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What happens when we rise a number to the power of i ?

  1. Jun 23, 2011 #1

    I have taken some math classes recently, but I feel that they opened some new questions.
    I am specifically unclear about complex numbers.
    I have understood how they can be added, divided, multiplied and subtracted algebraically, and then represented on the Argand diagram (is this the same as the complex plane?), after being converted into the cos (x)+ i sin (x) form, through Euler's Formula (e^ix= cos(x) + i sin (x)).
    We have also learned how to find roots of complex numbers with de Moivre's formula: (cos (x) + i sin (x))^n = cos (nx)+ i sin (nx).
    I seem to have all the bits of information, but what I am missing is the understanding what happens when we rise a number to the power of i. How (and mostly where,) do we represent something like i^i, which according to wolframalpha has actually a numerical value of 0.2078 (4 d.p.) and which can be represented as e^-pi/2. The fact that the latter does not have i in the exponent, indicates to me that I should treat it as a normal numerical value, but I am not even sure if this can be represented on the real axis, if e does not have i in the exponent.
    Can somebody please explain to me what rising a number to the power of i do to it in terms of representation on the Argand diagram, or how does one do it algebraically. (as I understand de Moivre's theorem only works for e as a base, but then I still don't understand what rising e^i actually means, apart from making it possible to represent it in the form of cos(x) +i sin (x))

    Thank you

    EDIT: ln(i)= i pi/2... it still does not make much sense...where does the pi/2 come from in this?
    Last edited: Jun 23, 2011
  2. jcsd
  3. Jun 23, 2011 #2


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    We can think of complex numbers as points in the "complex plane" (yes, that is the same as the Argand diagram). That is the 2 dimensional plane of points (x, y) with x interpreted as the real part of the complex number, y as the imaginary part: the point (x, y) corresponds to the complex number x+ iy.

    We can then convert to polar coordinates for the plane: [itex]x= r cos(\theta)[/itex], [itex]y= r sin(\theta)[/itex]. From that, [itex]x+ iy= r cos(\theta)+ r i sin(\theta)= r(cos(\theta)+ i sin(\theta)[/itex]. The imaginary unit itself, i= 0+ i, corresponds to the point (0, 1) on the vertical axis and so at angle [itex]\pi/2[/itex] radians to the horizontal axis. That is, in polar coordinates, the number i has r= 1, [itex]\theta= \pi/2[/itex].

    While it is relatively easy to add or subtract in "Cartesian Form", (a+ bi)+ (c+ di)= (a+ c)+ (b+ d)i, multiplying is easier in "polar form" and, in particular, powers: [itex](a+ bi)^n= (r(cos(\theta)+ i sin(\theta))^n= r^n(cos(n(\theta)+ i sin(n\theta))[/itex]

    Further, by comparing Taylor series for [itex]e^x[/itex], sin(x), and cos(x), we can see that [itex]e^{i\theta}= cos(\theta)+ i sin(\theta)[/itex] so we could also write [itex](a+ bi)^n= r^ne^{ni\theta}[/itex].

    To extend that to [itex]i^i[/itex] we would write [itex]\left(e^{i\pi/2}\right)^i= e^{i(i\pi/2)}= e^{-\pi/2}[/itex] which is a real number, about 0.20788
  4. Jun 23, 2011 #3
    Thank you, that makes it all perfectly clear :)
    If I understand it correctly, this also implies that if I wanted to express let's say 2^i, I would need to first express 2 in polar form and then rise it to the power of i?
    (I think so).
  5. Jun 23, 2011 #4
    Bear in mind that for arbitrary exponents, the operation of taking powers is multivalued. This is an extension of the familiar case of the square root: for example, [itex]1^{1/2}[/itex] could be plus or minus 1.

    Arbitrary powers are defined by [itex]a^b=\exp(b\log(a))[/itex]. But log is multivalued on the complex plane: this is because exp is no longer injective so it doesn't have an inverse. From Euler's formula, [itex]\exp(2n\pi i)=1[/itex] for arbitrary integer n, so given a logarithm you can always add any number of [itex]2\pi i[/itex]s to it and still have a logarithm. (By a logarithm of z here I mean any number w such that exp(w)=z).

    For example, [itex]i^i=\exp((2n+1/2)\pi i)^i=\exp(-(2n+1/2)\pi)[/itex] for arbitrary integer n. So you could legitimately say that [itex]i^i=\exp(-\pi/2)[/itex], or that [itex]i^i=\exp(3\pi/2)[/itex], or infinitely many other answers.
  6. Jun 23, 2011 #5
    I am still struggling to find what actually n^i (in this case, to make it easier 2^i) is.

    Wolfram Alpha gives me the complex number 0.76923... + 0.63896... i = 2^i

    But when I try to convert it into polar exp form, I get a different result

    2 can be represented as 2e^(i2pi) (at least I think so), so it should be possible to get the right result by rising this to the power if i : 2e^((i2pi)^i), which will give me 2e^(-2pi), which gives me the real number 0.003734...

    I find this confusing. Am I doing something wrong? Do I need to keep adding multiples of 2pi in the exponent? I have studied the standard thing of finding multiple complex roots of real numbers on the complex plane, but in this case I don't want to find a root, but the actual number.
    Or more, I want to know how was 0.76923... + 0.63896... i = 2^i arrived at.
  7. Jun 23, 2011 #6


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    Yes, this is wrong. If you raise [itex]2e^{2i\pi}[/itex], then you get [itex]2^ie^{-2\pi}[/itex], but you forgot to raise 2 to the power of i.

    Wolfram alpha calculates it as follows:

    First we need to know what Log(2) is. By definition Log(2) is such that [itex]e^{Log(2)}=2[/itex]. But this means that Log(2) has multiple values, indeed, if take k an integer, then setting

    [tex]Log(2)=log(2)+2\pi i k[/tex]

    satisfies the definition (where log(2) is the real log function). So Log(2) has an infinite number of values (normally, we set k=0 and call that value the principal branch of the logarithm, but let's not bother with this here).


    [tex]2^i=e^{Log(2^i)}=e^{iLog(2)}=e^{ilog(2)-2\pi k}=e^{-2\pi k}(\cos(log(2))+i\sin(log(2)))[/tex]

    So we see that there are infinitely many outcomes for [itex]2^i[/itex]. The wolfram alpha result is reached when taking k=0.
  8. Jun 23, 2011 #7
    Great, thank you micromass :)
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