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Why can exp function in complex plane be defind as e^x(cosy+i siny)

  1. Aug 29, 2009 #1
    Hi,

    I know this one is stupid, but i am still confused. why e^(iy) = cosy + i siny?

    thank you.
     
  2. jcsd
  3. Aug 29, 2009 #2
    You can take this as a definition or you can expand the power series for e^z in terms of the power series for cos z and sin z.
     
  4. Aug 29, 2009 #3

    mathman

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    It is not by definition. It is a theorem, most easily proven by matching power series.
     
  5. Aug 30, 2009 #4
    One absolutley can take this as a definition.
     
  6. Aug 30, 2009 #5

    Hootenanny

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    No, it's a theorem, i.e. it has been proved.
     
  7. Aug 30, 2009 #6
    it can be proved as a theorem but it can also be done the other way around as a definition. It is a matter of convention.
     
    Last edited: Aug 30, 2009
  8. Aug 30, 2009 #7
    If you make the right definitions and/or assumptions you can prove it, or you can define it that way. If you are sufficiently careful the different definitions imply each other.
     
  9. Aug 30, 2009 #8
    Look at it this way. I define exp(ia) = cos a + i sin a where cos and sin are defined over the real numbers. That just plain works. from this is is trivial to derive the entire complex exponential function.
     
    Last edited: Aug 31, 2009
  10. Aug 30, 2009 #9

    mathman

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    The exponential, sine, and cosine functions have precise definitions. The usual definitions are as follows:
    e=lim(n->oo) (1 + 1/n)sup[n][/sup]
    sin=ratio of side opposiste to hypotenuse of right triangle.
    cos=ratio of adjacent side to hypotenuse of right triangle.

    Given these definitions, the powers series for these functions can be derived and the relationship proved. It is NOT a definition of anything!
     
  11. Aug 30, 2009 #10
    ????? explain. There is no doubt that the complex exponential can be defined in terms of the real cosine and sine.
     
  12. Aug 30, 2009 #11
    You can also say that the definition:

    exp(x+i y) = exp(x)[cos(y) + i sin(y)]

    makes exp(z) a complex differentiable function (you can easily check that it satisfies the Cauchy-Riemann equations). The result from complex analysis that complex differentiable functions are analytic thus makes exp(z) an analytic function.

    Then the result from complex analysis that analytic continuations are unique makes the above definition the only possible complex analytic function that reduces to exp(x) on the real axis.
     
  13. Aug 31, 2009 #12

    HallsofIvy

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    Those are NOT the "usual definitions" and the power series for the functions cannot be derived from them. For one thing, those definitions do not define sin(x) for x greater than [itex]\pi/2[/itex] or less than 0.

    And, of course, "e=lim(n->oo) (1 + 1/n)n" does not define ex at all, it only defines e.

    It is, in fact, very common to define cos(x) to be
    [tex]\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} x^{2n}[/tex]
    and sin(x) to be
    [tex]\sum_{n=0}^\infty \frac{(1)^n}{(2n+1)!} x^{2n+1}[/tex]

    although, personally, I prefer
    "cos(x) is the function, y, satisfying y"= -y, y(0)= 1, y'(0)= 0"
    and
    "sin(x) is the function, y, satisfying y"= -y, y(0)= 0, y'(0)= 1"

    Similarly, ex can be defined to be
    [tex]\sum_{n=0}^\infty \frac{1}{n!} x^n[/itex]

    although it can also be defined as the inverse function to "ln(x)" where ln(x) is defined by
    [tex]ln(x)= \int_1^x \frac{1}{t}dt[/tex]
     
  14. Aug 31, 2009 #13

    arildno

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    Another way of proceeding, is by stating what properties one wish the exponential function to have, and in a non-rigorous manner, just display that cos(x)+isin(x) DOES, in fact, display those features!

    So, what do we wish the exponential function to display?

    First off, we want Exp(x+y)=Exp(x)*Exp(y) for all arguments.
    Note that this leads to a second requirement, namely Exp(0)=1.

    Furthermore, we wish Exp'(kx)=kExp(x) for all k's.

    Now, let us look at F(x)=cos(x)+isin(x)

    We have:
    [tex]F(x)*F(y)=(\cos(x)+i\sin(x))(\cos(y)+i\sin(y))=((\cos(x)\cos(y)-\sin(x)\sin(y))+i(\cos(x)\sin(y)+\cos(y)\sin(x))=\cos(x+y)+i\sin(x+y)=F(x+y)[/tex]

    The first requirement is therefore fulfilled!

    Secondly, we have: F'(x)=-sin(x)+icos(x)=i(cos(x)+isin(x))=iF(x).

