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A quick question on imaginary number

  1. Jan 26, 2016 #1
    1. The problem statement, all variables and given/known data
    Since i is defined by sq(-1),and we can also write it as (-1)^(1/2)
    Therefore,(-1)^(1/2) is equivalent to (-1)^(4/8),so it becomes [(-1)^4]^(1/8), so we have 1^(1/8) = 1,which is clearly absurd.....
    Besides,since (i^5)^3 = i^15 = -i,the multiplication of power rule still holds in complex number,so it is possible to write things like (-1)^(4/8) into [(-1)^4]^(1/8).....Therefore it should not be related to the applicability of this rule in C.
    I know this question is very short and dumb,but I really hope someone can help me..and to clear my concept...Thanks

    2. Relevant equations


    3. The attempt at a solution
    As shown in the problem statement
     
  2. jcsd
  3. Jan 26, 2016 #2

    Samy_A

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    This insight written by @micromass may help:
    Things Which Can Go Wrong with Complex Numbers Reference
     
  4. Jan 26, 2016 #3

    fresh_42

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  5. Jan 26, 2016 #4
    Thanks for the great post.But I remember my teacher had illustrated an example in my class: i^2016 = (i^4)^504 = 1^504 = 1
    According to "Rule 1" in your post,(x^a)^b = x^(ab) only when x>= 0 and a,b are real.Therefore,i^2016 =/=1 and since i is a complex number ?
     
  6. Jan 26, 2016 #5

    Mark44

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    Good question.
    No, ##i^{2016} = 1##, as your teacher showed. The problem comes when you have ##(x^a)^b##, when either a or b is smaller than 1, but both are positive. For example, with ##(-1)^{1/2}##. This can be rewritten as ##((-1)^1)^{1/2}## but that will give you a different answer if you rewrite it in what seems to be an equivalent form, as ##((-1)^2)^{1/4}##. The first simplies to i, but the second simplifies to 1. Obviously, they are not the same.
    @micromass might want to address this in his Insights article.
     
  7. Jan 26, 2016 #6

    Ray Vickson

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    There is nothing at all absurd about the statement 1^(1/8) = 1: it happens to be true. However, there are 7 other roots to the equation ##x^8 = 1##.
     
  8. Jan 26, 2016 #7
    1^(1/8) = 1,and 1^(1/8) = 1 is one of the roots of x^ 8 = 1.
    Assume the "multiplication rule of power" is correct, 1^(1/8) comes from (-1)^(4/8) = (-1)^(1/2),so i is a root of x^8 = 1,which is true.
    Now,i and 1 have one thing in common is that they both solve x^8 = 1,but 1 comes from i,we cannot put an equal sign between i and 1.It seems somewhat paradoxical to me since these 2 different numbers have a "mutual property" but at the same time they are equal?
     
  9. Jan 26, 2016 #8

    haruspex

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    No, it doesn't say they are equal only when x>= 0 and a,b (and x) are real. It says the rule is valid then. It may be valid under other conditions too, and they may happen to be equal (based on other info, perhaps) even when no rule guarantees it.
    As far as I can see, it would also be valid when x is complex (which includes negative reals), is nonzero, and a and b are integers.
    If x is complex and a and b are rationals, it will still be valid subject to certain constraints on the rationals.
    This should be familiar from reals. Given x2=1, there are two values possible for x. More generally, xn=1, nonzero integer n, yields 2 real solutions when n is even, but only one when n is odd. Allow complex solutions and you get n solutions always.
     
  10. Jan 26, 2016 #9

    Ray Vickson

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    I cannot figure out what you are trying to say. Here are the facts: a polynomial of degree ##n## always has ##n## roots, although they need not be distinct; that is, roots can be "repeated". So, the equation ##x^8 = 1## has 8 roots, and in this case they are all distinct. They are
    [tex] \pm 1, \: \pm i , \: \frac{1}{\sqrt{2}} \pm \frac{i}{\sqrt{2}}, \: -\frac{1}{\sqrt{2}} \pm \frac{i}{\sqrt{2}} [/tex]
    All eight "come from" 1, if that means anything.
     
  11. Jan 27, 2016 #10

    fresh_42

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    The example of ##x^8=1## in posts 6 and 9 also gives a hint how one can avoid uncertainties about what is allowed when dealing with complex numbers and what's not. Since ## ℂ ≅ ℝ(x) / (x^2+1) ## all calculations on complex numbers can be performed within ##ℝ(x)## with a variable (transcendental number, resp.) ##x## instead of ##i## and then taking the result modulo the ideal ##(x^2+1)##.

    This way ##i^2## would be ##x^2 = (x^2 +1) - 1 ≡ -1 \mod (x^2+1)## and there is no way to get ##1## as a possible result.
     
  12. Jan 27, 2016 #11

    Mark44

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    I'm nearly certain that this response is way over the OP's head...
     
  13. Jan 27, 2016 #12

    fresh_42

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    I've thought about it. Then I thought there might be other readers and it cannot harm to say where it all came from.
    Personally, I find this easier than to learn exceptions to rules I've been drilled to a whole school live long.
     
  14. Jan 27, 2016 #13

    Mark44

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    This is the Precalc section, so responses should be appropriate to that level. Your reply would be appropriate for someone who is taking an upper division (third or fourth year in university) course in Modern Algebra.
     
  15. Jan 27, 2016 #14
    Sorry about that,I am still a freshman;d so it might take me a bit more time to understand this...:nb)
    On the other hand,not sure about this but let's try to express i using the Euler's formula ------>
    √-1 = e^(iπ/2) ,while (-1)^(4/8) = (e^π i)^(4/8) = (e^4πi/8) = e^(πi/2) = √-1 .
    But if you compute 1^(1/8) by converting (-1)^4 into 1 in the midway through, then 1^(1/8) would become (e^ 2πi)^(1/8) = e^(π i/4) =/= e^(π i/2)..Not quite understand what they are going to reveal though.I guess "rule 1" might be linked to the above?
     
  16. Jan 27, 2016 #15

    fresh_42

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    Your calculation ##x^{\frac{1}{8}} = e^{i\pi/4}## (with ##x=1##) shows that there are more solutions to ##x^8 - 1 = 0## than the integer ##1##. Eight in total as listed in post #9. They are uniformly distributed on a circle of radius 1 around the origin (starting at (1,0)). Those solutions are called 8th unity roots or roots in general.
    It just happened that you calculated the solution ## \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}## which to the power of ##8## results in ##1##, too.
     
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