# Complex factor when finding derivative

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.

• I think that the connection actually happens to be circumstantial in this case; $\sin$ is a very special function after all. – abiessu Apr 10 '17 at 1:29
• 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 $a$. – Alfred Yerger Apr 10 '17 at 1:31

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.