I'm doing a little project on the $\zeta$ function, and I am at a complete loss of what it is actually doing. I understand it is way over my head, but when I am plugging say $\zeta(1 + i)$ into WolframAlpha, what is it even calculating? Wikipedia and .edu sites don't seem to have an answer, which is making me think there is no single answer.
Thanks, smart people of MathStack!
EDIT:
Why is everyone voting to close this? I understand it's similar to a different question, but this question might provide someone with some different intuition. Either way - this helped me a ton! Thanks @Mixed_Math.
The Riemann zeta function $\zeta(s)$ is a sum of reciprocals of powers of natural numbers,
$$\zeta(s) = \sum_{n \geq 1} \frac{1}{n^s}.$$
As written, this makes sense for complex numbers $s$ so long as $\text{Re } s > 1$. For these numbers, there is little more to be said.
But you've asked about an interesting number: $\zeta(1 + i)$, and $\text{Re }(1 + i) \not > 1$. What's happening there is a bit subtle, and a bit abusive in terms of notation.
It turns out there is another function (let's call it $Z(s)$) which makes sense for all complex numbers $s$ except for $s = 1$, and which exactly agrees with $\zeta(s)$ when $\text{Re } s > 1$. If you're familiar with some calculus or complex analysis, then you should also know that the function $Z(s)$ is also complex differentiable everywhere except for $s = 1$. This is a very special property that distinguishes $Z(s)$. The theory of complex analysis (in particular, the theory of "analytic continuation") gives that there can be at most one function that extends $\zeta(s)$ to a larger region, like $Z(s)$ does.
In this sense, we could realize that $Z(s)$ is uniquely determined by $\zeta(s)$. As it agrees with $\zeta(s)$ everywhere $\zeta(s)$ (initially) makes sense, it might even be reasonable to just use the name $\zeta(s)$ instead of $Z(s)$. That is, when I write $\zeta(s)$, what I'm really saying is $$\zeta(s) = \begin{cases} \zeta(s) & \text{if Re }s > 1 \\ Z(s) & \text{otherwise } \end{cases}$$ It is this function that W|A computes when you ask it for $\zeta(1 + i)$.
Although what I've written is true (and important), it doesn't answer one aspect of your question
I mentioned there exists this function $Z(s)$, or rather that it is possible to give meaningful values to $\zeta(s)$ for all $s \neq 1$. But how? Stated differently, yo're asking what is the analytic continuation of the Riemann zeta function?
The continuation is unique, but the steps to get there are not. I'll give a very short, incomplete proof that describes one way to calculate $\zeta(1+i)$.
We start by considering $\displaystyle h(s) = \sum_{n \geq 1} \frac{2}{(2n)^s}$. Performing some rearrangements, $$\begin{align} h(s) &= \sum_{n \geq 1} \frac{2}{(2n)^s} \\ &= \frac{1}{2^{s - 1}} \sum_{n \geq 1} \frac{1}{n^s} \\ &= \frac{1}{2^{s - 1}} \zeta(s) \end{align}$$ Let's subtract this from the regular zeta function. On the one hand, $$ \zeta(s) - h(s) = \zeta(s)(1 - \frac{1}{2^{s-1}}).$$
On the other hand, $$ \begin{align}\zeta(s) - h(s) &= \sum_{n \geq 1} \left( \frac{1}{n^s} - \frac{2}{(2n)^s} \right) \\ &= \sum_{n \geq 1} \frac{(-1)^{n+1}}{n^s}, \end{align}$$ and this last series makes sense for $\text{Re } s > 0$. (If you haven't looked at alternating series before, this might not be obvious. But the idea is that the sign changes cancel out a lot of the growth, so much that it converges for a larger region).
In total, this means that $$\zeta(s) = (1 - 2^{s - 1})^{-1} \sum_{n \geq 1} \frac{(-1)^{n+1}}{n^s},$$ and you can just "plug in" $1+i$ here. [Notice that the problem when $s = 1$ is apparent here, as you cannot divide by $0$.] In practice, it's an infinite sum, so you'll take the first very many terms to get the value of $\zeta(1+i)$ to any precision you want.
For completeness, it also turns out that $$\pi^{-s/2} \zeta(s) \Gamma(\tfrac{s}{2}) = \pi^{(s-1)/2} \zeta(1-s) \Gamma(\tfrac{1-s}{2}),$$ which lets us transform values of $\zeta(s)$ for $\text{Re } s > 0$ into values when $\text{Re } s < 1$. The $\Gamma(z)$ function here is called the "Gamma function" (it's an integral, a sort of generalization of a factorial) and this equation is called the symmetric functional equation of the zeta function.