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This was an answer provided to a question I asked previously. I followed the other approaches to the question; however, I couldn't seem to follow this one:

$\frac{1-z}{1+z}=\dfrac{1-e^{i\theta}}{1+e^{i\theta}}=\dfrac{e^{-\frac{i\theta}{2}}-e^{\frac{i\theta}{2}}}{e^{-\frac{i\theta}{2}}+e^{\frac{i\theta}{2}}}=\dfrac{-2i\sin\frac{\theta}{2}}{2\cos\frac{\theta}{2}}=-i\tan\frac{\theta}{2}$

How do you progress from the 2nd term to the 3rd term and then from the 3rd term to the 4th? What identity/logic is used etc.?

2 Answers 2

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To get from the second term to the third, multiply numerator and denominator by $e^{-\frac{i\theta}{2}}$:

$ e^{-\frac{i\theta}{2}}(1 \pm e^{i\theta}) = e^{-\frac{i\theta}{2}} \pm e^{\frac{i\theta}{2}} $

Then to get to the fourth term from the third, use the following identities:

$ \sin x = \frac{e^{ix}-e^{-ix}}{2i} $

$ \cos x = \frac{e^{ix}+e^{-ix}}{2} $

These can be derived straightforwardly from $e^{ix} = \cos x + i \sin x$.

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    @stariz77: Yes, that's correct.2012-03-19
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To get from the second term to the third, multiply numerator and denominator by $e^{-i\theta/2}$; to get from the third term to the fourth, use the identities $\cos \theta=\frac12(e^{i\theta}+e^{-i\theta})$ and $\sin \theta=\frac{i}2(e^{-i\theta}-e^{i\theta})=\frac{e^{i\theta}-e^{-i\theta}}{2i}\;,$ which are derivable from $e^{i\theta}=\cos\theta+i\sin\theta$ and $e^{-i\theta}=\cos(-\theta)+i\sin(-\theta)=$ $\cos\theta-i\sin\theta$.

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    Be$a$t me to it. +12012-03-19