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Imaginary number's definition misunderstanding

  1. Aug 5, 2013 #1
    1. The problem statement, all variables and given/known data

    I'm in this self-learning course. I came on this problem I thought of.

    So, i^2=-1.
    But, isn't i=sqroot of -1?
    If so, the product of the two minus -1 and the square root of that should give 1.

    Am I not getting something?
    I searched the web with the keywords of my question, but I couldn't find any result...
  2. jcsd
  3. Aug 5, 2013 #2


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    Is i positive or negative?

    The relation i^2 = -1 is the converse of saying i = SQRT(-1). If a = SQRT(b), then a^2 = b.

    Then, there are these relations:

    i = SQRT(-1)
    i^2 = -1
    i^3 = -i
    i^4 = +1
  4. Aug 5, 2013 #3


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    I do not understand what you mean by "The product of two minus -1 and the square of that". Literally interpreted, "two minus -1" is 2- (-1)= 3 but surely that is not what you mean. Did you mean "the product of -1, twice"? Does "that" refer to i? Yes, [itex]i^2= -1[/itex]. The product of "-1, twice" is [itex](-1)^2= 1[/itex]. What does that have to do with [itex]i^2= -1[/itex]?

    If there is something you are "not getting", perhaps it is the difference between the "square" and the "square root". The square of -2 is (-1)(-1)= 1. But the square root of -1 is i.

    (There is a "technical" point here: while every number has a single square, every complex number has two square roots. The two square roots of -1 are i and -i.)
  5. Aug 5, 2013 #4
    Oh boy, sorry. I'm a second language speaker so speaking English messes my train of thoughts. :P
    Let me repair the situation:
    So, I have been taught that:''the product of two radicals is the same as the radical of the product, and vice versa.''-purplemath
    But, now, dealing with the imaginary number:
    We know that i=sqroot(-1).
    So, i^2 should equal 1 since:

    But, that makes no sense because it is said that i^2=-1

    Is the property that I have been taught wrong? If so, when can I not use it?
  6. Aug 5, 2013 #5


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    The square root has an ambiguity in the complex numbers. You cannot use all the regular operations possible with positive real numbers.
  7. Aug 5, 2013 #6


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    @alingly1: You may find this post in the FAQ section by micromass interesting:

    https://www.physicsforums.com/showthread.php?t=637214 [Broken]

    You will find a discussion of some of the "paradoxes" arising from mis-using the usual rules of arithmetic with complex numbers.
    Last edited by a moderator: May 6, 2017
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