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

Option 1)
- 1

Option 2)
- 2

Option 3)
- 4

Option 4)
- 5

Answers (2)

best_answer

As we have learned

Polar Form of a Complex Number -

$z=r(cos\theta+isin\theta)$

- wherein

r= modulus of z and $\theta$ is the argument of z

Euler's Form of a Complex number -

$z=re^{i\theta}$

- wherein

r denotes modulus of z and $\theta$ denotes argument of z.

$z= x+iy = r( \cos \theta+ i\sin \theta )$

$= re^{i\theta }$

So, $Im z^5= Im(re^{i\theta })^5$

$= Im(r^5e^{i\theta5 })$

$= r^5 \sin 5 \theta$

$(imz)^5 = (r\sin \theta)^5$

$= (r^5\sin^5 \theta)$

So, $\frac{Imz^5}{(Imz)^5}= \frac{\sin 5 \theta}{\sin^5 \theta}$

for minimum value , diffrentiating w.r.t $\theta$

So, $\frac{\sin^5 \theta \cdot 5 \cos \theta - 5 \sin 5\theta \sin ^4\theta \cos \theta }{\sin ^{10} \theta }$

$\Rightarrow \sin \theta \cdot \cos 5\theta - \sin 5 \theta \cos \theta = 0$

$\Rightarrow \sin 4\theta \cdot = 0$

$4 \theta = n \pi$

$\theta = n \pi /4$

for n= 1

$\frac{\sin 5 \theta}{\sin ^5 \theta }= \frac{-VV_2}{1/4V_2}= -4$

Option 1)

- 1

Option 2)

- 2

Option 3)

- 4

Option 4)

- 5

Posted by

Himanshu

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