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Need a review on Complicated explonential math

  1. Sep 23, 2007 #1
    I completely forget how to do this... can someone give me a review on how to do this?

    For example:
    z=8e^(i(pi)/3)
    w=4e^(i(pi)/6)

    z+w= ?

    This is an example in my book but no steps are given for this, the answer they give is
    (4 + 2sqrt[3]) + (4sqrt[3] + 2) i

    I have virtually no memory of this...
    I'd greatly appreciate any guidance I can get with this.
    Thanks!
     
  2. jcsd
  3. Sep 23, 2007 #2
    this isn't log rules, this is euler's identity : [itex]e^{i\theta}=cos{\theta}+isin{\theta}[/itex]

    i don't think you can generally add thing with different exponents, i mean you can't simplify [itex] x^2 + x^3[/itex] right
     
  4. Sep 23, 2007 #3

    mathman

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    Use Euler's identity to get z=8cos(pi/3)+8isin(pi/3) and w=4cos(pi/6)+4isin(pi/6).

    cos(pi/3)=sin(pi/6)=1/2, cos(pi/6)=sin(pi/3)=[sqrt(3)]/2. You can finish up to get the book answer.
     
  5. Sep 24, 2007 #4
    Can you assist me in how to do this step?
    I can't seem to 'get', for example, z=8cos(pi/3)+8isin(pi/3) using that identity.
     
  6. Sep 24, 2007 #5
    How did you manage to get rid of the pesky e using that identity?
     
  7. Sep 24, 2007 #6

    learningphysics

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    z=8e^(i(pi)/3)

    From ice109's post: Euler's identity is

    [itex]e^{i\theta}=cos{\theta}+isin{\theta}[/itex]

    directly applying this integral to z=8e^(i(pi)/3)... here theta = pi/3

    so z = 8(cos(pi/3) + isin(pi/3))
     
  8. Sep 24, 2007 #7
    I'm blind.

    Ok, that's not hard. What about the next step of getting the pair of three equalities?

    cos(pi/3)=sin(pi/6)=1/2, cos(pi/6)=sin(pi/3)=[sqrt(3)]/2
     
  9. Sep 24, 2007 #8

    learningphysics

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    pi/3 = 60. pi/6=30

    cos(60) = sin(30) (for any angle x<90, cos(x) = sin(90-x), sin(x) = cos(90-x)

    so, cos(60) = sin(30) = 1/2, cos(30) = sin(60) = sqrt(3)/2
     
    Last edited: Sep 24, 2007
  10. Sep 24, 2007 #9
    Ok I can see they are equal.

    So, is the logic that you simplify each to 1/2 and sqrt[3]/2 and then add those together?
     
  11. Sep 24, 2007 #10

    learningphysics

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    gather the complex terms (the terms that are multiplied by i) and the real terms together.
     
  12. Sep 24, 2007 #11
    I have z+w

    z+w = 8cos(pi/3) + 8isin(pi/3) + 4cos(pi/6) + isin(pi/6)

    Gather them.....
     
  13. Sep 24, 2007 #12

    learningphysics

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    yup... gather the "i" terms together... I mean like this:

    z+w = 8cos(pi/3) + 4cos(pi/6) + i[8sin(pi/3) + sin(pi/6)]

    now substitute in the values for cos(pi/3) etc...
     
  14. Sep 24, 2007 #13
    is...

    4 + 4sqrt[3]/2 + i[8sqrt[3]/2 + 1/2]
     
  15. Sep 24, 2007 #14

    learningphysics

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    That should be:

    z+w = 8cos(pi/3) + 8isin(pi/3) + 4cos(pi/6) + 4isin(pi/6)
     
  16. Sep 24, 2007 #15
    Yep your right.
    That gives me the right answer now :P

    (4 + 2sqrt[3]) + i(4sqrt[3] + 2)
     
  17. Sep 24, 2007 #16

    learningphysics

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    yeah, except that should have been 4isinpi/6... so the answer is actually:

    4 + 4sqrt[3]/2 + i[8sqrt[3]/2 + 4(1/2)]

    so when you simplify... you get

    4 + 2sqrt[3] + i(4sqrt[3] + 2)
     
  18. Sep 24, 2007 #17
    yeah I saw your correction. I was right wasnt i?
     
  19. Sep 24, 2007 #18

    learningphysics

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    yeah, sorry. I replied before seeing your reply. :tongue:
     
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