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I Constant raised to complex numbers

  1. May 22, 2017 #1
    It's not a homework question. I just thought up a method of finding answers to problems where a number is raised to a complex number and I need to know if I am right. If we have to find e^(i), can we do it by; first squaring it to get, e^(-1) which is 1/e and then taking its square root to get 1/e^(0.5), which is e^(-0.5). Is my method correct? I don't know if the answer is right or even how to find answers to such questions where we raise a number to a complex number but my method seems correct to me.
    p.s. Applying this method to e^(i × pi) gives e^(-pi /2).
    Mr R
  2. jcsd
  3. May 22, 2017 #2


    Staff: Mentor

    If you use the Euler formula

    ##e^{i\theta} = cos(\theta) + i sin(\theta) ##

    To evaluate your answers you'll see they don't match.

    Hence you've found an exception to your method.

    Taking the square root of anything means there are now two solutions.

    ##i ^ 2 = (-i)^2 ##
  4. May 22, 2017 #3
    But where exactly was I wrong in my method or I'm not wrong in all the steps I have shown but the last step, that of taking the square root of a number will be wrong somehow. And can't we have two solutions here like we get for the roots of a quadratic equation?
  5. May 22, 2017 #4


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    Unfortunately for your method:

    ##(e^{i})^2 = e^{2i} \ne e^{-1}##
  6. May 22, 2017 #5
    Such a silly mistake! Sorry for wasting your( jedishrfu & PeroK ) time and thanks for clearing my doubt. Cheers!
    Mr R
  7. May 22, 2017 #6


    Staff: Mentor

    Its not a silly mistake, its a subtlety of mathematics that you must always take into account.

    Its great that you have discovered it and now know to avoid it.
  8. May 22, 2017 #7
    Thank you for your words and time.:smile:
    Mr R
  9. May 22, 2017 #8


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    Also, in general, squaring an equation will usually generate an additional solution. This is most simply illustrated by:

    ##x = 1 \ \Rightarrow \ x^2 = 1 \ \Rightarrow \ x = \pm 1##
  10. May 22, 2017 #9


    Staff: Mentor

  11. May 22, 2017 #10
    Yeah, I am aware of that. That's why in post #3, I had raised a question about why would it not be okay to get two solutions in this case when it's okay in some other cases like when finding roots of an algebraic equation.
    Mr R
  12. May 22, 2017 #11
  13. May 22, 2017 #12


    Staff: Mentor

    Squaring both sides of an equation, and later taking the square root is generally not valid.

    For example, suppose x = -3
    Squaring both sides gives another equation, ##x^2 = 9##
    If I take the square root of both sides, I get ##\sqrt{x^2} = \sqrt 9##, or |x| = 3, and equation that has two solutions, one of which is different from the equation we started with.

    If the operation you apply to both sides is one-to-one, such as adding the same number to both sides, multiplying both sides by the same number, etc., then this problem doesn't arise. However, squaring a number is not one-to-one, as both 3 and -3 have the same squares; namely, 9.
    If you're working with an equation involving radicals (specifically, square roots), the standard technique is to square both sides. In doing so, though, there's the possibility that you are introducing an extraneous root, one that is not a solution of the original equation.
    Another example: ##\sqrt{x} = -2##
    If you square both sides, you get x = 4. This is not a solution of the original equation, which in fact has no solutions in the real numbers.
  14. May 22, 2017 #13
    Wow, the example makes it seem so obvious but I didn't know this property of squaring.Thanks for telling me about it.
    Mr R
  15. Jun 12, 2017 #14
    Another way to look at it is that as a function in a formula, you generally always take the principle root (the positive value) of a square root. But, as an operation in algebra, taking a square root requires you to acknowledge two possible solutions. This is why, for example, the quadratic formula has the ##\pm## symbol in it, to let you know that you have 2 solutions.
  16. Jun 12, 2017 #15


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    I think one way to see it is through the use of Complex logarithm/exponentiation: Under " reasonable conditions" *

    ## z^w ## = ## e^{wlogz} ##.

    * Sorry for the cop-out, I am exhausted and cannot give you a full answer at the moment.
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