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Complex exponential to trigonometric simplification

  1. Apr 15, 2013 #1
    1. The problem statement, all variables and given/known data
    Given (e^(ix) - 1)^2 , show that it is equal to 2-2cosx

    2. Relevant equations
    e^ix = cosx + isinx

    3. The attempt at a solution
    After subbing in Euler identity and expanding I get:

    cos(x)^2+sin(x)^2-2cosx-2jsinx+2jcosxsinx + 1

    after using the addtion formulas I get

    cos(2x)+jsin(2x)-2cosx-2jsinx+1

    seems like i'm missing a fundamental trig ID... Please help, and let me know if my approach is correct.

    Thanks
     
  2. jcsd
  3. Apr 15, 2013 #2

    SammyS

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    Please state the problem, word for word as it was given to you.

    It's obvious that (eix-1)2 ≠ 2 - 2cos(x) .

    Also, please stick with one symbol for the imaginary unit: either i or j . Don't switch back and forth.
     
  4. Apr 15, 2013 #3
    Sorry EE student here.

    Hm, maybe I'm missing the bigger picture then. This is part of my digital systems homework, regarding filter design. The question I posed is of my own formulation, as that is where I am stuck with the with problem. If the question is nonsensical i must be missing a cancellation. I attached a picture so you can hopefully help me figure it out. ixC4zZN.png
     
  5. Apr 15, 2013 #4

    SammyS

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    That makes things clearer.

    It does look like ##\displaystyle \ \left|\left(e^{\,j\,\theta}-1 \right)^2\right|=2-2\cos(\theta) \,, \ ## where j is the imaginary unit.
     
  6. Apr 15, 2013 #5
    Ok. Could you go through the steps?
     
  7. Apr 15, 2013 #6

    SammyS

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    Let's help you go through the steps.

    One thing that may help: (a + bj)(a - bj) = a2 + b2 = |a + bj|2 .

    The other is the Euler identity, so that [itex]\displaystyle \ e^{\,x\,j}-1=(\cos(x)-1)+j\sin(x)\ .[/itex]

    See what you can do with that.
     
  8. Apr 15, 2013 #7
    Well, I got that far on my own. The final simplification is the issue.

    so after expansion I get:

    cos(x)^2-2cosx+1-sin(x)^2

    I use Pythagorean identity to get:

    2cos(x)^2 - 1 + 2cosx + 1

    so 2cos(x)^2 + 2cosx

    What's next? Or where's the error?
     
  9. Apr 15, 2013 #8

    Mark44

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    Starting with |(ejx - 1)2|, show us what you did.
     
  10. Apr 15, 2013 #9
    I did...

    So from the start:

    |(e^jx - 1)^2| = | (cosx +jsinx -1)^2|

    using |a+jb|^2=a^2+b^2

    we get

    (cosx - 1)^2 + (jsinx)^2

    = cos^2(x)-2cosx +1 - sin^2(x)

    using sin^2(x)=1-cos^2(x)

    = 2cos^2(x) - 2cos(x)
     
  11. Apr 15, 2013 #10

    Mark44

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    The last term above has the wrong sign. In the formula you used, b is the coefficient of j, and doesn't include it. You should get (cosx - 1)2 + sin2x
     
  12. Apr 15, 2013 #11
    Oh I found my mistake. I applied the identity incorrectly.

    So it is just (cosx -1)^2 + (sin^2(x))

    which will come to 2 - 2cos(x)

    Thanks Sammy!
     
  13. Apr 15, 2013 #12
    Yup just saw it myself! Thank yoU!
     
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