    Thus, we see why it is convenient, and justified, to DEFINE F(x)=Exp(ix).
     
  15. Aug 31, 2009 #14
    As other people have stated, we start with some basic facts about the functions e^x, sin x, and cos x. Namely, we write them out as a power series:

    [tex]\cos x = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} x^{2n}[/tex]

    [tex]\sin x = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} x^{2n+1}[/tex]

    [tex]e^x = \sum_{n=0}^\infty \frac{1}{n!} x^n[/tex]

    So just for this conversation, we're agreeing that these are true statements.

    To show the relation, we simply evaluate e^{ix} and expand out the series.

    [tex]e^{ix} = \sum_{n=0}^\infty \frac{1}{n!} {(ix)}^n = \sum_{n=0}^\infty \frac{1}{n!} i^n x^n[/tex]

    You see? All we did was substitute xi for x and did a teeny bit of algebra. But Something interesting happens with the [tex]i^n[/tex] expression.

    When n is divisible by 4, (so n = 0, 4, 8, 12, 16, etc), [tex]i^n = 1[/tex].
    When n is one greater than a number divisible by 4 (n = 1, 5, 9, 13, 17, etc), then [tex]i^n = i[/tex].
    When n is two greater than a number divisible by 4 (n = 2, 7, 10, 14, etc), then [tex]i^n = -1[/tex].
    When n is three greater than a number divisible by 4 (n = 3, 8, 11, 15, etc), then [tex]i^n = -i[/tex].

    The notation isn't great for this next step, but if you think about it logically, you can split the sum into four sums: a positive real part, a negative real part, a positive imaginary part, and a negative imaginary part. If you put the real parts with the other real parts, you get a sum which alternates between positive and negative terms with even powers. Play with it a bit and you'll notice this is exactly the series for cosine! Then take the imaginary terms. They too alternate signs. But they are all odd powers! This is exactly the series for sine, multiplied by i.

    When we put them together, we see the result is in fact,

    [tex]e^{ix} = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} x^{2n} + i \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} x^{2n+1} = \cos x + i \sin x
    [/tex]

    Note, lastly, that this only works when x is a real number. Otherwise, the imaginary part of x would interfere with the argument above and mess everything up! This isn't a big problem though. If z = [tex]a + bi[/tex] is a complex number, and a and b are the real and imaginary parts respectively, then [tex]e^z = e^{a + bi} = e^a e^{bi} = e^a (\cos b + i \sin b)[/tex]. In fact, it's quite convenient because the imaginary part rotates in the complex plane and the real part scales it as an exponent.
     
  16. Aug 31, 2009 #15
  17. Aug 31, 2009 #16

    mathman

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    Once e is defined, ex can be defined just like any other power of a positive number, first by defining integer powers, then defining integer roots, then all rational powers, and finally all powers.
     
  18. Sep 2, 2009 #17
    This thread is funny. I thought that if four statements were equivalent and one was a definition, any of them could be taken as the definition.
     
  19. Sep 2, 2009 #18

    mathman

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    (repeating myself) The sine, cosine, and exponential functions are all defined separately. Once they are defined, their relationships are theorems. The simplest one of course is sin2x + cos2x = 1, which even holds for complex x. Euler's formula is also a theorem.

    A lttle history (from Wikipedia)
    Euler's formula was proven for the first time by Roger Cotes in 1714 in the form:

    ln(cosx + isinx) = ix

    (where "ln" means natural logarithm, i.e. log with base e).[4]

    It was Euler who published the equation in its current form in 1748, basing his proof on the infinite series of both sides being equal. Neither of these men saw the geometrical interpretation of the formula: the view of complex numbers as points in the complex plane arose only some 50 years later (see Caspar Wessel). Euler considered it natural to introduce students to complex numbers much earlier than we do today. In his elementary algebra text book, Elements of Algebra, he introduces these numbers almost at once and then uses them in a natural way throughout.
     
  20. Sep 3, 2009 #19

    Landau

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    Mathman, I don't know why you are so stubborn. There are no absolute, universal, mathematical 'laws' which dictate that everyone must use a certain definition. There are several ways to define the sine, cosine, and exp. Which one you use is a matter of taste (or convention), and given some definition, you can proceed and prove theorems.

    Five common definitions of the exponential function, and their being equivalent, are nicely displayed here. Your geometric definitions of sine and cosine are obviously not useful for doing serious mathematical analysis. Moreover, they aren't even defined for angles larger than pi/2 as already pointed out.
     
  21. Sep 4, 2009 #20

    HallsofIvy

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    How is he being stubborn? He did not say that we MUST use any specific definition, only that "The sine, cosine, and exponential functions are all defined separately. Once they are defined, their relationships are theorems. "
     
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