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Homework Help: Value of i^15

  1. Aug 10, 2010 #1
    1. The problem statement, all variables and given/known data
    find the value of i15

    2. Relevant equations
    i = √-1
    i2= -1

    3. The attempt at a solution
    method a
    i15 = (i2)7 x i = -i

    method b
    i15 = ((-1)1/2)15= ((-1)15)1/2 = (-1)1/2 = i

    method a ≠ method b ??????????
  2. jcsd
  3. Aug 10, 2010 #2
  4. Aug 10, 2010 #3


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    It should be no surprise they're different. Unless, of course, you haven't internalized the fact that exponentiation is multi-valued.

    Having a sensible exponential operation is a peculiarity of the positive real number bases with real number exponents. (or some other special cases)
  5. Aug 10, 2010 #4


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    (-1)^(1/2) is really either +i or -i. Since i^2=(-i)^2=(-1). You can't play fast and loose with the rules of exponents if you are dealing with fractional powers. Your first method is safe.
  6. Aug 10, 2010 #5
    I have checked it and I do not understand it. I did not see any paragraphs that explain the difference between my methods; the explanation is just about i4n = 1, etc which I think just some formulas we should memorize. I am trying to understand things although I am still in 9th grade.I like to study things outside school subjects but I do not have strong concepts.

    I haven't internalized the fact that exponentiation is multi-valued. What I understand now is the value of exponentiation is always a single-valued, like 24 = 16 and does not have another value. Can you please tell me a simple example of multi-valued of exponentiation? I will try to understand it even though I am in 9th grade.

    Why is (-1)^(1/2) either +i or -i ? We can write (-1)^(1/2) = √-1 = i by definition of i. I understand method (a) is safer but I do not understand why method (b) is wrong

    thank you very much for the reply and I hope my post will be replied again. thank you again
  7. Aug 10, 2010 #6
    May be this will be helping and interesting to you:

    Euler's Formula:

    exp(i*x) = cos(x) + i*sin(x)

    a complex number can be represented as coordinate in a plane (complex plane) where the first coordinate is the "Real" value and the second coordinate is the "Imaginary" value of the complex number.

    (check: http://en.wikipedia.org/wiki/Complex_number)

    we can write then:

    i = cos(pi/2) + i*sin(pi/2)

    cos(pi/2) + i*sin(pi/2) = exp(i*(pi/2))


    i^15 = [exp(i*(pi/2))]^15 = exp(i*(15/2)*pi) = exp(i*[6*pi+(3/2)*pi])

    cos(x) + sin(x) is a periodic function with a period of 2 pi (it has the same values for any x + 2*pi*n):

    z = cos(x) + sin(x) = cos(x + 2*pi*n) + sin(x + 2*pi*n) ; n = 0,1,2,3,...

    then when you turn your complex vector in the complex plane by 3/2 pi it will be at (0, -i)


    i^15 = -i

    *it came out to be more complicated then i thought ! :)
    just try to read the link about complex numbers I posted, good luck !
  8. Aug 10, 2010 #7
    Okay, it's good to have a good foundation with complex numbers before you tackle this problem. First:

    i1 = i or -11/2

    i4 = 1 ( -11/2 x -11/2 .... =1, if you do it in steps)

    i3 = -i ( -11/2 x -11/2 is -1, x i = -i)

    Once you simplify it, you'll get i12 x i3.

    i12 is equal to i4 x i4 x i4. Which, is 1 x 1 x 1. Hence, i15 = i3 = -i

    Much easier when you just break it down. Absolutely no need to even touch base on Euler's identity, whatsoever...
    Last edited: Aug 10, 2010
  9. Aug 10, 2010 #8
    @theJorge551 I assumed harimakenji wants to learn new things and understand them. and thus, I introduced an interesting notion about complex numbers.
    the question could be fully answered through the way I described above.
    basically there is a need to talk about Euler's Formula, you can just check what the answer to i^15 on the net and never understanding it.
  10. Aug 10, 2010 #9


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    Viewed as complex exponentiation, the expression 41/2 has as its possible values both 2 and -2. The expression [itex]4^{1 / (2 \pi)}[/itex] has as its possible values (where this time exponentiation is real exponentation):
    [tex]2^{1 / \pi} \left( \cos n + i \sin n \right)[/tex]​
    where n can be any integer.

