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Math Help - trig functions over complex values

  1. #1
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    trig functions over complex values

    Usually sin and cos are defined as being over the domain of angles. However, what is to stop the following:
    (1) if x is in radians, we have the infinite series expansions of each one, and if we put in "i" for x, we get a value.
    (2) since, again with x in radians, sin x = (e^ix - e^-ix)/2i and similarly for cos x, if we now put in "i" for x, we again get values.

    This would seem to make it appear that the trig functions could indeed be expanded to complex values. True, what it would be interpreted as is another question, but otherwise, is there anything wrong with this idea?
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    Your (1) and (2) definitely confirm that the trigonometric functions (and hyperbolic) can be expanded to complex values. It all comes from Euler's Formula...

    Consider these MacLaurin series, which can have complex values substituted...

    \sin{z} = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \frac{z^7}{7!} + \dots - \dots

    \cos{z} = 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \frac{z^6}{6!} + \dots - \dots

    e^{z} = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!}+ \frac{z^4}{4!} + \frac{z^5}{5!} + \frac{z^6}{6!} + \frac{z^7}{7!} + \dots


    Notice that means that

    e^{i\theta} = 1 + i\theta + \frac{(i\theta)^2}{2!} + \frac{(i\theta)^3}{3!} + \frac{(i\theta)^4}{4!} + \frac{(i\theta)^5}{5!} + \frac{(i\theta)^6}{6!} + \frac{(i\theta)^7}{7!} + \dots

     = 1 + i\theta - \frac{\theta^2}{2!} - \frac{i\theta^3}{3!} + \frac{\theta^4}{4!} + \frac{i\theta^5}{5!} - \frac{\theta^6}{6!} - \frac{i\theta^7}{7!} + \dots - \dots

     = \left(1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} + \dots - \dots\right) + i\left(\theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \frac{\theta^7}{7!} + \dots - \dots \right)

     = \cos{\theta} + i\sin{\theta}.


    So e^{i\theta} \equiv \cos{\theta} + i\sin{\theta}.

    That means with the combination of other trigonometric identities, the trigonometric functions can be defined for complex values through the exponential function.
    Last edited by Prove It; August 13th 2010 at 03:17 AM.
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  3. #3
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    In fact, there is nothing wrong with complex-valued trig functions. In fact, in complex analysis, many functions are allowed to take complex arguments, and in return they produce complex numbers. If you're interested in these things, I would highly recommend that you study complex analysis. It's a great subject, anyway, and extremely powerful.
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  4. #4
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    Quote Originally Posted by Ackbeet View Post
    In fact, there is nothing wrong with complex-valued trig functions. In fact, in complex analysis, many functions are allowed to take complex arguments, and in return they produce complex numbers. If you're interested in these things, I would highly recommend that you study complex analysis. It's a great subject, anyway, and extremely powerful.
    One thing that makes complex analysis so powerful is the fact that the manipulation of complex functions can actually aid manipulation of real functions.

    A perfect example is trigonometric identities...

    Suppose we had

    (\cos{x} + i\sin{x})^3.

    A binomial expansion leads to

    (\cos{x} + i\sin{x})^3 = \cos^3{x} + 3i\cos^2{x}\sin{x} + 3i^2\cos{x}\sin^2{x} + i^3\sin^3{x}

     = \cos^3{x} - 3\cos{x}\sin^2{x} + i(3\cos^2{x}\sin{x} - \sin^3{x})

     = \cos^3{x} - 3\cos{x}(1 - \cos^2{x}) + i[3(1-\sin^2{x})\sin{x} - \sin^3{x}]

     = \cos^3{x} - 3\cos{x} + 3\cos^3{x} + i(3\sin{x} - 3\sin^3{x} - \sin^3{x})

     = 4\cos^3{x} - 3\cos{x} + i(3\sin{x} - 4\sin^3{x}).


    But we also have

    (\cos{x} + i\sin{x})^3 = \cos{3x} + i\sin{3x}

    by DeMoivre's Theorem.


    So that means

    4\cos^3{x} - 3\cos{x} + i(3\sin{x} - 4\sin^3{x}) = \cos{3x} + i\sin{3x}.


    By equating real and imaginary parts, that means

    \cos{3x} = 4\cos^3{x} - 3\cos{x} and \sin{3x} = 3\sin{x} - 4\sin^3{x}.


    So using Complex Analysis, we have been able to prove two Real Trigonometric identities.



    Also, if you appreciate the beauty of mathematics, look at Euler's formula

    e^{i\theta} = \cos{\theta} + i\sin{\theta}.


    Notice that if \theta = \pi we have

    e^{i\pi} = \cos{\pi} + i\sin{\pi}

    e^{i\pi} = -1 + 0i

    e^{i\pi} = -1

    e^{i\pi} + 1 = 0.


    This is considered to be the most beautiful equation in existence, because it links the five fundamental constants of mathematics, as well as the symbol for balance and the symbol for incrementing...
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    continuation of complex values of trigonometric functions

    First, Akbeet and Prove It, thank you for your answers. Comments on your answers:
    The first answer by Prove It merely restates my question by putting down the formulas that I simply referred to by name.
    The second answer, by Akbeet, with the recommendation to study complex analysis: I have (albeit a number of years ago). Of course I know that many functions are defined over the set of complex numbers. I simply do not remember any instance which explicitly uses, or even needs, complex values in the domain and/or the codomain of sin or cos. Could you provide me with an example with a clear interpretation of the domain and range? I would imagine that we would be talking about some periodic function where the equivalence to the schoolbook definition of the argument of sin as representing the angle measurements, usually in Radians, and the range values as representing the corresponding y-value on the unit circle, is clear. However, all the periodic functions that I can think of that use a trig function uses real values in its domain. I would be grateful for an actual, not contrived, application of complex-valued trig functions.
    The following post, by Akbeet, presents a few equations which, although of course very pretty, are not relevant to my question, since the trig functions you use are standardly defined over the real numbers. The exponential function is, of course, defined over the complex numbers, but I am referring to sin, cos, tan, etc.
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    PS to last post

