1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Gold Member

    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

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    yeah, sorry. I replied before seeing your reply. :tongue:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Need a review on Complicated explonential math
  1. Idea needed on maths (Replies: 0)

  2. I need help with Math (Replies: 14)

Loading...