JEE Main 2015 — Mathematics Algebra

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

$${z}^{5} ={r}^{5} (\text{cos}5\theta +i\text{sin}5\theta )$$z5=r5(cos5θ+isin5θ)

$$Im({z}^{5})={r}^{5}\mathrm{sin}5\theta$$Im(z5)=r5sin5θ

and ${(Im z)}^{5}={r}^{5}{\mathrm{sin}}^{5}\theta$(Imz)5=r5sin5θ

$\frac{Im ({z}^{5})}{{(Im z)}^{5}}= \frac{\mathrm{sin}5\theta }{{\mathrm{sin}}^{5}\theta }=A$(Imz)5Im(z5)​=sin5θsin5θ​=A(Let)

$$\Rightarrow \frac{dA}{d\theta }=\frac{5{\mathrm{sin}}^{5}\theta \mathrm{cos}5\theta -5\mathrm{sin}5\theta {\mathrm{sin}}^{4}\theta \mathrm{cos}\theta }{{({\mathrm{sin}}^{5}\theta )}^{2}}$$⇒dθdA​=(sin5θ)25sin5θcos5θ−5sin5θsin4θcosθ​

$$\Rightarrow \frac{dA}{d\theta }=\frac{5{\mathrm{sin}}^{4}\theta (\mathrm{sin}\theta \mathrm{cos}5\theta -\mathrm{sin}5\theta \mathrm{cos}\theta )}{{\mathrm{sin}}^{10}\theta }$$⇒dθdA​=sin10θ5sin4θ(sinθcos5θ−sin5θcosθ)​

$$\Rightarrow \frac{ dA}{ d\theta } =\frac{5}{ { \mathrm{sin}}^{6} \theta } [ \mathrm{sin}( -4\theta )]=0$$⇒dθdA​=sin6θ5​[sin(−4θ)]=0

$$\Rightarrow \theta =\frac{\pi }{4}$$⇒θ=4π​

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

$$\Rightarrow A=\frac{\mathrm{sin}5\frac{\pi }{4}}{{(\mathrm{sin}\frac{\pi }{4})}^{5}}$$⇒A=(sin4π​)5sin54π​​

$$=\frac{\frac{-1}{\sqrt{2}}}{{(\frac{1}{\sqrt{2}})}^{5}}$$=(2​1​)52​−1​​

$$=-{(\sqrt{2})}^{4}=-4$$=−(2​)4=−4