Almost everything interesting about the Riemann zeta function is controlled by where it vanishes. The Prime Number Theorem, the size of the error in counting primes, the Riemann Hypothesis itself — all of them are, at heart, statements about the zeros of \zeta(s). So this page does one thing carefully: it maps out exactly where \zeta can be zero, proves as much of that map as is elementary, and marks off the one region where the map is still a conjecture.

The zeros split cleanly into two families. There are the tame, fully-understood trivial zeros strung along the negative real axis, and the mysterious non-trivial zeros hiding in a thin vertical band called the critical strip. The story of modern number theory is almost entirely the story of that second family.

First: where zeta is not zero

The best way to corner the zeros is to rule out everywhere they can't be. Start on the right, where \zeta is given by the Euler product. For \Re(s) > 1,

The middle bullet is the crux. An infinite product \prod (1 - a_p)^{-1} with \sum |a_p| < \infty converges to a value that is zero only if one of its factors is zero — and none of ours is. Intuitively: multiplying together numbers all close to 1 can never sneak you down to 0. So the entire half-plane to the right of \Re(s) = 1 is zero-free.

The functional equation carries the news leftward

The functional equation stitches the value at s to the value at 1-s:

Now sweep into the left half-plane, \Re(s) < 0. Look at the four factors on the right, one at a time:

• 2^{s}\,\pi^{s-1} is a non-zero exponential — never zero;
• \Gamma(1-s) is the Gamma function, which is never zero anywhere;
• \zeta(1-s): since \Re(s) < 0 we have \Re(1-s) > 1, and we just proved \zeta is non-zero there;
• that leaves \sin(\pi s / 2) as the only factor that can possibly vanish.

So in the left half-plane, \zeta(s) = 0 happens exactly when \sin(\pi s/2) = 0. That sine is zero at every even integer s = 0, \pm 2, \pm 4, \dots; restricting to \Re(s) < 0 leaves the negative even integers. (The point s = 0 is a special case where \Gamma(1-s) has a pole that cancels the sine's zero — and indeed \zeta(0) = -\tfrac12 \neq 0.)

Family one: the trivial zeros

Worked example — \zeta(-2) = 0. Put s = -2 in the functional equation. The sine factor is

and every other factor is finite, so the whole right-hand side collapses to 0. Hence \zeta(-2) = 0. The identical calculation gives \zeta(-4) = \zeta(-6) = \dots = 0. They are called "trivial" only because they are easy to locate — one sine, done — not because they aren't real zeros. (They very much are; see the Watch out! below.)

Family two: the non-trivial zeros live in the critical strip

We have now pinned every zero except those in the middle band. To the right of \Re(s) = 1: none. To the left of \Re(s) = 0: only the trivial ones. A separate, deep theorem — equivalent to the Prime Number Theorem — rules out the boundary line \Re(s) = 1 as well, and the functional equation then clears \Re(s) = 0 too. What remains is a single vertical corridor.

The open region 0 < \Re(s) < 1 is the critical strip, and the vertical line \Re(s) = \tfrac12 running down its centre is the critical line. The first few non-trivial zeros climb the imaginary axis at heights

Every one computed so far has real part exactly \tfrac12 — sitting bang on the critical line. Whether that holds for all of them is the Riemann Hypothesis.

The whole landscape at a glance

Here is the complete picture. The horizontal axis is \Re(s), the vertical axis \Im(s). Step through it: the critical strip that must contain every non-trivial zero, the critical line down its middle, the trivial zeros marching off to the left along the negative real axis, and finally the first non-trivial zeros — every one landing exactly on the line, in conjugate pairs above and below the real axis.

Notice how different the two families look: the trivial zeros are evenly spaced and boring, marching away to -\infty; the non-trivial zeros huddle in the strip at irregular heights, mirror-imaged across the real axis.

The zeros come in symmetric quadruples

The non-trivial zeros are not scattered at random — they obey two reflection symmetries that lock them into groups. Both come from identities you already know.

Worked example — from \rho to 1-\rho. Suppose \rho is a non-trivial zero, so 0 < \Re(\rho) < 1 and \zeta(\rho) = 0. Evaluate the functional equation at s = 1 - \rho:

Every factor in front is finite (here \Gamma(\rho) is finite because \Re(\rho) > 0), and the last factor \zeta(\rho) = 0. A finite number times zero is zero, so \zeta(1-\rho) = 0. The zero at \rho has produced a partner at 1-\rho, its mirror image across the line \Re(s) = \tfrac12.

Seeing the quadruple

Suppose, hypothetically, a zero \rho sat off the critical line. Watch the two symmetries generate its three companions: first a reflection across the real axis to \bar\rho, then a reflection across the critical line to 1-\rho and 1-\bar\rho. Four zeros, one rectangle.

Here is the punchline. If \rho happens to lie on the critical line, \rho = \tfrac12 + it, then 1-\rho = \tfrac12 - it = \bar\rho — the "reflect in the line" partner and the "reflect in the real axis" partner are the same point. The quadruple collapses to a simple conjugate pair \tfrac12 \pm it. That is exactly why the zeros in the landscape plot appeared as pairs, not fours: on the line, symmetry does double duty.

The edge of the strip, and the conjecture

Two facts about the strip's boundary deserve to be stated precisely, because they mark the frontier between what is proven and what is merely believed.

By the functional equation, no zeros on \Re(s) = 1 means no zeros on \Re(s) = 0 either. So the non-trivial zeros live in the open strip 0 < \Re(s) < 1 — never on its walls. That much is settled. What is not settled is where inside they sit:

In the language of quadruples: RH says the quadruple always degenerates to a pair — that no genuine four-cornered rectangle of zeros ever occurs. Prove that, and you win a million dollars and pin down the primes; find a single off-line zero, and you disprove it. In over 165 years, neither has happened.