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From a small simple calculation , we get the following formulas: $ \begin{cases} e^x = \Big( \displaystyle \sum_{n \geq 0} \frac{x^{3n}}{(3n)!} \Big) + \Big( \displaystyle \sum_{n \geq 0} \frac{x^{3n+1}}{(3n+1)!} \Big) + \Big( \displaystyle \sum_{n \geq 0} \frac{x^{3n+2}}{(3n+2)!} \Big) = A_0 (x) + A_1 (x) + A_2 (x) \\ e^{jx} = \Big( \displaystyle \sum_{n \geq 0} \frac{x^{3n}}{(3n)!} \Big) + j \Big( \displaystyle \sum_{n \geq 0} \frac{x^{3n+1}}{(3n+1)!} \Big) + j^2 \Big( \displaystyle \sum_{n \geq 0} \frac{x^{3n+2}}{(3n+2)!} \Big) = A_{0} (x) + j A_{1} (x) + j^2 A_2 (x) \\ e^{j^{2} x} = \Big( \displaystyle \sum_{n \geq 0} \frac{x^{3n}}{(3n)!} \Big) + j^2 \Big( \displaystyle \sum_{n \geq 0} \frac{x^{3n+1}}{(3n+1)!} \Big) + j \Big( \displaystyle \sum_{n \geq 0} \frac{x^{3n+2}}{(3n+2)!} \Big) = A_{0} (x) + j^2 A_{1} (x) + j A_2 (x) \end{cases} $ with : $ j = e^{ i \dfrac{2 \pi}{3} } $.

That means : $ \begin{cases} A_0 (x) = \displaystyle \sum_{n \geq 0} \frac{x^{3n}}{(3n)!} = \frac{1}{3} ( e^{x} + e^{jx} + e^{j^{2} x } ) \\ A_1 (x) = \displaystyle \sum_{n \geq 0} \frac{x^{3n+1}}{(3n+1)!} = \frac{1}{3} ( e^{x} + j^2 e^{jx} + j e^{j^{2} x } ) \\ A_2 (x) = \displaystyle \sum_{n \geq 0} \frac{x^{3n+2}}{(3n+2)!} = \frac{1}{3} ( e^{x} + j e^{jx} + j^2 e^{j^{2} x } ) \end{cases} $

My question is the following:

Could you tell me a method to find $ a $ and $ b $ in $ \mathbb {C} $ according to $ x $ such that : $ e^{ jx } = ( A_0 (a) + j A_1 ( a) ) ( A_0 (b) + j A_1 ( b) ) $

Thank you very much for your help.

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    $\begingroup$ I might note that perhaps the best answer is to steer clear of these big summations and simply to stick with Euler form: $$e^{ix}=\cos(x)+i\sin(x)$$Where $i=\sqrt{-1}$ $\endgroup$ Jan 31, 2016 at 15:22

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