    In general, the complex exponential xy is defined to be
    [tex]x^y = \exp(y \log x)[/tex]​

    However, log x is multivalued -- if L is one of the values of log x, then so is [itex]L + 2 \pi i n[/itex] for any integer n.
  11. Aug 11, 2010 #10
    Yes I want to learn new things and thanks to both of you for the explanation. Now I understand euler's form for complex number. I have understood how to obtain the value of i15 and now I have a new problem. I do not understand why my 2nd method (method b) is wrong. Can you please explain it to me?

    OK, I want to ask several questions (hope you will not feel annoyed)

    1. I do not understand how 41/2 can change to [itex]4^{1 / (2 \pi)}[/itex]

    2. I do not understand how [itex]4^{1 / (2 \pi)}[/itex] can change to [tex]2^{1 / \pi} \left( \cos n + i \sin n \right)[/tex]. From what I understand, [tex]2^{1 / \pi} \left( \cos n + i \sin n \right)[/tex] = [tex]2^{1 / \pi} e^{in}[/tex] and how can this be equal to [itex]4^{1 / (2 \pi)}[/itex] ?

    3. about log x is multivalued. Can I take x = 2 for example? If can, then log 2 = 0.3010.... and this will be equal to 0.3010... + 2π i n, n is integer. I do not see 0.3010.... = 0.3010... + 2π i n for any values of n

    thank you very much for the reply.
  12. Aug 11, 2010 #11
    It's because this identity:

    [tex] (a^b)^c = (a^c)^b [/tex]

    while true for reals, just isn't true for complex numbers.

    This is a good question. The log function with which you are familiar, which maps the reals to the reals, is not multi-valued. The extended log function which maps the complex plane to the complex plane, is multi-valued. So if 2 is a real number, then log 2 is single-valued. If it's really 2+0i because you're working in the complex plan, then it's multi-valued.
  13. Aug 11, 2010 #12


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    Here's my attempt to explain it simply:
    -When you raise a number to the 1/2 power, you're taking its square root
    -By definition, the square root of any number x is the number that, when multiplied by itself, gives you x
    -Both √x and -√x satisfy this condition
    -Therefore, both √x and -√x are square roots of x
    -Therefore, x1/2 has two values, √x and -√x

    In any given problem, you may have to choose one of the two values, depending on the conditions of the problem. It's just like solving a quadratic equation; you get two solutions, and you may have to pick one or the other, depending on the problem.

    So anyway, if you accept what I've said so far, you'll agree that by your method b, i15 is either +i or -i. How to tell which? Well, you can't, really, at least not without using method a. The thing is, in your first step of method b, you change
    i into (-1)1/2. But if it were (-i), you could just as well have changed that into the same thing, (-1)1/2. So you've just lost some information: once you write (-1)1/2, you no longer know whether it started out as +i or -i. Whenever that happens, it's a sign that the method you're using isn't adequate, and you'll need to bring in some other information or some other method to truly solve your problem. In this case, the other method would be your method a.

    Again, it's like solving a quadratic equation: it's often the case in physics that you need to solve a quadratic equation to find the time at which something happens, and you'll get two answers, for example +5s and -2s. For physical reasons, you might know that the thing could not have happened before t=0s; that's the extra information you have to bring in to solve the problem.
  14. Aug 11, 2010 #13


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    Those were two different examples.

    Everything, I think, is best understood in terms of logarithms. The complex logarithm is equal to:
    [tex]\log(x + i y) = \ln(r) + i \theta[/tex]​
    where ln is the ordinary real logarithm, and [itex](r, \theta)[/itex] are polar coordinates for the point (x,y) in the Euclidean plane.

