Differentiably redundant functions. I am looking for a differentiably redundant function of order 6 from the following.
(a) $e^{-x} + e^{-x/ 2} \cos({\sqrt{3x} \over 2})$
(b) $e^{-x} + \cos(x)$
(c) $e^{x/2}\sin({\sqrt{3x} \over 2})$
I know that (b) has order 4, but I cannot solve for (a) and (c).
It would be a huge waste of time if I took the derivatives and calculate them, so there must be a simple way to solve this.
According to the book, it is related to $1/2 \pm i\sqrt{3} /2$, but why is that?

"Differentiably redundant function of order $n$" is not a standard mathematical term: this is something that GRE Math authors made up for this particular problem. 

Define a function $f(x)$ to be differentiably redundant of order $n$ if the $n$th derivative $f^{(n)}(x)=f(x)$ but $f^{(k)}(x)\ne f(x)$ when $k<n$. Which of the following functions is differentiably redundant of order $6$? 

By the way, this is not a shining example of mathematical writing: "when $k<n$" should be "when $0<k<n$" and, more importantly, $\sqrt{3x}$ was meant to be $\sqrt{3}x$ in both (A) and (C). This looks like a major typo in the book.
If you are familiar with complex numbers, the appearance of both $-1/2$ and $\sqrt{3}/2$ in the same formula is quite suggestive, especially since both exponential and trigonometric functions appear here. Euler's formula  $e^{it}=\cos t+i\sin t$ should come to mind. Let $\zeta=-\frac12+i\frac{\sqrt{3}}{2}$: then 
$$e^{-x/2}\cos \frac{\sqrt{3}x}{2} = \operatorname{Re}e^{\zeta x},\qquad e^{-x/2}\sin \frac{\sqrt{3}x}{2} = \operatorname{Im}\, e^{\zeta x}$$
Differentiating $n$ times, you get the factor of $\zeta^n$ inside of $\operatorname{Re}$ and $\operatorname{Im}$. Then you should ask yourself: what is the smallest positive integer $n$ such that   $\zeta^n=1$? Helpful article.
