1. Limited time only! Sign up for a free 30min personal 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!

Complex exponential function

  1. Mar 16, 2007 #1
    good evening all!

    1. The problem statement, all variables and given/known data

    Determine the exact values of

    2. Relevant equations

    j = sauare root of -1

    3. The attempt at a solution

    stuck :cry: :cry: :cry:
    Last edited by a moderator: Jan 3, 2014
  2. jcsd
  3. Mar 16, 2007 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    What do you know about the complex exponential function? Surely your text has some information you can apply.
  4. Mar 16, 2007 #3
  5. Mar 16, 2007 #4
    i'm still stuck
  6. Mar 16, 2007 #5
    okay this is what i did:


    i know this looks crazy but i still don't know how to use latex properly :redface: ...if you copy it out on a paper you'll understand what i'm trying to write...then i multiplied -1 with the 1/2...i can do that right?
  7. Mar 16, 2007 #6

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    YOu won't get the answer that way (since it doesn't deal with branches properly).

    Complex numbers have representations as r*exp(jt) for r, j in R. Use it.
  8. Mar 16, 2007 #7
    does exp(-2pi) = -1
  9. Mar 16, 2007 #8
    no it doesn't

    wait no one give me any tips i think i can solve it
  10. Mar 16, 2007 #9

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    No. If you must, put it in your calculator, but e^{-2pi} is obviously a positive real number, and something in the region of 1/2^64.
  11. Mar 16, 2007 #10

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    solve what? exp(-2pi) is right. It just isn't anything nicer than that.
  12. Mar 16, 2007 #11
    okay...i was about to say i'm stuck again but now i know that my teacher is crazy...but just to make sure [tex]j^j[/tex] doesn't equal -1...does it?

    ( [tex]j^j[/tex] <--that is almost the only thing i can write in latex form)
    Last edited: Mar 16, 2007
  13. Mar 16, 2007 #12
    yes it does
  14. Mar 16, 2007 #13

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    no it doesn't. j (or i) is exp(j*pi/2) so j^j is..... You can even google the phrase

    i to the i



    and you'll be given the answer by google calculator.

    Interesting googling j^j doesn't yield the answer. Google obviously doesn't employ engineering grads.
  15. Mar 16, 2007 #14
    my teacher is very very clever man and he said that the answer is -1 but he didn't use any of the complex number rules...he did something else! and he got -1

    sorry i used this formula to show that it does equal -1:

    r[cos(O) + jsin(O)] = rexp(jO)

    and in my desperate want to get this question right i didn't see the j next to the O, and so after i substituted the numbers i'm obviously going to get -1...never mind!

    ( O = theta)

    using your advice i went on google and thank god for wikipedia i got the answer...and i understand this time (for a change)

    you know i'm still stuck on that bernoulli number question...for B_3 you're supposed to get zero i actually got a number
  16. Mar 16, 2007 #15

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    Well keep trying 'til you do get zero. x/(e^x-1) differentiate it three times... eugh, no thanks. Do it the smarter way - sub in e^x first, then do the 1/(1-s) expansion. collect terms and you get

    -1/4! + 2/2!3! - (1/2!)^3 = -1/24+1/6-1/8=0

    (and when did 0 cease to be a number?)
  17. Mar 16, 2007 #16

    [tex]j^j=e^{j\cdot jx}=e^{-x}[/tex]

    so what should be the answer? (certainly not -1)
    Last edited: Mar 16, 2007
  18. Mar 16, 2007 #17
    i took your advice ( i kinda knew it anyway) but i'm going to write out in full what i did:

    [tex]e^x[/tex] = 1 + x + [tex]x^2[/tex]/2! + [tex]x^3[/tex]/3! + ...

    [tex]e^x[/tex] - 1 = x + [tex]x^2[/tex]/2! + [tex]x^3[/tex]/3! + ...

    x/([tex]e^x[/tex] - 1) = x/(x + [tex]x^2[/tex]/2! + [tex]x^3[/tex]/3! + ...)

    divid top and bottom by x => 1/(1 + x/2! + [tex]x^2[/tex]/3! + ...)

    (1+ x/2! + [tex]x^2[/tex]/3! + [tex]x^3[/tex]/4! +...)^(-1)

    let x/2! + [tex]x^2[/tex]/3! + [tex]x^3[/tex]/4! +.. = X

    (1 + X)^(-1) = 1- X + [((-1)(-2))/2!] [tex]X^2[/tex] +...

    so (1+ x/2! + [tex]x^2[/tex]/3! + [tex]x^3[/tex]/4! +...)^(-1) = 1 - x/2! - [tex]x^2[/tex]/3! - [tex]x^3[/tex]/3! -...((-1)(-2))/2!(X)^2 +...

    so B_1 =-1/2, B_2= -1/3!, B_3 = -1/4!.

    B_2 and B_3 are obviously wrong but i don't know where i'm going wrong???
    Last edited: Mar 16, 2007
  19. Mar 16, 2007 #18
    yes, okay, cheers, i get the picture :rolleyes:

    :biggrin: (<-- this is my favourite smiley)
  20. Mar 16, 2007 #19

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    Why does B_2=-1/3? What happend to the higher powers of X (a bad choice. Y would be better). You have only take the contribution from X^1, not X^2 or higher.

    and 1/(1+s) is 1-s+s^2-s^3+s^4+...

    Now put s = x/2!+x^2/3!+x^3/4!+...
    Last edited: Mar 16, 2007
  21. Mar 16, 2007 #20
    i still get a negative for B_2 i know what the answer is uppossed to be but i just can't see how...if i don't see why i don't sleep well!

    i'm tired, you know i'm just going to live with the fact that i'm never going to get this question...

    the thing is there is a minus outside the s and s^2 and the rest don't matter...


    thank you very much for your time and help!!!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook