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Let $f$ and $g$ be two non-zero functions $ \mathbb R \to \mathbb R $ which are continuous and differentiable everywhere. Furthermore, say that for all integer $n$:

$$ (f(\theta) + i g(\theta))^n = f(n\theta) + ig(n\theta) \tag{de Moivre's Theorem}$$

Does it follow that $f(\theta) = e^{bx} \cos a\theta$ and $g(\theta) = e^{bx} \sin a\theta$ for some $a$ and $b$? If not, can we characterize all pairs of functions $f$ and $g$ satisfying this identity? (edit: changed conjectured forms for $f$ and $g$)

I have derived some facts about $f$ and $g$:

$$ f(0) = 1 \text{ and } g(0) = 0 $$ $$ f(\theta)^2 + g(\theta)^2 = 1 $$ $$ f(n\theta) = f(\theta)f((n-1)\theta) - g(\theta)g((n-1)\theta)$$ $$ g(n\theta) = g(\theta)f((n-1)\theta) + f(\theta)g((n-1)\theta)$$

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  • $\begingroup$ Hint: Let $f(\theta)+ig(\theta)=a(\theta)e^{i\phi(\theta)}$ then, since $f$ and $g$ are non-zero,$\frac{a(\theta)^n}{a(n\theta)}=e^{i[\phi(n\theta)-n\phi(\theta)]}\in \mathbb{R}\,\forall n$. $\endgroup$ – lemon Sep 1 '14 at 16:54
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Well, this is actually a lot easier than I thought after getting the (correct) conjectured form. Thanks to @lemon for helping with that.

Let $q(x) = f(x) + i g(x) $. Now we have that:

$$ q(x)^n = q(nx) $$

From which $q(x)$ is the exponential function:

$$q(x) = e^{cx} \text{ with } c\in \mathbb C$$

The conjectured form follows.

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