# How to solve this equation in $\mathbb{C}$?

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.

• 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 $i=\sqrt{-1}$ Jan 31, 2016 at 15:22