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First time user of Stack Exchange and this is my first post. This question is likely to be more trivial and vague but if someone gets the idea behind this I would be grateful.

During the time I was learning the basics of the frequency response of a resonant circuit, I have been presented with such a case, where:

  1. $f(x) = \sin(x) $

  2. $a = i = \sqrt{-1}$

  3. $\frac{d}{dx}\sin(x) = \cos(x) = \sin(x + 90^\circ) = i\,\sin(x) = a\,f(x)$

It is evident that the derivative of the function $f(x)$ is a multiplication of a complex number $a$ and the original function $f(x)$.

With this in mind, what would be the method of finding the complex factor/function '$a$' for any given function $f(x)$ such that the product of '$a$' and $f(x)$ results in the derivative of $f(x)$?

I am familiar with basic Laplace transforms and Fourier series but do not see the connection otherwise.

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  • $\begingroup$ I think that the connection actually happens to be circumstantial in this case; $\sin$ is a very special function after all. $\endgroup$ – abiessu Apr 10 '17 at 1:29
  • $\begingroup$ This doesn't have anything to do with Laplace or Fourier, but just how the trigonometric functions work. The use of complex anything here seems overkill though. Also, I don't know why anyone would ever denote $i$ by $a$. $\endgroup$ – Alfred Yerger Apr 10 '17 at 1:31
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Only some functions are solutions to $f'(x)=a\cdot f(x)$ (for complex $a$).

Rearranging it gives: $\frac{f'(x)}{f(x)}=a$

Integrating gives: $\ln f(x) = ax+c,\ c\in\mathbb{C}$

$$f(x)=e^{ax+c}$$

As $a$ is complex then $f(x)=e^{ax+c}=e^{nx+mi+c}=e^{nx+c}(\cos mx + i\sin mx),\ n,m\in \mathbb{R}$

If we let $e^c=A,\ A\in\mathbb{C}$ and $e^n=B,\ B\in\mathbb{R}$ then this becomes:

$$f(x)=A\cdot B^x(\cos mx + i\sin mx)$$

So only functions of this form will have that property.

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