Question

If $z$ is a non-real complex number, then the minimum value of $\frac{Im\,z^{5}}{\left(Im\,z\right)^{5}}$ is :

• A. $-1$
• B. $-2$
• C. $-4$
• D. $-5$

Answer 1

Answer: (C)

Let $z=r(\cos \theta+i \sin \theta)$
$\therefore \frac{Im\left(z^{5}\right)}{(Im z)^{5}}=\frac{Im\left(r e^{i \theta}\right)^{5}}{\left(Im\left(r e^{1 \theta}\right)\right)^{5}}$
$=\frac{Im\left(r^{5} e^{s i \theta}\right)}{(r \sin \theta)^{5}}=\frac{r^{5} \sin ^{5 \theta}}{r^{5} \sin ^{5 \theta}}=\frac{\sin 5 \theta}{(\sin 5 \theta)}$...(1)
$\because \cos 5 \theta+i \sin 5 \theta=\left(e^{\prime} \theta\right)^{5}=(\cos \theta +i \sin \theta)^{5}$
$=^{5} C_{0}(\cos \theta)^{5}+{ }^{5} C_{1}(\cos \theta)^{4}(i \sin \theta)+{ }^{5} C_{2}(\cos \theta)^{3}(i \sin \theta)^{2}+{ }^{5} C_{3}(\cos \theta)^{2}(i \sin \theta)^{3}+{ }^{5} C_{4}(\cos \theta)^{1}(i \sin \theta)^{4}+{ }^{5} C_{5}(\cos \theta)^{\circ} \sin \theta)^{5}$
$\Rightarrow \sin 5 \theta={ }^{5} C_{1} \cos ^{4} \theta \sin \theta$
$\Rightarrow { }^{-5} C_{3} \cos ^{2} \theta \sin ^{3} \theta+{ }^{5} C_{5} \sin ^{5} \theta$
$\Rightarrow \frac{\sin 5 \theta}{(\sin \theta)^{5}}=\frac{5 \cos ^{4} \theta}{\sin ^{4} \theta}-\frac{10 \cos ^{2} \theta}{\sin ^{2} \theta}+1$
$=5 \cot ^{4} \theta-10 \cot ^{2} \theta+1=5\left[\cot ^{4} \theta-2 \cot ^{2} \theta+1\right]-4$
$=5\left[\cot ^{2} \theta-1\right]^{2}-4 \geq-4$
$\Rightarrow $ Least value is $- 4$