Question

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

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

Answer 1

Answer: (C)

Let $z=r(\cos \theta +i\sin \theta )$
$\therefore \frac{Im\left(z^5\right)}{(Imz{)}^5}=\frac{Im{\left(r{e}^{i\theta }\right)}^5}{{\left(Im\left(r{e}^{1\theta }\right)\right)}^5}$
$=\frac{Im\left(r^5{e}^{si\theta }\right)}{(r\sin \theta {)}^5}=\frac{r^5{\sin }^{5\theta }}{r^5{\sin }^{5\theta }}=\frac{\sin 5\theta }{(\sin 5\theta )}$...(1)
$\because \cos 5\theta +i\sin 5\theta ={\left(e^{\prime }\theta \right)}^5=(\cos \theta +i\sin \theta {)}^5$
${=}^5{C}_0(\cos \theta {)}^5+{}^5{C}_1(\cos \theta {)}^4(i\sin \theta )+{}^5{C}_2(\cos \theta {)}^3(i\sin \theta {)}^2+{}^5{C}_3(\cos \theta {)}^2(i\sin \theta {)}^3+{}^5{C}_4(\cos \theta {)}^1(i\sin \theta {)}^4+{}^5{C}_5(\cos \theta {)}^{\circ }\sin \theta {)}^5$
$\Rightarrow \sin 5\theta ={}^5{C}_1{\cos }^4\theta \sin \theta$
$\Rightarrow {}^{-5}{C}_3{\cos }^2\theta {\sin }^3\theta +{}^5{C}_5{\sin }^5\theta$
$\Rightarrow \frac{\sin 5\theta }{(\sin \theta {)}^5}=\frac{5{\cos }^4\theta }{{\sin }^4\theta }-\frac{10{\cos }^2\theta }{{\sin }^2\theta }+1$
$=5{\cot }^4\theta -10{\cot }^2\theta +1=5\left[{\cot }^4\theta -2{\cot }^2\theta +1\right]-4$
$=5{\left[{\cot }^2\theta -1\right]}^2-4\ge -4$
$\Rightarrow$ Least value is $-4$