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Q.
If $z = \left(\frac{\sqrt{3}}{2} + \frac{i}{2}\right)^{5} + \left(\frac{\sqrt{3}}{2} - \frac{i}{2}\right)^{5} $, then
Binomial Theorem
Solution:
On simplification, we get $z = 2 \left[^{5}C_{0} \left(\frac{\sqrt{3}}{2}\right)^{2} + ^{5}C_{2} \left(\frac{\sqrt{3}}{2}\right)^{3} \left(\frac{i}{2}\right)^{2} + ^{5}C_{4} \left(\frac{\sqrt{3}}{2}\right)\left(\frac{i}{2}\right)^{4}\right] $
Since $i^2 = - 1$ and $i^4 = 1, z$ will not contain any i and hence $I_m (z) = 0$