    Postscript: in my post of a few minutes ago, I typed in "codomain" too quickly. Of course if your circle is in the complex plane you are going to have a range of the sin function of imaginary numbers, and so forth. So please concentrate on my question about the domain. Also, note I am talking about the trig functions, not the inverse trig functions.
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  7. #7
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    I think you are talking specifically about say \cos(z) = \cos(x+i y). In this case

    [LaTeX ERROR: Convert failed]

    [LaTeX ERROR: Convert failed]

    [LaTeX ERROR: Convert failed] " alt="\displaystyle 2\cos(x+i y) = e^{(x+i y)i}+e^{-(x+i y)i} [/Math]

    [LaTeX ERROR: Convert failed]

    [LaTeX ERROR: Convert failed]

    [LaTeX ERROR: Convert failed] " />[LaTeX ERROR: Convert failed] [LaTeX ERROR: Convert failed]

    [Math]\displaystyle = 2\cos(x)\cosh(y)-2i\sin(x)\sinh(y)" alt="

    [LaTeX ERROR: Convert failed]

    [Math]\displaystyle = 2\cos(x)\cosh(y)-2i\sin(x)\sinh(y)" />

    Now for a complex argument, you get this interpretation. If x = 0 you have a purely imaginary argument which leads to \cos(i y) = \cosh(y), which is kind of neat and shows that the hyberbolic functions are very tightly together with the trig functions.
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    Exellent

    Thank you, Vlasev. Finally a reply that directly addressed the question. That trig functions with complex arguments can be directly tied to hyperbolic functions with real arguments is the point that I had overlooked, and leads to the real interpretations that I was looking for. Thanks again.
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    Quote Originally Posted by nomadreid View Post
    Thank you, Vlasev. Finally a reply that directly addressed the question. That trig functions with complex arguments can be directly tied to hyperbolic functions with real arguments is the point that I had overlooked, and leads to the real interpretations that I was looking for. Thanks again.
    Perhaps the reason why you did not get specific answers which "addressed the question" is because your "question" was not clear enough. You asked if trig functions could be extended to complex values. Every post in this thread has shown that yes, they can. If you had wanted to know how to evaluate them, then you should have asked that specifically.
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    Point taken

    Right you are, Prove It. I re-read my original post, and realize that it was too narrowly stated. What I meant by "is there anything wrong with this idea" was that for a mathematical concept to be meaningful, it needs to have some interpretation, so that my fault lay in the caveat about the interpretation in the original post. Anyway, mea culpa, and thanks for the input.
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    Postscript

    I really have to stop pressing the Submit button so quickly. I forgot to say that one of the implicit questions in my first post is that I mentioned that the equalities used assumed that the argument is in Radians, but then I used a complex argument, and Radians are defined only over real values. Hence, part of the "is anything wrong with this?" implied "does the condition that the argument be in Radians have any sense?" So my real fault was not being explicit enough.
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  12. #12
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    Quote Originally Posted by nomadreid View Post
    I really have to stop pressing the Submit button so quickly. I forgot to say that one of the implicit questions in my first post is that I mentioned that the equalities used assumed that the argument is in Radians, but then I used a complex argument, and Radians are defined only over real values. Hence, part of the "is anything wrong with this?" implied "does the condition that the argument be in Radians have any sense?" So my real fault was not being explicit enough.
    I think the answer to that question lies in the fact that some functions change "meaning" when they are extended from the being a real-valued function to a complex-valued function.
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    Thanks again, Prove It. Good answer. By the way, I add to my apologies: being new to this Forum, I didn't really pay much attention to the "Thank You" button that apparently I should have been clicking. That goes for everyone else who has been contributing. Thanks all around, and sorry if I have been a bit sloppy.
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  14. #14
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    Quote Originally Posted by nomadreid View Post
    Usually sin and cos are defined as being over the domain of angles.
    This is an incorrect statement. In "elementary" mathematics, sine and cosine are typically used to solve "triangle problems" and so their argument is typically an angle, however, for more advanced mathematics, (calculus and beyond) they are used primarily to model periodic functions and they are defined over numbers, like any function, not angles.

    However, what is to stop the following:
    (1) if x is in radians, we have the infinite series expansions of each one, and if we put in "i" for x, we get a value.
    (2) since, again with x in radians, sin x = (e^ix - e^-ix)/2i and similarly for cos x, if we now put in "i" for x, we again get values.

    This would seem to make it appear that the trig functions could indeed be expanded to complex values. True, what it would be interpreted as is another question, but otherwise, is there anything wrong with this idea?
    Any function that can be defined for real numbers can be extended to complex numbers. Those that are "analytic" (are equal to their Taylor's series in some domain) like sine and cosine are particularly easy to extend.
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    sine and cosine are ... used primarily to model periodic functions and they are defined over numbers, like any function, not angles.
    Off the cuff, I would say that if one replaced numbers by real numbers, there would be no distinction, since any periodic function is isomorphic to a function which winds around a center, and so the real numbers in the arguments could be interpreted as angles. It's the inclusion of complex numbers in the domain that throws this interpretation to the wolves. Right?

    Any function that can be defined for real numbers can be extended to complex numbers.
    Interesting statement, and intuitively would seem to be correct, but is there an easy proof? Yes, analytical ones are easy enough, but to paraphrase a character of Hanns Johst, when I hear the word "all", I reach for my gun. Er, at least for a pen.
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