Question

Why is \[\ Im(\frac{e^{i \theta}-re^{-i \theta}}{1+r^2-r(e^{i2 \theta}+re^{-i 2\theta)}) = Im \frac{sin( \theta )+rsin( \theta)}{1+r^2-2rcos( 2\theta)} \]?

Answer 1

i cant see latex :)

Answer 2

\[\Im (\frac{ e^{ix}-r e^{-ix} }{ 1+r^2-r(e^{i2x}+e^{-i2x}) })=\frac{sinx+rsinx}{ 1+r^2-2rcos(2x) }\]

Answer 3

\[\Im(\frac{a}{b}) \neq \frac{\Im(a)}{\Im(b)}\]

Answer 4

denum is a real function

Answer 5

\[e^{2ix}+e^{-2ix}=2\cos 2x\]

Answer 6

Surely it has an imaginary part?\[\frac{ e^{ix} }{ 1-re^{i2x} }\]

Answer 7

Thought it was more complicated than it really was, thanks!