As far as I learned from the literature, the non-trivial zeros of the zeta function are symmetric about the critical line Re(s) = 1/2, because xi(s) = xi(1-s). But this equation does not give zeros which are the mirror reflection of each other. Instead the zeros are symmetric about Re(s) = 1/2 AND Im(s) = 0.
Shahid, it might help to understand this by using a different form of the zeta function, an expanded form. I have one of these in my preprint featured on my profile. Take equation (6) and set it equal to the second equation on page 7, which are both assumed solutions for the zeros of the zeta function. That is, you will need to check if the argument s really gives zeta equal to zero. Next, replace s on the left hand side with an arbitrary value, and then solve for s on the right hand side. You will get two solutions that are reflections through the real point 1/2. The first will be the value you gave for s on the left hand side and the second will be 1-s.
This calculation can be done on Wolfram Alpha (handles it well). If you would like me to send you a link to an example, I can email it.
let me know if I can clarify it further. Happy to help.
Shahid, here's an example. I am applying the argument 1/3 + i 1/2 on the left hand side (which is equation 4 or 6 from my preprint). If this were a zero of the zeta function, the second solution would be equal to 1- (1/3+i 1/2). But as you see below, it is not a zero of the zeta function, as the result 7/2 - i 1/2 's distance across the critical line is not symmetric.
https://www.wolframalpha.com/input/?i=%28i+%28%28%281%2F3%2B+i+1%2F2%29%5E*+%29%5E2-+%28%281%2F3%2Bi+1%2F2%29+-+1%29%5E2+%29%29%2F%282+%28%281%2F3%2Bi+1%2F2%29+-+1%29++%281%2F3%2Bi+1%2F2%29%5E*+%29+%3D+%28-%E2%88%9A%28%282+Im%28s%29%2B+i%29%5E2+%29++%7C1+-+2+Re%28s%29%7C%2B+%28s%5E*%2B+s+-+1%29%282+Im%28s%29%2B+i%29%29%2F%284+%28%7C%28s%29%5E2+%7C-+s%5E*%29+%29
Now using an actual non-trivial zero (a rounded critical zero)...
https://www.wolframalpha.com/input/?i=%28i+%28%28%281%2F2%2B+i+14.1347%29%5E*+%29%5E2-+%28%281%2F2%2Bi+14.1347%29+-+1%29%5E2+%29%29%2F%282+%28%281%2F2%2Bi+14.1347%29+-+1%29++%281%2F2%2Bi+14.1347%29%5E*+%29+%3D+%28-%E2%88%9A%28%282+Im%28s%29%2B+i%29%5E2+%29++%7C1+-+2+Re%28s%29%7C%2B+%28s%5E*%2B+s+-+1%29%282+Im%28s%29%2B+i%29%29%2F%284+%28%7C%28s%29%5E2+%7C-+s%5E*%29+%29
We see that the real part of its pair must be 1/2, and since this equality will always return the minus imaginary part of the value on the left hand side, we see that we have its complex conjugate.
Hope that helps.
Here's a link to the preprint.
Preprint A Direct Proof of the Riemann Hypothesis
Dear Jeffrey N Cook , Thank you for adding the answers to my question, I'll take look at your paper. It must be an interesting paper as the title indicates.
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