Question

$Prove \ \dfrac{w-1}{w+1} = i \tan \theta \ where \ w \in \mathbb{C},|w|=1,arg(w) = 2\theta$

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

Let ω = cos 2θ + i sin 2θ

Given,

$\frac{w - 1}{w + 1} = \frac{cos 2\theta - 1 + i sin 2\theta}{cos 2\theta + 1 + i sin 2\theta}$

On rationalization, we get

$\Rightarrow \frac{(cos 2\theta - 1 + i sin 2\theta)(cos 2\theta + 1 - i sin 2\theta)}{(cos 2\theta - 1 + i sin 2\theta)(cos 2\theta + 1 - i sin 2\theta)}$

$\Rightarrow \frac{cos^2 2\theta + sin^2 2\theta - 1 + 2i sin 2\theta}{cos^2 2\theta + sin^2 2\theta + 1 + 2 cos 2\theta}$

$\Rightarrow \frac{2isin2\theta}{2 + 2cos2\theta}$

$\Rightarrow \frac{2isin\theta cos\theta}{1 + 2cos^2 \theta - 1}$

$\Rightarrow \frac{isin\theta cos\theta}{cos^2\theta}$

$\Rightarrow \bold {i tan \theta}$

Hope it helps!