Finding a general solution to $\frac{d^n u}{dx^n} = -u$, for someone with limited knowledge of Complex Analysis I understand that for $a\in \mathbb{R},$ the function $\,u(x)=ae^{-x}\,$ is a solution to the differential equation,
$$\frac{d u}{dx} = -u,$$
and that for $b,c \in \mathbb{R},\,$ the function $u=b\cos(x) + c\sin(x)$ satisfies
\begin{equation}
\qquad \;\frac{d^2u}{dx^2} = -u. \quad (1)
\end{equation}
I also understand that Euler's formula gives $e^{ix} = \cos(x)+i\sin(x).$ This is where the first obvious hole in my knowledge appears: I want to say that $e^{ix}$ must be a solution to $(1)$, but I have no concept of the derivative for complex-valued functions. Does the chain rule hold for functions like this?
Anyways, I want to conclude that the pattern here continues; in other words, can we say that $u(x)=e^{\omega x}$ satisfies 
$$\frac{d^n u}{dx^n} = -u $$
whenever $\omega$ is a $2n^{\text{th}}$ root of unity? From here, how can we construct a general solution consisting only of real valued functions? This last part is what I'm mainly interested in, so if there is a less complex way to approach this problem, (pun not intended) that would be nice to see as well. 
 A: The complex solutions are of the form:
$$u(x) = e^{\omega x}$$
Whenever $\omega$ is a complex number satisfying $\omega^{n} = -1$. If $\omega = \alpha + \beta i$, then we can get real solutions:
$$u(x) = e^{\alpha x}\cos(\beta x); \qquad u(x) = e^{\alpha x}\sin(\beta x)$$
We can find all complex numbers $\omega$ with $\omega^{n} = -1$: They are $e^{i\pi(2k + 1)/n} = \cos(\frac{(2k + 1)\pi}{n}) + i\sin(\pi\frac{(2k + 1)\pi}{n})$, for $k = 0,1,2,\ldots,n-1$. However, you can see that $\omega = \alpha + \beta i$ and $\overline{\omega} = \alpha - \beta i$ give you the same pair of real solutions. So when $n$ is even, you only need to consider $k = 0,1,\ldots,n/2 - 1$. When $n$ is odd, notice that $k = (n-1)/2$ gives $\omega = e^{i \pi} = -1$, which gives the already-real solution $e^{-x}$. In conclusion, the general solution is:
($n$ even):
$$y = \sum_{k = 0}^{n/2 - 1} A_ke^{\alpha_kx}\cos(\beta_k x) + B_ke^{\alpha_k x}\sin(\beta_k x); \qquad \alpha_k = \cos\left(\frac{(2k + 1)\pi}{n}\right), \beta_k = \sin\left(\frac{(2k + 1)\pi}{n}\right)$$
($n$ odd):
$$y = Ce^{-x} + \sum_{k = 0}^{(n - 1)/2 - 1} A_ke^{\alpha_kx}\cos(\beta_k x) + B_ke^{\alpha_k x}\sin(\beta_k x)$$
A: First, a short 'proof' of Euler's formula,
$$y=\frac{\cos(x)+i\sin(x)}{e^{ix}}$$
$$\frac{dy}{dx}=\text{insert quotient rule here}=0$$
$$\frac{dy}{dx}=0\implies y=C$$
Now plug in $x=0$ to find $C=1$, so we have
$$1=\frac{\cos(x)+i\sin(x)}{e^{ix}}\implies e^{ix}=\cos(x)+i\sin(x)$$

Now, notice that if $f(x)$ can be broken into a real part and imaginary part, then the real and imaginary parts will themselves be solutions to the differential equation.
So, if we have
$$\frac{d^n}{dx^n}u=-u$$
$$u(x)=e^{\omega x}=e^{(a+bi)x}=e^{ax}e^{bix}=e^{ax}(\cos(bx)+i\sin(bx))$$
Then another solution is given by
$$v_R(x)=e^{ax}\cos(bx),\ v_I(x)=e^{ax}\sin(bx)$$
And sum up these solutions (with constant in front) as @AlexZorn has shown:
$$f(x)=\sum R_ke^{a_kx}\cos(b_kx)+I_ke^{a_kx}\sin(b_kx)$$
where $R_k$ and $I_k$ are arbitrary constants.
