Return to Answer

This is a continuation of the argument above, which completes the argument.

Let $C_n$ denote the square with vertices $[n \pm 1/2, \pm 4 I]$ for a positive integer $n$. We have the following inequalities for $z \in C_n$ and $n \ge 15$: $$|\sin(\pi z)| \ge 1, \quad z \in C_n.$$ $$|\Gamma(z)| \ge \frac{1}{2} \Gamma(n - 1/2),$$ $$|\Gamma(1-z)| \le \frac{\pi}{\Gamma(n - 1/2)} \le 1,$$ $$|\psi(1-z)|, |\psi(z)| \le 2 \log(n), $$

The first is easy, the second follows from Stirling's formula (this requires $n$ to be big enough, and also requires $z$ to have imaginary part at most $4$), the third follows from the previous two by the reflection formula for $\Gamma(z)$, the last follows by induction and by the formula $\psi(z+1) = \psi(z) + 1/z$. It follows that $$\left| \frac{1}{2 \pi i} \oint_{C_n} \frac{\Gamma'(z)}{\Gamma(z)} - \frac{d/dz (\Gamma(z) + \theta \cdot \Gamma(1-z))}{\Gamma(z) + \theta\cdot \Gamma(1-z)} \right|$$ $$= \left| \frac{1}{2 \pi i} \oint_{C_n} \frac{\theta \Gamma(1-z) (\psi(1-z) + \psi(z))} {\Gamma(z) + \theta \cdot \Gamma(1-z)} \right|$$ $$ \le \frac{8 |\theta| \cdot \log(n) \pi}{2 \pi \cdot \Gamma(n - 1/2)} \oint_{C_n} \frac{1} {|\Gamma(z) + \theta \cdot \Gamma(1-z)|}$$ $$ \le \frac{8 |\theta| \cdot \log(n) \pi}{2 \pi \cdot \Gamma(n - 1/2)} \cdot \frac{1}{1/2 \Gamma(n - 1/2) + 1} \ll 1,$$ where $\theta = \pm 1$ (or anything small) and $n \ge 15$, where the final inequality holds by a huuuge margin. It follows that $\Gamma(z) + \theta \cdot\Gamma(1-z)$ and $\Gamma(z)$ have the same number of zeros minus the number of poles in $C_n$. Since $\Gamma(z)$ has no zeros and poles in $C_n$, it follows that $\Gamma(z) + \theta\cdot\Gamma(1-z)$ has the same number of zeros and poles. It has exactly one pole, and thus exactly one zero. If $\theta = \pm 1$ (and so in particular is real), by the Schwarz reflection principle, this zero is forced to be real. By symmetry, the same argument applies in the region $z = s + i t$ with $|t| \le 4$ and $s \le -15$. Combined with the above argument, this reduces the claim to $z = s + i t$ with $|s| \le 15$ and $|t| \le 4$ where the claim can be checked directly.

Hence all the zeros are either in $\mathbf{R}$, or lie on the line $1/2 + i \mathbf{R}$.

This is a continuation of the argument above, which completes the argument.

Let $C_n$ denote the square with vertices $[n \pm 1/2, \pm 4 I]$ for a positive integer $n$. We have the following inequalities for $z \in C_n$ and $n \ge 15$: $$|\sin(\pi z)| \ge 1, \quad z \in C_n.$$ $$|\Gamma(z)| \ge \frac{1}{2} \Gamma(n - 1/2),$$ $$|\Gamma(1-z)| \le \frac{\pi}{\Gamma(n - 1/2)} \le 1,$$ $$|\psi(1-z)|, |\psi(z)| \le 2 \log(n), $$

The first is easy, the second follows from Stirling's formula (this requires $n$ to be big enough, and also requires $z$ to have imaginary part at most $4$), the third follows from the previous two by the reflection formula for $\Gamma(z)$, the last follows by induction and by the formula $\psi(z+1) = \psi(z) + 1/z$. It follows that $$\left| \frac{1}{2 \pi i} \oint_{C_n} \frac{\Gamma'(z)}{\Gamma(z)} - \frac{d/dz (\Gamma(z) + \theta \cdot \Gamma(1-z))}{\Gamma(z) + \theta\cdot \Gamma(1-z)} \right|$$ $$= \left| \frac{1}{2 \pi i} \oint_{C_n} \frac{\theta \Gamma(1-z) (\psi(1-z) + \psi(z))} {\Gamma(z) + \theta \cdot \Gamma(1-z)} \right|$$ $$ \le \frac{8 |\theta| \cdot \log(n) \pi}{2 \pi \cdot \Gamma(n - 1/2)} \oint_{C_n} \frac{1} {|\Gamma(z) + \theta \cdot \Gamma(1-z)|}$$ $$ \le \frac{8 |\theta| \cdot \log(n) \pi}{2 \pi \cdot \Gamma(n - 1/2)} \cdot \frac{1}{1/2 \Gamma(n - 1/2) + 1} \ll 1,$$ where $\theta = \pm 1$ (or anything small) and $n \ge 15$, where the final inequality holds by a huuuge margin. It follows that $\Gamma(z) + \theta \cdot\Gamma(1-z)$ and $\Gamma(z)$ have the same number of zeros minus the number of poles in $C_n$. Since $\Gamma(z)$ has no zeros and poles in $C_n$, it follows that $\Gamma(z) + \theta\cdot\Gamma(1-z)$ has the same number of zeros and poles. It has exactly one pole, and thus exactly one zero. If $\theta = \pm 1$ (and so in particular is real), by the Schwarz reflection principle, this zero is forced to be real. By symmetry, the same argument applies in the region $z = s + i t$ with $|t| \le 4$ and $s \le -15$. Combined with the above argument, this reduces the claim to $z = s + i t$ with $|s| \le 15$ and $|t| \le 4$ where the claim can be checked directly.

