Find the real and imaginary parts

Question

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

Answer 1

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)}=$

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

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

Answer 2

Here is the answer. $$ \frac {e^{iθ}} {1-λe^{iΦ}}=\frac {e^{iθ}(1-λe^{-iΦ})} {(1-λe^{iΦ)}(1-λe^{-iΦ})} =\frac {e^{iθ}-λe^{i(\theta-Φ)}} {(1-λe^{iΦ)}(1-λe^{-iΦ})}= \frac {cos θ - λ cos(θ-Φ)+i \left(sin θ - λ sin(θ-Φ)\right)}{1-2λ cos Φ + λ^2}. $$ You can write the real and imaginary parts separately.