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Question: If \(\theta - \pi\)and \(z = \sin\alpha + i(1 - \cos\alpha)\), then \(2\sin\frac{\alpha}{2}\)is equa...

If θ−π\theta - \piand z=sin⁡α+i(1−cos⁡α)z = \sin\alpha + i(1 - \cos\alpha), then 2sin⁡α22\sin\frac{\alpha}{2}is equal to.

A

0

B

Purely imaginary

C

Purely real

D

None of these

Answer

0

Explanation

Solution

We have ∣z∣∣ω∣=1|z||\omega| = 1

⇒ arg⁡(zω)=π2⇒zω=i\arg\left( \frac{z}{\omega} \right) = \frac{\pi}{2} \Rightarrow \frac{z}{\omega} = i

⇒ ∣zω∣=1\left| \frac{z}{\omega} \right| = 1

Let ∣z∣=∣ω∣=1|z| = |\omega| = 1, then zˉω=−zωˉ=−zωωˉω\bar{z}\omega = - z\bar{\omega} = \frac{- z}{\omega}\bar{\omega}\omega

∴ zˉω=−i∣ω∣2=−i.z=x+iy\bar{z}\omega = - i|\omega|^{2} = - i.z = x + iy

and z2z_{2}z2=π−ar⥂g(z)z_{2} = \pi - ar ⥂ g(z)

Hence 12∣z∣2=18\frac{1}{2}|z|^{2} = 18.