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Curious about imaginary solution

  1. Oct 26, 2007 #1
    It's clear that [tex]f(x)=e^x[/tex] has no solution for [tex]f(x)\le0[/tex]. So can there be imaginary solutions for values not in the range of the exponential function? Just as an example, can this have imaginary solutions, and how would one go about solving it: [tex]e^x=-e[/tex]?
     
  2. jcsd
  3. Oct 26, 2007 #2

    mjsd

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    do yon mean something like [tex]e^{y} = -1[/tex]
    and so
    [tex]y = i \pi[/tex] is a solution ?
     
  4. Oct 26, 2007 #3
    The equation [tex]e^x = 0[/tex] still have no soltuons in the complex plane even.
     
  5. Oct 26, 2007 #4
    Well, if you just assume [itex]e^{i\pi} = -1[/itex] holds**, then you simply have [itex]e^{i\pi + 1} = -e[/itex]. Since [itex]e^{x} = n[/itex] is defined in [itex]\mathbb{R}^2[/itex] for any [itex]n > 0[/itex], you can come up with any [itex]z[/itex] such that [itex]e^{i\pi + z} = - n[/itex] for any positive [itex]n[/itex].


    ** Of course, you don't have to assume anything. :biggrin: The proofs for the formula [itex]e^{i\pi} + 1 = 0[/itex], using Euler's Formula, [itex]e^{ix} = \cos(x) + i\sin(x)[/itex], are online somewhere.
     
    Last edited: Oct 26, 2007
  6. Oct 26, 2007 #5

    Gib Z

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    However for a negative, non-zero value, the logarithm can be analytically extended. Assuming the primary branch, [tex]e^{i\pi}=-1[/tex] as mjsd hints, so using some log properties, [tex]\ln (-b) = \ln b + {i\pi}[/tex].

    So for your example, [tex]x= \ln (-e) = \ln e + \ln (-1) = i\pi +1[/tex]. Note this is all only in the primary branch cut, the functions in discussion here are multivalued in the complex plane.

    EDIT: Didn't see Battousii's post, his one seems better explained than mine so ignore this =]
     
    Last edited: Oct 26, 2007
  7. Oct 26, 2007 #6

    mjsd

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    yeah.
    unless you count "complex infinity" as a possible solution. [tex]e^{-\infty}\rightarrow 0[/tex]
     
  8. Oct 26, 2007 #7
    No you cannot count it. Maybe in physics it is okay, in math no such thing. In complex analysis there is no such thing. (Unless you are talking about the Riemann sphere, but that still infinity is not used in arithmetic. Additionally, infinity in complex plane is simplify one point it does not have directions like on the real line).
     
  9. Oct 27, 2007 #8
    Thanks a lot guys.
    I completely forgot about Euler's Identity as a prime example... it just slipped my mind.
    However, the answer to this question brings up another question: do all functions have imaginary solutions for values that are not in their range? For example: [tex]f(x)=|x|[/tex] does not have negative numbers in its range. However, even imaginary numbers seem to produce positive magnitudes (since it gives the distance from the origin to a value on the real-imaginary plane).
     
  10. Oct 27, 2007 #9

    Gib Z

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    In general it is not possible to extend the range of a real function, in order to do so you must find a method of analytic continuation, which is usually done by defining a function that agrees with all values of the original function for where the original is defined, and also defines values elsewhere. The absolute value function has no analytic continuation that I know of.
     
  11. Oct 27, 2007 #10

    CompuChip

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    Don't go mixing things up here.
    You have a function f(x) which has a range. For any number y in the range you can find (at least) one x such that f(x) = y.

    It is often possible to extend a function to the complex plane. This means, we define a new function g(z) for z a complex number, such that g(z) = f(z) if z happens to lie on the real line. By convention, we also denote this function g(z) as f(z) and call it the analytic continuation of f. It is not necessary that an analytic continuation exists (in fact, it is quite special for a function to be analytic) and there are certain theorems that state when it is possible to find such a function. But if it exists, then you have a new function f(z) defined on a bigger set, with a new range. Of course the old range is a subset of the new range, but the new range can be larger or not. For example, whereas exp(x) only has a range of exp(x) > 0, the function exp(z) has as range the entire complex plane, except for the origin. So the range has increased considerably. On the other hand, the absolute value function has range |z| > 0, whether z is real or complex, so the range doesn't increase for the analytic continuation. Therefore, going complex will not help you in finding a solution for |z| = -1, but it will for solving exp(z) = -e.
     
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