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

In summary, the conversation is about simplifying and finding the real and imaginary parts of a complex number. The person provides their work and it is pointed out that there is a typo and a mistake in the final result. They then make corrections and discuss a simpler way to solve the problem.
  • #1
danik_ejik
18
0
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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  • #2
seems fair to me
 
  • #3
Show your work in detail. The final result is wrong. How did you get cos x in the denominator?

ehild
 
  • #4
here's my work in details,
http://img811.imageshack.us/img811/7444/exsol.jpg"
 
Last edited by a moderator:
  • #5
It is better if you type in your work.

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

ehild
 
  • #6
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:
  • #7
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.
 
  • #8
oh, forgot about that.
thanks
 
  • #9
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
 

1. What are complex numbers?

Complex numbers are numbers that are expressed in the form a + bi, where a and b are real numbers and i is the imaginary unit, which is defined as the square root of -1.

2. How do you simplify complex numbers?

To simplify complex numbers, you need to combine like terms. This means adding or subtracting the real parts and adding or subtracting the imaginary parts. For example, to simplify (3 + 2i) + (4 - 5i), you would combine the real parts (3 + 4 = 7) and the imaginary parts (2i - 5i = -3i), giving you a simplified form of 7 - 3i.

3. Can you simplify a complex number with a negative square root?

Yes, you can simplify a complex number with a negative square root. The key is to rewrite the square root as i and then combine like terms as usual. For example, to simplify 3 + 2√-1, you would rewrite √-1 as i, giving you 3 + 2i.

4. What is the difference between simplifying and evaluating complex numbers?

Simplifying complex numbers involves combining like terms to get a more simplified form, while evaluating complex numbers involves finding the numerical value of a complex expression. For example, simplifying (3 + 2i) + (4 - 5i) would give you 7 - 3i, while evaluating this expression at x = 2 would give you 11 - 3i.

5. Can you simplify an expression with multiple complex numbers?

Yes, you can simplify an expression with multiple complex numbers by combining like terms, just as you would with a single complex number. For example, to simplify (3 + 2i) + (-4 + 5i) + (2 - i), you would combine the real parts (3 - 4 + 2 = 1) and the imaginary parts (2i + 5i - i = 6i), giving you a simplified form of 1 + 6i.

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