    The interesting point is that there are many different values of [itex]\theta[/itex] for a particular (x,y). For example, the polar angles -4pi, -2pi, 0, 2pi, 4pi, 6pi, et cetera all define the same ray in the Euclidean plane.

    So, the values of the expression [itex]\log(4)[/itex] are the complex numbers of the form [itex]\ln 4 + 2 \pi i n[/itex] where n is an integer.

    The other thing you need to know is that
    [tex]\exp(x + i y) = e^x (\cos(y) + i \sin(y))[/tex]​
    where the exponentiation on the right hand side is the ordinary real one.

    Complex exponentiation is defined by
    [tex]x^y = \exp(y \log(x))[/tex]​

    The point of being a multi-valued function is that, to the same input value, more than one output value is associated to it -- multi-valued functions are not functions. They're a different type of expression that violates the vertical line test. However, they come up often enough that it's worth trying to reason with them.
  15. Aug 11, 2010 #14


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    And yet another explanation...

    As Hurkyl noted, the definition of exponentiation is

    [tex]x^y = e^{y \log x}[/tex].

    Applying this to z=(-1)15, you get

    [tex]z = e^{15 \log (-1)} = e^{15\pi i}[/tex]

    Applying the definition again to calculate z1/2, you get

    [tex]z^{1/2} = e^{(1/2) \log z}[/tex]

    Since [itex]z=e^{15\pi i}[/itex], [itex]\log z = 15\pi i[/itex], and you get

    [tex]z^{1/2} = e^{15\pi i/2} = -i[/tex]

    which matches the answer you got with your first method.

    In your second method, however, what you did was say [itex]z=e^{15\pi i}=-1[/tex], so [itex]\log z = \log (-1) = i\pi[/itex]. Now you get

    [tex]z^{1/2} = e^{i\pi/2} = i[/tex]

    You can see the difference in the two cases arises because you have different values for log z. When you wrote [itex](-1)^{15}=-1[/tex], you lost the information about the complex phase, which led to getting the incorrect answer.

    The upshot is you need to be careful when working with complex numbers and exponents. The rules that apply when you're restricted to real numbers don't always hold with complex numbers.
  16. Aug 11, 2010 #15
    So the rule is not true for complex. I did not know it before. Thank you.

    I tried to change -i:
    -i = -√-1 = -(-1)1/2
    How that can be equal to (-1)1/2? Dick has explained that in post 4 that i2=(-i)2=(-1). I get it when I see it that way, but I do not get it when trying to start from what I have written above.

    Shouldn't [tex]x^y = \exp(y \ln(x))[/tex]​
    because ln and log have different bases?

    I know that [tex]\exp(x + i y) = e^x (\cos(y) + i \sin(y))[/tex], but I do not understand why the RHS is ordinary real one. It contains i so it should be imaginary instead of real?

    Is "log" in your post the same as "ln" ?

    Isn't e15 π i equal to eπ i, so eπ i = cos π + i sin π = -1 ?

    thank you very much for the reply
  17. Aug 11, 2010 #16


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    Yes. Despite what they told you in school, people often write "log" to mean natural log.
    Yes and no. They're both equal to -1, but they're not equivalent ways of expressing -1 when it comes to finding roots.
  18. Aug 11, 2010 #17


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    Because (-1)1/2 is equal to both +i and -i. So -(-1)1/2 is equal to both -(+i) and -(-i), which is -i and +i, which is the same set of numbers.

    To clarify, here's what I meant by my notation:
    (-1)1/2 = any number which, when squared, gives you -1
    √-1 = the principal square root of -1, which is +i
  19. Aug 13, 2010 #18

    I guess I understand better now. I really appreciate all responses and explanation here. Thank you very much for your time and passion
  20. Aug 13, 2010 #19


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    Typically in Calculus or higher, there is no need for the common logarithm so that "log", without any base given, is used for the natural logarithm.

    (It's just as well, with all those people thinking "ln" is "In". I have never figured out why!)
  21. Sep 6, 2010 #20
    I think part of the problem is that the square root of a number, say x^2, is +/- x depending on whether x is +/-ve. That of use?
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