# Find the real and imaginary parts

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

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

Well, just do it...

$\frac {e^{iθ}} {1-λe^{iΦ}}=$

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

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

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

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

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

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

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.