# Usage of Complex Numbers in the Riemann Hypothesis.

I don't have a very good understanding of the Riemann Hypothesis, however that being said, could someone explain to me why complex numbers are used, instead of just using real numbers? Everything I've read talks about absolute convergence, among other things, and it seems as if the real part of the complex number is all that ends up being important. Is it written in terms of complex numbers simply so that the structure of the complex numbers can be used, or is there something that led Riemann to naturally choose the function to take in complex numbers?

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The Riemann hypothesis is a statement about the complex zeros of a certain complex function... – PVAL Sep 1 '14 at 1:32

Very often understanding a property that is purely of real numbers or real functions elegantly passes through the complex numbers. Why that may be the case is perhaps a somewhat philosophical question, but it proves to be a very powerful technique. More technically, the complex numbers exhibits better behaviour than the reals do (e,g,. every polynomial has a root).

A general principle in analysis is that any function that is analytic (i.e., power-series expansion of non-zero radius at every point of its definition) is automatically well-defined as a holomorphic function, and extensions of it to the complex numbers give a lot of information about the original function. This is (perhaps), in a very small nut-shell, the answer to your question. The following is a baby example illustrating the advantages of thinking about complex numbers even when you are really just interested in real numbers.

Consider the function $f(x)=\frac {1}{1-x}$. Its Taylor expansion at $x=0$ is $\sum x^k$. The radius of convergence is $1$, and this is obvious since at $x=1$ the function explodes. We like this kind of understanding where we see exactly why the radius of convergence is what it is, rather than deducing it from some formula.

Now consider the function $f(x)=\frac {1}{1+x^2}$. The radius of convergence of its Taylor series at $x=0$ is also $1$. But now, there is no obvious reason why that is the case. The function does not explode at either $x=1$ or $x=-1$. Well, the reason does become obvious when you consider the function as a complex function. Indeed, for $x=i$, the function explodes again, so now we again understand the reason why the radius of convergence is $1$.

I hope this answers your question somewhat, even though I did not give any details of the $\zeta$ function (doing so would require a much more technical elaboration).

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While this is a great introduction to the aspect of complex analysis helping real analysis, I do not think it applies to the $\zeta$ function. Namely, we do not start asking questions about the real $\zeta$ function, extend it to the complex $\zeta$ function, and then get results about the real $\zeta$ function. For the Riemann $\zeta$ function, the complex part is what we're interested in! – RghtHndSd Sep 1 '14 at 2:17
@RghtHndSd yes, but what I meant is that we are interested in the combinatorics of the primes. The zeta function is a clever way to pack that combinatorics (which is essentially about real numbers) into the coefficients of an analytic function. We study this function in the complex plane to understand the combinatorics of real things. – Ittay Weiss Sep 1 '14 at 2:46
I'm not sure I understand what you mean. Is there a way to think about the $\zeta$ function as just being a function of the real numbers that gives information about the primes? Again - it seems as if your answer here is saying that we are interested in the real values of the $\zeta$ function and pass to the complex version only to learn about the real values. – RghtHndSd Sep 1 '14 at 3:01
@RghtHndSd I am trying to say that we are interested in the behaviour of the zeta function in order to learn things about its coefficients and that even when the coefficients are real, it is profitable to consider the function as a complex one. I also stated that my answer does not really answer anything particular about the zeta function and I agree with you that it's a somewhat weak answer, not addressing anything particular about the zeta function, but only very general ideas. – Ittay Weiss Sep 1 '14 at 3:05

The function $\pi(x)$ counts the numbers of primes that are less than or equal to $x$. For example, $\pi(5) = 3$ because of the primes 2,3,5. Define

$$\Pi(x) = \sum_{i=1}^\infty \frac{\pi(x^{1/i})}{i}.$$

Then we have $\pi(x) = \sum_{n=1}^\infty \frac{\mu(n)}{n} \Pi(x^{1/n})$. Finally define

$$\Pi_0(x) = \lim_{\varepsilon \rightarrow 0} \frac{\pi(x-\varepsilon) + \pi(x + \varepsilon)}{2}.$$

This is the same function as $\Pi$ except at the points where $\Pi$ is discontinuous. At those points, it is the average of the left hand and the right hand limits.

Riemann proved the formula

$$\Pi_0(x) = \text{Li}(x) - \sum_{\rho} \text{Li}(x^\rho) - \log(2) + \int_{x}^\infty \frac{dt}{t(t^2-1)\log(t)}$$

where $$\text{Li}(x) = \int_{0}^x \frac{dt}{\log(t)}.$$

All terms are well understood except for the summation. The summation in the middle is over the nontrivial zeros of the Riemann zeta function. These are necessarily not real.

Bottom line: The prime numbers are determined by complex values of the Riemann zeta function.

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The RH is about non-real zeros of the zeta-function, and moreover in a region where the initial definition of the zeta-function does not make sense, so it is completely natural to expect complex numbers, and moreover complex analysis, to play a role.

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One simple reason is the statement of the hypothesis: All non trivial zeroes of the zeta function are on the line $Re(z)=\frac{1}{2}$. If you consider only real numbers, this conjecture reduces to check if $x=\frac{1}{2}$ is a zero or not of the zeta function. Way easier problem!

Another possible reason is that complex analysis (rather than real analysis) is the good context to study power series. For example, using real analysis, the radius of $\frac{1}{1-x}$ at $0$ is intuitively $1$ since $1$ is a pole of this function, so it "explodes" to $\infty$ at that point. Same thing for $\frac{1}{1+x}$ with its pole at $-1$ ($|-1|=1$ and the interior of the interval of convergence has to be symmetric about $0$). But why is the radius of $\frac{1}{1+x^2}$ equal to $1$? This function is bounded on ${\mathbb R}$! If you go to the complex analysis side, it is clear since $i$ and $-i$ are poles of this function.

A last possible reason is that the zeta function is not well defined if you don't consider complex analytic functions. The zeta function is the unique complex analytic extension of the series $\displaystyle{\sum_{n\geqslant 0}\frac{1}{n^s}}$ (convergent for $Re(s)>1$) to the set ${\mathbb C}\setminus{\mathbb Z}_{\leqslant 0}$. This extension is unique by property of complex analytic functions. Extensions of real analytic functions are not unique, especially in this case, since the domain (${\mathbb R}\setminus{\mathbb Z}_{\leqslant 0}$) would not be connected.

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