How do I simplify complex numbers and find real and imaginary parts?

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danik_ejik
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Hello,
I need to simplify and find Re and I am of
gif.latex?\frac{1+e^{ix}}{1+e^{i2x}}.gif
.

I've multiplied by
gif.latex?\frac{1+e^{-i2x}}{1+e^{-i2x}}.gif
and got that Re=
gif.latex?\frac{1}{2}\cdot(1+\frac{1}{cos(x)}).gif
and Im=
gif.latex?\frac{-1}{2}tan(x).gif
.

Is that the correct way and a correct result ?
 
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Show your work in detail. The final result is wrong. How did you get cos x in the denominator?

ehild
 
here's my work in details,
http://img811.imageshack.us/img811/7444/exsol.jpg"
 
Last edited by a moderator:
It is better if you type in your work.

I found a typo: You wrote that e-ix=cos (2x)-isin(2x).

ehild
 
it should have been
[URL]http://latex.codecogs.com/gif.latex?e^{-i2x}[/URL]
so the latter expansion is correct.
 
Last edited by a moderator:
OK, I see now. But you should exclude x=(2k+1)pi/2 before multiplying both numerator and denominator by 1+exp(-2ix) which is 0 if x=(2k+1)pi/2.
 
oh, forgot about that.
thanks
 
It would have been simpler to factor out exp(ix) from the denominator:

[tex]\frac{1+e^{ix}}{1+e^{2ix}}=\frac{1+e^{ix}}{e^{ix}(e^{-ix}+e^{ix})}=\frac{e^{-ix}(1+e^{ix})}{2\cos{x}}[/tex]

ehild