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Math Help - e^ix

  1. #1
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    e^ix

    didnt know where to post this sorry

    how do we know that e^ix is a circle when plotted??
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    Quote Originally Posted by hmmmm View Post
    didnt know where to post this sorry

    how do we know that e^ix is a circle when plotted??
    e^{ix} = \cos x + i \sin x, thus if we let x = \cos x and y = \sin x then the coordinates of the points we use to plot the graph will be of the form (\cos x, \sin x) which results in a circle of radius 1
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    Quote Originally Posted by hmmmm View Post
    didnt know where to post this sorry

    how do we know that e^ix is a circle when plotted??
    Show that |e^{ix}|=1.
    This is true because e^{ix} = \cos x + i \sin x \implies |e^{ix}| = \cos^2 x + \sin^2 x = 1
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  4. #4
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    If \cos x + i \sin x = e^{ix} = a + ib, then a^2 + b^2 = 1. But then a^2 + b^2 = 1 is the equation of a circle with unit radius and origin as center in a-b plane.
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  5. #5
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    thanks

    thanks but the part i dont understand is how do you know that e^ix can be represented as cosx+isinx?? sorry
    thanks for the help
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  6. #6
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    Quote Originally Posted by hmmmm View Post
    thanks but the part i dont understand is how do you know that e^ix can be represented as cosx+isinx?? sorry
    thanks for the help
    Euler's formula - Wikipedia
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  7. #7
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    Quote Originally Posted by Jhevon View Post
    e^{ix} = \cos x + i \sin x, thus if we let x = \cos x and y = \sin x then the coordinates of the points we use to plot the graph will be of the form (\cos x, \sin x) which results in a circle of radius 1
    It may help to be able to derive Euler's formula from the polar form of a complex number \cos{x} + i\sin{x}. We can do this using derivatives.


    Let z = \cos{x} + i\sin{x}.

    \frac{dz}{dx} = -\sin{x} + i\cos{x}.

    But since i^2 = -1

    \frac{dz}{dx} = i^2\sin{x} + i\cos{x}

    \frac{dz}{dx} = i(\cos{x} + i\sin{x})

    \frac{dz}{dx} = iz.


    If we divide both sides by z, we get

    \frac{1}{z}\frac{dz}{dx} = i

    and so integrating both sides with respect to x gives

    \int{\frac{1}{z}\frac{dz}{dx}\,dx} = \int{i\,dx}

    \int{\frac{1}{z}\,dz} = ix + C

    \ln|z| = ix + C

    |z| = e^{ix + C}

    |z| = e^Ce^{ix}

    z = \pm e^Ce^{ix}

    z = Ae^{ix}.

    \cos{x} + i\sin{x} = Ae^{ix}.


    Substituting x = 0 into the equation gives A = 1.


    So z = \cos{x} + i\sin{x} = e^{ix}.

    And since we know \cos{x} + i\sin{x} is a circle of radius 1, so must be e^{ix}.
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  8. #8
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    Quote Originally Posted by Prove It View Post
    It may help to be able to derive Euler's formula from the polar form of a complex number \cos{x} + i\sin{x}. We can do this using derivatives.


    Let z = \cos{x} + i\sin{x}.

    \frac{dz}{dx} = -\sin{x} + i\cos{x}.

    But since i^2 = -1

    \frac{dz}{dx} = i^2\sin{x} + i\cos{x}

    \frac{dz}{dx} = i(\cos{x} + i\sin{x})

    \frac{dz}{dx} = iz.


    If we divide both sides by z, we get

    \frac{1}{z}\frac{dz}{dx} = i

    and so integrating both sides with respect to x gives

    \int{\frac{1}{z}\frac{dz}{dx}\,dx} = \int{i\,dx}

    \int{\frac{1}{z}\,dz} = ix + C

    \ln|z| = ix + C

    |z| = e^{ix + C}

    |z| = e^Ce^{ix}

    z = \pm e^Ce^{ix}

    z = Ae^{ix}.

    \cos{x} + i\sin{x} = Ae^{ix}.


    Substituting x = 0 into the equation gives A = 1.


    So z = \cos{x} + i\sin{x} = e^{ix}.

    And since we know \cos{x} + i\sin{x} is a circle of radius 1, so must be e^{ix}.
    Proof by calculus does require making some assumptions about complex derivatives. However, it can be proved that convenient properties (such as the chain rule) carry over from R to C.

    Personally I prefer the proof by Maclaurin series.

    You can also work with e^{ix} = \lim_{n \rightarrow \infty} \left( 1 + \frac{ix}{n} \right)^n. It's not too hard to show that the modulus is equal to 1 and that the argument is equal to x (I leave this for others who are interested).
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  9. #9
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    Quote Originally Posted by mr fantastic View Post
    Proof by calculus does require making some assumptions about complex derivatives. However, it can be proved that convenient properties (such as the chain rule) carry over from R to C.

    Personally I prefer the proof by Maclaurin series.

    You can also work with e^{ix} = \lim_{n \rightarrow \infty} \left( 1 + \frac{ix}{n} \right)^n. It's not too hard to show that the modulus is equal to 1 and that the argument is equal to x (I leave this for others who are interested).
    I KNEW you'd point that out Mr F. I tend to prefer this way just because it's more pleasing to the eyes (at least to my eyes). But you are right - I deliberately left out that piece of information to keep it simple for the OP
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  10. #10
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    thanks

    thanks but how then do we prove that the attributes of real numbers and thier derivatives can be applied to complex numbers??
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  11. #11
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by hmmmm View Post
    thanks but how then do we prove that the attributes of real numbers and thier derivatives can be applied to complex numbers??
    haha, short answer, but probably not the answer you want, is to study complex analysis. a complex analysis text will go through all the proofs and show what things carry over from real analysis and what things don't. you will find that complex analysis is a lot like working in \mathbb{R} \times \mathbb{R}. and so derivatives are defined in the same way, but the limits are taken differently. this leaves a lot of familiar properties intact though. you can look these things up if you're interested.
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  12. #12
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    thanks

    thanks
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