Results 1 to 5 of 5
Like Tree4Thanks
  • 1 Post By Plato
  • 1 Post By abender
  • 2 Post By ILikeSerena

Math Help - Complex number on a+ib form?

  1. #1
    Newbie
    Joined
    Jan 2013
    From
    Oslo
    Posts
    2

    Complex number on a+ib form?

    Really not sure if this is the right place to post, but how can i write (-e)^{i\pi} on a+ib form?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,639
    Thanks
    1592
    Awards
    1

    Re: Complex number on a+ib form?

    Quote Originally Posted by komodekork View Post
    Really not sure if this is the right place to post, but how can i write (-e)^{i\pi} on a+ib form?
    This is not an easy problem. Recall that complex exponentiation does follow the usual rules that we use in the real numbers.

    If each of z~\&~w is a complex number then w^z=\exp\(z\text{Log}(w))

    Here \text{Log}(-e)=\ln|-e|+i\pi=1+i\pi. That is the principal value.

    So you want [\exp[(i\pi)(1+i\pi)]=~?
    Thanks from komodekork
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Mar 2008
    From
    Pennsylvania, USA
    Posts
    269
    Thanks
    37

    Re: Complex number on a+ib form?

    (-e)^{i\pi} = (-1\cdot e)^{i\pi} = (-1)^{i\pi} \cdot e^{i\pi} = \left((e^{i\pi})^{i\pi}\right) \cdot e^{i\pi} =  e^{i^2\pi^2} \cdot e^{i\pi} =  e^{-\pi^2} \cdot e^{i\pi} = -e^{-\pi^2}

    Note that I used this identity: e^{i\pi}=-1.

    Does this help?


    To to hopefully more directly answer the OP's question, recall that the real numbers are a subset of the complex numbers. So, the answer I detailed above is written in the complex notation a=-e^{-\pi^2} and b=0.
    Last edited by abender; January 27th 2013 at 10:48 AM.
    Thanks from komodekork
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member ILikeSerena's Avatar
    Joined
    Dec 2011
    Posts
    733
    Thanks
    121

    Re: Complex number on a+ib form?

    Quote Originally Posted by abender View Post
    (-e)^{i\pi} = (-1\cdot e)^{i\pi} = (-1)^{i\pi} \cdot e^{i\pi} = \left((e^{i\pi})^{i\pi}\right) \cdot e^{i\pi} =  e^{i^2\pi^2} \cdot e^{i\pi} =  e^{-\pi^2} \cdot e^{i\pi} = -e^{-\pi^2}

    Note that I used this identity: e^{i\pi}=-1.

    Does this help?
    This is the so called principal value.

    More accurately you have -1 = e^{i(\pi + 2k\pi)} where k is an integer (slight modification of your identity ;)).

    If you substitute this, you'll find:
    (-e)^{i\pi} = - e^{-\pi^2(1+2k)}
    which is multi valued (all real) with the principal value you already found.
    Last edited by ILikeSerena; January 27th 2013 at 10:53 AM.
    Thanks from abender and komodekork
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jan 2013
    From
    Oslo
    Posts
    2

    Re: Complex number on a+ib form?

    Thank you all, i was really stuck there for a while.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. The exponential form of complex number
    Posted in the Advanced Math Topics Forum
    Replies: 4
    Last Post: December 21st 2012, 11:41 AM
  2. Complex Number in its simplest form
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: July 21st 2010, 10:28 AM
  3. Complex number(exponential form)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 8th 2009, 02:35 AM
  4. Replies: 3
    Last Post: July 12th 2009, 09:53 AM
  5. complex number polar form
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: February 23rd 2009, 10:26 AM

Search Tags


/mathhelpforum @mathhelpforum