Question

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

• A. -2
• B. -4
• C. -5
• D. -1

Answer 1

Answer: (B)

-4

$z$ is a non-real complex number.

let $z=r{e}^{i\theta }$

$\frac{Im{z}^5}{(Imz{)}^5}=\frac{r^5\sin 5\theta }{r^5{\sin }^5\theta }$

say, $f\left(\theta \right)=\frac{\sin 5\theta }{{\sin }^5\theta }$

equating first derivative to $0$

$f^{\prime }\left(\theta \right)=0$

$\Rightarrow {\sin }^5\theta \cdot 5\cos 5\theta -\sin 5\theta \cdot 5{\sin }^4\theta \cdot \cos \theta$

$\Rightarrow \sin \theta \cdot \cos 5\theta =\cos \theta \cdot \sin 5\theta$

$\Rightarrow \tan \theta =\tan 5\theta$

$\Rightarrow \theta =0,\pi /4...$

but $f\left(0\right)$ is indeterminate

$f\left(\pi /4\right)=-4$

$\therefore$ Minimum value of $\frac{Im{z}^5}{(Imz{)}^5}$ is $-4$

Hence, option $B$.