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Prove the identity: $\operatorname{Im}\left(\dfrac{1+re^{i\theta}}{1-e^{i\theta}}\right) = \dfrac{2r\sin\theta}{1-2r\cos\theta + r^2}$

My only lead is to assume $r$ and $\theta$ are real.

Then,

$\operatorname{Im}\left(\dfrac{1+re^{i\theta}}{1-e^{i\theta}}\right) = \dfrac{r\sin\theta + \sin\theta}{\sin^2+(1-\cos\theta)^2}$

Em.
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David
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2 Answers2

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I suspect there is a missing part in the expression, because the following identity holds:
$ \operatorname{Im}\left(\dfrac{1+re^{i\theta}}{1-re^{i\theta}}\right) = \dfrac{2r\sin\theta}{1-2r\cos\theta + r^2} $

VictorZurkowski
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RTP: $Im\bigg(\dfrac{1+re^{i\theta}}{1-e^{i\theta}}\bigg) = \dfrac{2r\sin\theta}{1-2r\cos\theta + r^2}$=$\frac{2r\sin\theta}{(r-\cos\theta)^2+1-\cos^2\theta}$

$$Im\bigg(\dfrac{1+re^{i\theta}}{1-e^{i\theta}}\bigg) = Im\bigg(\frac{(1+r\cos \theta+ri\sin\theta )(1-\cos\theta +i\sin\theta)}{(1-\cos\theta-i\sin\theta)(1-\cos\theta+i\sin\theta)}\bigg)$$

$$=Im\bigg(\frac{1-\cos\theta+i\sin\theta+r\cos\theta-r\cos^2\theta+r\sin\theta\cos\theta \cdot i +r\sin\theta \cdot i-r\sin\theta\cos\theta\cdot i-r\sin^2\theta}{1-\cos\theta +i\sin\theta -\cos\theta+\cos^2\theta-\cos\theta\sin\theta\cdot i -i\sin\theta +\sin\theta\cos\theta\cdot i +\sin^2\theta}\bigg)$$

$$= \frac{\sin\theta+r\sin\theta}{1-2\cos\theta+\cos^2\theta+\sin^2\theta}$$

$$= \frac{(r+1)\sin\theta}{2(1-\cos\theta)}$$

I don't see how to arrive at the identity from here, could someone comment if they see an error/ way to proceed?

mrnovice
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  • I guess you didn't notice that the OP posted essentially the same expression in the original question: $$\operatorname{Im}\left(\dfrac{1+re^{i\theta}}{1-e^{i\theta}}\right) = \dfrac{r\sin\theta + \sin\theta}{\sin^2+(1-\cos\theta)^2}$$ – dxiv Mar 30 '17 at 05:40
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    @dxiv Oh, yeah I may have just attempted the question without reading further xD – mrnovice Mar 30 '17 at 05:42