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Let $A_1,A_2,A_3$ and $\alpha_1,\alpha_2,\alpha_3$ be real constants.

Suppose the equation $A_1e^{i\alpha_1x}+A_2e^{i\alpha_2x}=A_3e^{i\alpha_3x}$ holds $\forall x\in\mathbb{R}$. (where $i^2=-1$)

Can we find the relation between the constants $\alpha_1,\alpha_2$ and $\alpha_3$?

2 Answers 2

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Suppose we have $a_1e^{i\alpha_1x} + A_2e^{i\alpha_2x} = A_3e^{i\alpha_3x}$. Scale the $\alpha_i$ such that their absolute values are each greater than $1$. Differentiating $k$ times we get and then dividing out by $i^k$ we get: $A_1\alpha_1^ke^{i\alpha_1x} + A_2\alpha_2^ke^{i\alpha_2x} = A_3\alpha_3^ke^{i\alpha_3x}$ Plugging in $x=0$ we get the following holds for all nonnegative integers $k$: $A_1\alpha_1^k + A_2\alpha_2^k = A_3\alpha_3^k$ Suppose not all three of the $\alpha_i$ were equal. Then let $k$ go towards infinity. But then whichever side has the $\alpha_i$ with the highest modulus will be by far the largest term unless two of the $\alpha_i$ are equal, thus we must have two of the $\alpha_i$ are equal. But then it is easy to deduce the corresponding $A_i$ are equal. Thus we can deduce two solutions:

Either two of the $\alpha_i$ are equal and the corresponding $A_i$ are equal, and the third $A_i$ is $0$ OR all the $\alpha_i$ are equal and we have $A_1 + A_2 = A_3$.

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Since the equation \begin{gather}A_1e^{i\alpha_1x}+A_2e^{i\alpha_2x}=A_3e^{i\alpha_3x} \tag{*} \end{gather} holds $\forall x\in\mathbb{R}$ and $e^{i\alpha x}$ is infinitely differentiable function, differentiation $(*)$ yields $A_1 {i\alpha_1}e^{i\alpha_1x}+A_2{i\alpha_2}e^{i\alpha_2x}=A_3{i\alpha_3}e^{i\alpha_3x}.$ Then we can put $x=0:$ \begin{gather} A_1 {i\alpha_1}+A_2{i\alpha_2}=A_3{i\alpha_3}, \\ A_1 {\alpha_1}+A_2{\alpha_2}=A_3{\alpha_3}. \end{gather}