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Math Help - Euler's identity

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    Euler's identity

    Euler's identity basically states that e^{x\imath}=-1 which is derived from e^{x\imath}=\cos(x)+{\imath}\sin(x). (I'm using \imath= i. Which I think is the correct use, though am unsure.)

    I don't understand how they are getting to the latter form of the equation e^{x\imath}=\cos(x)+{\imath}\sin(x). If someone could explain this to me it would be greatly appreciated.
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    Re: Euler's identity

    Quote Originally Posted by Xeritas View Post
    Euler's identity basically states that e^{x\imath}=-1 which is derived from e^{x\imath}=\cos(x)+{\imath}\sin(x). (I'm using \imath= i. Which I think is the correct use, though am unsure.)

    I don't understand how they are getting to the latter form of the equation e^{x\imath}=\cos(x)+{\imath}\sin(x). If someone could explain this to me it would be greatly appreciated.
    It's actually \displaystyle e^{i\pi} = -1 which you get from letting \displaystyle x = \pi in the equation \displaystyle e^{ix} = \cos{x} + i\sin{x}.

    Anyway, in order to derive \displaystyle e^{ix} = \cos{x} + i\sin{x}, start by defining a complex number \displaystyle z = \cos{x} + i\sin{x}. Then

    \displaystyle \begin{align*} z &= \cos{x} + i\sin{x} \\ \frac{dz}{dx} &= -\sin{x} + i\cos{x} \\ \frac{dz}{dx} &= i^2\sin{x} + i\cos{x} \\ \frac{dz}{dx} &= i\left(\cos{x} + i\sin{x}\right) \\ \frac{dz}{dx} &= i\,z \\ \frac{1}{z}\,\frac{dz}{dx} &= i \\ \int{\frac{1}{z}\,\frac{dz}{dx}\,dx} &= \int{i\,dx} \\ \int{\frac{1}{z}\,dz} &= i\,x + C_1 \\ \ln{|z|} + C_2 &= i\,x + C_1 \\ \ln{|z|} &= i\,x + C \textrm{ where }C = C_1 - C_2 \\ |z| &= e^{i\,x + C} \\ |z| &= e^Ce^{i\,x} \\ z &= \pm e^Ce^{i\,x} \\ z &= A\,e^{i\,x} \textrm{ where } A = \pm e^C \end{align*}

    We know that when \displaystyle x = 0, z = \cos{0} + i\sin{0} = 1, so \displaystyle 1 = Ae^0 \implies A = 1

    Therefore \displaystyle z = e^{i\,x}, and so \displaystyle e^{i\,x} = \cos{x} + i\sin{x}.
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