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Find the real and imaginary parts of : $$ \frac {e^{iθ}} {1-λe^{iΦ}} $$

Here i=iota

I have used $ e^{iθ} = \cos θ +i \sin θ $ but I am not able to separate real and imaginary parts. I am not getting any clue how to proceed.

The answer given in my textbook: Real: $ \frac {cos θ - λ cos(θ-Φ)} {1-2λ cos Φ + λ^2} $

Imaginary: $ \frac {sin θ - λ sin(θ-Φ)} {1-2λ cos Φ + λ^2} $

Thank you

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  • $\begingroup$ Is $\lambda$ a real number? $\endgroup$ – user491874 Feb 1 '18 at 19:17
  • $\begingroup$ $i = \iota$?? Shouldn;t $i$ eqaul the imaginary unit $i$ so that $i^2 = -1$? What's iota? $\endgroup$ – fleablood Feb 1 '18 at 19:18
  • $\begingroup$ What does $z = x+iy$ have to do with.... anything. $\endgroup$ – fleablood Feb 1 '18 at 19:19
  • $\begingroup$ In general, we have $$\frac{z}{w}=\frac{z\bar w}{w\bar w}=\frac{z\bar w}{|w|^2}$$ $\endgroup$ – Dave Feb 1 '18 at 19:21
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Well, just do it...

$\frac {e^{iθ}} {1-λe^{iΦ}}=$

$\frac {\cos \theta + i\sin \theta}{1 - \lambda (\cos \Phi + i\sin \Phi)}=$

$\frac {\cos \theta + i\sin \theta}{1 - \lambda \cos \Phi - \lambda i\sin \Phi)}=$

$ \frac {(\cos \theta + i\sin \theta)(1 - \lambda \cos \Phi + \lambda i\sin \Phi)}{(1 - \lambda \cos \Phi - \lambda i\sin \Phi)(1 - \lambda \cos \Phi + \lambda i\sin \Phi)}=$

$\frac {(\cos \theta + i\sin \theta)(1 - \lambda \cos \Phi + \lambda i\sin \Phi)}{(1 - \lambda\cos\Phi)^2 - \lambda^2 \sin^2\Phi}=$

$\frac{\cos \theta(1 - \lambda \cos \Phi)- \sin \theta\lambda \sin \Phi}{(1 - \lambda\cos\Phi)^2 - \lambda^2 \sin^2\Phi}+ i \frac {\sin \theta(\lambda\cos\Phi -1)-\lambda \cos\theta\sin\Phi}{(1 - \lambda\cos\Phi)^2 - \lambda^2 \sin^2\Phi}$

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  • $\begingroup$ thanks for answering I have added answer given in my textbook could you take a look at that? $\endgroup$ – user527235 Feb 1 '18 at 19:47
  • $\begingroup$ Just do trig identities. $\endgroup$ – fleablood Feb 1 '18 at 20:00

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