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About the complex number​

About the complex number​-example-1
User ErsatzRyan
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We're given


z = \mathrm{cis}(\theta) = \cos(\theta) + i \sin(\theta)

By de Moivre's theorem,


z^n = \cos(n\theta) + i \sin(n\theta)

so that


(z^2 - 1)/(z^2 + 1) = (\cos(2\theta) + i \sin(2\theta) - 1)/(\cos(2\theta) + i \sin(2\theta) + 1)

Multiply uniformly by the conjugate of the denominator.


(z^2 - 1)/(z^2 + 1) = (\cos(2\theta) + i \sin(2\theta) - 1)/(\cos(2\theta) + i \sin(2\theta) + 1)\cdot(\cos(2\theta) - i \sin(2\theta) + 1)/(\cos(2\theta) - i \sin(2\theta) + 1)


(z^2 - 1)/(z^2 + 1) = (\cos^2(2\theta) - 1 + 2i\sin(2\theta) + \sin^2(2\theta))/((\cos(2\theta)+1)^2 + \sin^2(2\theta))


(z^2 - 1)/(z^2 + 1) = (2i\sin(2\theta))/(\cos^2(2\theta) + 2\cos(2\theta) + 1+ \sin^2(2\theta))


(z^2 - 1)/(z^2 + 1) = (2i\sin(2\theta))/(2\cos(2\theta) + 2)


(z^2 - 1)/(z^2 + 1) = (i\sin(2\theta))/(\cos(2\theta) + 1)


(z^2 - 1)/(z^2 + 1) = (2i\sin(\theta)\cos(\theta))/(2\cos^2(\theta) - 1 + 1)


(z^2 - 1)/(z^2 + 1) = (i\sin(\theta))/(\cos(\theta))


(z^2 - 1)/(z^2 + 1) = i\tan(\theta)

QED

User Megaetron
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