Question 619152
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Yes


Since all complex numbers can be expressed in polar coordinates, for some real numbers *[tex \LARGE r] and *[tex \LARGE \varphi] depending on *[tex \LARGE x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^{ix}\ =\ r\left(\cos(\varphi)\ +\ i\sin(\varphi)\right)]


Since *[tex \LARGE \frac{d}{dx}e^{ix}\ =\ ie^{ix}]


Differentiate both sides using the chain rule in the RHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ie^{ix}\ =\ r\left(\cos(\varphi)\ +\ i\sin(\varphi)\right)\frac{dr}{dx}\ +\ r\left(-\sin(\varphi)\ +\ i\cos(\varphi)\right)\frac{d\varphi}{dx}]


Substitute *[tex \LARGE r\left(\cos(\varphi)\ +\ i\sin(\varphi)\right)] for *[tex \LARGE e^{ix}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ir\left(\cos(\varphi)\ +\ i\sin(\varphi)\right)\ =\ r\left(\cos(\varphi)\ +\ i\sin(\varphi)\right)\frac{dr}{dx}\ +\ r\left(-\sin(\varphi)\ +\ i\cos(\varphi)\right)\frac{d\varphi}{dx}]


Equating the real and imaginary parts in the above gives *[tex \LARGE \frac{dr}{dx}\ =\ 0] and *[tex \LARGE \frac{d\varphi}{dx}\ =\ 1]


Since *[tex \LARGE e^{i0}\ =\ 1] we have the initial values of *[tex \LARGE r(0)\ =\ 1] and *[tex \LARGE \varphi(0)\ =\ 0], hence *[tex \LARGE r\ =\ 1] and *[tex \LARGE \varphi\ =\ x] and therefore:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^{ix}\ =\ \left(\cos(x)\ +\ i\sin(x)\right)]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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