Why is $\cos(i)>1$? I always thought that cosine only ranges from $-1$ to $1$. But, I found out that 
$$\cos(i)=\frac12\left(e+\frac{1}{e}\right)$$ 
which is certainly greater than $1$. Why is that?
 A: The general definition of $\cos(z)$ is 
$$\cos(z)=\frac{{e^{iz}}+e^{-iz}}{2}$$
When you plug in complex numbers into $\cos(z$), you can get values greater than $1$ or less than $-1$
A: The function $\cos z$ belongs in the interval $\left[-1,1 \right]$ when $z$ is a real number, not necessarily when $z$ is a complex number. An example of this is $\cos i > 1$, as you correctly pointed out.
A: Hint
Use the identity$$\cos(x+iy)=\cos x\cosh y-i\sin x\sinh y$$
A: Here's what happening visually: cosh is essentially cosine with domain of imaginary numbers, as I said in the comments. Cosine and sine trace the unit circle, and COSH (the specific one you asked about) and sinh for the unit hyperbola. The circle parametrization functions (sin and cos) have range between $-1$ and $1$, whereas the unit hyperbola ($x^2-y^2=1$) is unbounded. You could look up these graphs to see this or even deeper see their shape through their relation to conic sections. So that's some visual intuition if want a reason whilst skipping all the knarly mindless algebra. 
*If you continue the graph of the unit circle, putting in outputs outside the unit circles range and getting out imaginary numbers, it traces the unit hyperbola. And it sort of makes sense that if a real angle gives the circle, a hyperbola might come from something imaginary. Anyways, there are also other reasons this particular hyperbola is considered "unit" too. cosh and sinh parametrize the unit hyperbola because subbing the cos and sin for a function of their hyperbolic counterparts in the identity $cos^2(x)+sin^2(x)=1$ you get $cosh^2(x)-sinh^2(x)=1$, which is why the hyperbolic sine and cosine parametrize the unit hyperbola, which again is $x^2-y^2=1$, letting x=cosh(t) and y=sinh(t). 
A: It's true that the cosine of a real number must be between $-1$ and $1$, but this is not true for the cosine of a complex number. In fact, complex-differentiable functions can never be bounded (unless they are constant). 
Here is an analogy, if you like. Let $f(x) = x^2$. Then we learn some rule that $f(x) \geq 0$ for all $x$. But wait a second, $f(i)$ is negative. There's no scandal, since $i$ is not a real number.
