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Question

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


A
-2
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B
-4
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C
-5
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D
-1
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Solution

The correct option is A -4
$$z$$ is a non-real complex number.

let $$z=re^{i\theta}$$

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

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

equating first derivative to $$0$$

$$f'(\theta)=0$$

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

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

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

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

but $$f(0)$$ is indeterminate

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

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

Hence, option $$B$$.

Mathematics

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