a i^{2}+(1-a) i+5$
i.e $32^{2} i-2 a(-1)+(1-a) i+5$
i.e 3i + 2a + (1 – a) i + 5
i.e 2a + 5 + i(1 – a – 3)
i.e 2a + 5 + i (–a – 2)
Since the above expression is given to be real
⇒ –a – 2 = 0
⇒ a = – 2
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