# Does this disprove the Riemann Hypothesis?

I found interesting results when I put in certain values for z into zeta(z). I believe my findings do not disprove the Riemann hypothesis, but I am not so sure.

When I input z = 1/3 + i into the function, I got a result with a positive real component and a negative imaginary component, which can be written as a - bi.

When I input z = 1/3 - i, I got a result with a positive real component and a positive imaginary component, which can be written as c + di.

For z = 2/5 - i/2, I got -f + gi.

For z = 2/5 + i/2, I got -h - li.

Because the Riemann Zeta Function is continuous, these results imply that there is a number in this range z for which zeta(z) has a real component of zero. In fact, there would be infinitely many of these numbers in an unbroken chain. It also implies that there is a number in this range z for which zeta(z) has a complex component of zero and there there are an infinitely many of these in a broken chain. Also, these two chains should meet with each other at some point. This seems like it would disprove the Riemann Hypothesis.

• The intermediate value theorem does not hold for complex-valued functions. Example: $e^0 = 1$, $e^{ \pi i} = -1$, but $e^z$ has no zeros. Mar 18, 2021 at 20:35
• What are all these downvotes for? Is it a sin to be wrong? Mar 18, 2021 at 20:38
• I take the function $e^z$; when I input $z=1/2+ 2n\pi i \pm 1$ I get results with positive real part and plus minus imaginary part; if I input same but with an odd multiple of pi over two in the imaginary part and a little more I get results with negative real part and plus minus imaginary part; I conclude by continuity and reasoning above that $e^z$ has infinitely many zeroes which is of course absurd Mar 18, 2021 at 20:38
• @uniquesolution, plus, OP recognized that something must be wrong and only sought clarification. Mar 19, 2021 at 17:12

Yes, you can draw a curve separating $$\{\frac13 + i, \frac13 - i\}$$ from $$\{\frac25 + \frac i2, \frac25 - \frac i2\}$$ where the real component of the zeta function is $$0$$.
Yes, you can draw a curve separating $$\{\frac13 + i, \frac25 + \frac i2\}$$ from $$\{\frac13 - i, \frac25 - \frac i2\}$$ where the imaginanary component of the zeta function is $$0$$.
But you have no control over where these curves intersect; they do not have to intersect inside the trapezoid whose corners are at $$\{\frac13 + i, \frac13 - i, \frac25 + \frac i2, \frac25 - \frac i2\}$$, because the curves we drew may leave that trapezoid. So the zero you find may be a trivial zero, or it may be a nontrivial zero with real part $$\frac12$$, neither of which would disprove the Riemann hypothesis.
In fact, here is the picture we get if we actually draw these curves: in blue is the curve with real component $$0$$, and in orange is the curve with imaginary component $$0$$. (There's more going on outside the range $$[-3,3] \times [-3,3]$$, of course.) We see them intersecting once at $$z=-2$$, and we see them pretending to intersect at $$z=1$$. (But they actually don't, because the zeta function has a singularity at $$1$$, corresponding to the harmonic series $$1 + \frac12 + \frac13 + \dots$$.)