Mathematics - Complex Number Question with Solution | TestHub

Let, $z=r(\text{cos}\theta +i\text{sin}\theta )$

$z^5=r^5(\text{cos}5\theta +i\text{sin}5\theta )$

$Im\left(z^5\right)=r^5\sin 5\theta$

and ${\left(Im z\right)}^5=r^5{\sin }^5\theta$

$\frac{Im \left(z^5\right)}{{\left(Im z\right)}^5}= \frac{\sin 5\theta }{{\sin }^5\theta }=\mathrm{A }$(Let)

$\Rightarrow \frac{dA}{d\theta }=\frac{5{\sin }^5\theta \cos 5\theta -5\sin 5\theta {\sin }^4\theta \cos \theta }{{\left({\sin }^5\theta \right)}^2}$

$\Rightarrow \frac{dA}{d\theta }=\frac{5{\sin }^4\theta \left(\sin \theta \cos 5\theta -\sin 5\theta \cos \theta \right)}{{\sin }^{10}\theta }$

$\Rightarrow \frac{dA}{d\theta }=\frac{5}{{\sin }^6\theta }\left[\sin \left(-4\theta \right)\right]=0$

$\Rightarrow \theta =\frac{\pi }{4}$

The minimum value of $A$ will be at  $\theta =\frac{\pi }{4}.$

$\Rightarrow A=\frac{\sin 5\frac{\pi }{4}}{{\left(\sin \frac{\pi }{4}\right)}^5}$
$=\frac{\frac{-1}{\sqrt{2}}}{{\left(\frac{1}{\sqrt{2}}\right)}^5}$

$=-{\left(\sqrt{2}\right)}^4=-4$