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Problem proving Euler's formula?

  1. Aug 2, 2014 #1
    1. The problem statement, all variables and given/known data
    Prove ei[itex]\theta[/itex] = cos(θ) + isin(θ)

    2. Relevant equations

    3. The attempt at a solution
    Let g = cos(θ) + isin(θ)
    [itex]\frac{dg}{dθ}[/itex] = -sin(θ) + icos(θ)
    => [itex]\frac{dg}{dθ}[/itex] = ig
    => [itex]\frac{dθ}{dg}[/itex] = [itex]\frac{1}{ig}[/itex]
    => [itex]\frac{dθ}{dg}[/itex] = -i[itex]\times[/itex][itex]\int[/itex][itex]\frac{1}{g}[/itex]
    => θ = -i[itex]\times[/itex]ln|g| + c
    => iθ - ic = ln|g|
    => |g| = e[itex]^{iθ - ic}[/itex]
    g(0)=1, => c = 0
    When g=1,θ=0
    => |g| = e[itex]^{iθ}[/itex]

    Here's where the problem arises. Wouldn't that mean that g = [itex]\pm[/itex]e[itex]^{iθ}[/itex]
    Help is appreciated. Thanks.
    Last edited: Aug 2, 2014
  2. jcsd
  3. Aug 2, 2014 #2


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    Use the series definition of [itex]e^x[/itex]. Set [itex]x = i\theta[/itex] and compare the real and imaginary parts of the result to the taylor series of cosine and sine.
  4. Aug 2, 2014 #3

    I'm afraid I have to use straight calculus to solve.
  5. Aug 2, 2014 #4


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    Then show that
    1) [itex]e^{i\theta}[/itex] and [itex] cos(\theta)+ isin(\theta)[/itex]
    have the same second derivative

    ii) The two functions have the same value at [itex]\theta= 0[/itex]

    iii) The derivatives of the two functions have the same value at [itex]\theta= 0[/itex].
  6. Aug 2, 2014 #5
    Finding the second derivative gives the same problem with the modulus. I also can't really use the value of θ as proof (for the first or second derivative), as I subbed it in to find the value of c from the integral.

    Edit: would simply stating +-[itex]e^{i\theta}[/itex] = [itex] cos(\theta)+ isin(\theta)[/itex], subbing in θ = 0 and stating therefore [itex]e^{i\theta}[/itex] = [itex] cos(\theta)+ isin(\theta)[/itex] be sufficient?
    Last edited: Aug 2, 2014
  7. Aug 2, 2014 #6

    Ray Vickson

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    You have been mislead by the "|.|" notation; although the symbol is the same, the meaning is very different in the real and complex cases. Your equation ##\int dg/g = \ln |g|## is false when ##g## is complex. In fact ##g = \cos(\theta) + i \sin(\theta)## satisfies ##|g| = 1 ## for all ##\theta##.

    I cannot understand why you refuse to use the series expansion approach. Has somebody told you that you are not allowed to do that?
  8. Aug 2, 2014 #7


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    I should have said "(1) Show that they satisfy the same second order differential equation" rather than have the same second derivative (they do but you have to use the statement you want to prove to show that).

    The first derivative of [itex]y= e^{i\theta}[/itex] is [itex]y'= ie^{i\theta}[/itex] and the second derivative is [itex]y''= -e^{i\theta}= -y[/itex].

    The first derivative of [itex]y= cos(\theta)+ isin(\theta)[/itex] is [itex]y'= -sin(\theta)+ icos(\theta)[/itex] and the second derivative is [itex]y''= -cos(\theta)- isin(\theta)= -y[/itex].

    Since both functions satisfy [itex]y''= -y[/itex], it is sufficient to show that they have the same value and first derivative at [itex]\theta= 0[/itex].
  9. Aug 2, 2014 #8
    I apologize if I am mistaken but I am doing advanced high school calculus and we are supposed to prove it using calculus and by treating i as a constant. Integrating complex functions is not part of the curriculum in any way but these are the instructions I have been given.
  10. Aug 2, 2014 #9

    Ray Vickson

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    You do not need to apologize for making a mistake, but you should take the opportunity of learning from your mistakes.

    You say you are required to "use calculus". What, exactly, does that mean? For example, the series expansion method is using calculus---that is where the series expansions come from. The other methods suggested are also using calculus.

    Before going farther you really need to understand something rather fundamental: what, exactly, do you mean by ##e^{i \theta}\,##? How is it defined? Depending on the type of answer you give, the subsequent steps may be straightforward or difficult; it all depends. That is why it is important for you to answer the question.
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