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

AI Thread Summary
To simplify complex numbers and find their real and imaginary parts, the discussion emphasizes the importance of showing detailed work. A participant initially presented a result but was corrected for a typo regarding the exponential form of complex numbers. The correct expression for e^{-i2x} was highlighted, and it was noted that certain values of x should be excluded before simplification. Additionally, factoring out e^{ix} from the denominator was suggested as a more straightforward approach. Overall, clarity and accuracy in mathematical expressions are crucial for correct results.
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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seems fair to me
 
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:

\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}}

ehild
 
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