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Does e^x = 1

  1. Sep 26, 2011 #1
    This is not a homework question.

    I know that the equation [itex] e^x=1[/itex] has infinitly many solutions in form of [itex]2i\pi n[/itex] but I don't understand how and why.

    Will appreciate any help, thanks!
     
  2. jcsd
  3. Sep 26, 2011 #2
    Re: e^x=1

    You should know that a complex number can be represented as a vector in the gauss plane, where te x-axis is the Real axes and the y-axis is the Imaginary axis. Take for example the complex number z=a+ib (where a and b are real), it can be represented by the vector (a,b). You can use also a polar coordinate system where r is the modulus of the vector (a,b) and [itex]\phi[/itex] is the so-called anomaly, meaning the angle between the vector and the positive side of the x-axis. So you have:

    [itex]z=r\left(\cos(\phi)+i\sin(\phi)\right)[/itex]

    If you consider the taylor expansion of the function sine, cosine and , [itex]e^{x}[/itex]you can observe that:

    [itex]\cos(x)=\frac{e^{ix}+e^{-ix}}{2}[/itex]

    [itex]\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}[/itex]

    and so:

    [itex]e^{ix}=\cos(x)+i\sin(x)[/itex]

    So you can write a complex number with Euler's notation:

    [itex]z=re^{i\phi}[/itex]

    Since in your equation z=1, i.e. the vector (1,0), in Euler's notation you have r=1 and [itex]\phi[/itex] can be 0, [itex]2\pi i[/itex], [itex]4\pi i[/itex] and so on, since they represent the same angle (i.e sine and cosine are periodic and the period is [itex]2\pi[/itex] and so also the exponential is periodic, and the period is [itex]2\pi i[/itex]
     
    Last edited by a moderator: Sep 26, 2011
  4. Sep 26, 2011 #3
    Re: e^x=1

    Thank you for the detailed explanation!
     
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