Hence all the zeros are either in $\mathbf{R}$, or lie on the line $1/2 + i \mathbf{R}$.

EDIT To clarify, I didn't actually check that there were no ``exceptional'' zeros in the box $\pm 15 \pm 4 I$, since I presumed that the original poster had done so. If $F(z) = \Gamma(z) - \Gamma(1-z)$, then computing the integral $$\frac{1}{2 \pi i} \oint \frac{F'(z)}{F(z)} dz$$ around that box, one obtains (numerically, and thus exactly) $1$. There are (assuming the OP at least computed the critical line zeros correctly) $2$ zeros in that range on the critical line. Along the real line in that range, there are $30$ poles and $25$ zeros. This means that there must be $1 + 30 - 25 = 6$ unaccounted for zeros. For such a zero $\rho$ off the line, by symmetry one also has $\overline{\rho}$, $1 - \rho$ and $1 - \overline{\rho}$ as zeros. Hence there must be either $1$ or $3$ pairs of zeros on the critical line, and either $1$ or $0$ quadruples of roots off the line. Varying the parameters of the integral, one can confirm there is a zero with $\rho \sim 2.7 + 0.3 i$, which is one of the four conjugates of the root found by joro. A similar argument applies for $\Gamma(z)+\Gamma(1-z)$. Hence:

Any zero of $\Gamma(z) - \Gamma(1-z)$ is either in $\mathbf{R}$, on the line $1/2 + i \mathbf{R}$, or is one of the four exceptional zeros $\{\rho,1-\rho,\overline{\rho},1-\overline{\rho}\}$. A similar calculation implies the same for $\Gamma(z) + \Gamma(1-z)$, except now with an exceptional set $\{\mu,1-\mu,\overline{\mu},1-\overline{\mu}\}$.

This is a continuation of the argument above, which completes the argument.

Let $C_n$ denote the square with vertices $[n \pm 1/2, \pm 4 I]$ for a positive integer $n$. We have the following inequalities for $z \in C_n$ and $n \ge 15$: $$|\sin(\pi z)| \ge 1, \quad z \in C_n.$$ $$|\Gamma(z)| \ge \frac{1}{2} \Gamma(n - 1/2),$$ $$|\Gamma(1-z)| \le \frac{\pi}{\Gamma(n - 1/2)} \le 1,$$ $$|\psi(1-z)|, |\psi(z)| \le 2 \log(n), $$

The first is easy, the second follows from Stirling's formula (this requires $n$ to be big enough, and also requires $z$ to have imaginary part at most $4$), the third follows from the previous two by the reflection formula for $\Gamma(z)$, the last follows by induction and by the formula $\psi(z+1) = \psi(z) + 1/z$. It follows that $$\left| \frac{1}{2 \pi i} \oint_{C_n} \frac{\Gamma'(z)}{\Gamma(z)} - \frac{d/dz (\Gamma(z) + \theta \cdot \Gamma(1-z))}{\Gamma(z) + \theta\cdot \Gamma(1-z)} \right|$$ $$= \left| \frac{1}{2 \pi i} \oint_{C_n} \frac{\theta \Gamma(1-z) (\psi(1-z) + \psi(z))} {\Gamma(z) + \theta \cdot \Gamma(1-z)} \right|$$ $$ \le \frac{8 |\theta| \cdot \log(n) \pi}{2 \pi \cdot \Gamma(n - 1/2)} \oint_{C_n} \frac{1} {|\Gamma(z) + \theta \cdot \Gamma(1-z)|}$$ $$ \le \frac{8 |\theta| \cdot \log(n) \pi}{2 \pi \cdot \Gamma(n - 1/2)} \cdot \frac{1}{1/2 \Gamma(n - 1/2) + 1} \ll 1,$$ where $\theta = \pm 1$ (or anything small) and $n \ge 15$, where the final inequality holds by a huuuge margin. It follows that $\Gamma(z) + \theta \Gamma(1-z)$ and $\Gamma(z)$ have the same number of zeros minus the number of poles in $C_n$. Since $\Gamma(z)$ has no zeros and poles in $C_n$, it follows that $\Gamma(z) + \Gamma(1-z)$ has the same number of zeros and poles. It has exactly one pole, and thus exactly one zero. If $\theta = \pm 1$ (and so in particular is real), by the Schwarz reflection principle, this zero is forced to real. By symmetry, the same argument applies in the region $z = s + i t$ with $|t| \le 4$ and $s \le -15$. Combined with the above argument, this reduces the claim to $z = s + i t$ with $|s| \le 15$ and $|t| \le 4$ where the claim can be checked directly.

Hence all the zeros are either in $\mathbf{R}$, or lie on the line $1/2 + i \mathbf{R}$.

This is a continuation of the argument above, which completes the argument.

Let $C_n$ denote the square with vertices $[n \pm 1/2, \pm 4 I]$ for a positive integer $n$. We have the following inequalities for $z \in C_n$ and $n \ge 15$: $$|\sin(\pi z)| \ge 1, \quad z \in C_n.$$ $$|\Gamma(z)| \ge \frac{1}{2} \Gamma(n - 1/2),$$ $$|\Gamma(1-z)| \le \frac{\pi}{\Gamma(n - 1/2)} \le 1,$$ $$|\psi(1-z)|, |\psi(z)| \le 2 \log(n), $$

The first is easy, the second follows from Stirling's formula (this requires $n$ to be big enough, and also requires $z$ to have imaginary part at most $4$), the third follows from the previous two by the reflection formula for $\Gamma(z)$, the last follows by induction and by the formula $\psi(z+1) = \psi(z) + 1/z$. It follows that $$\left| \frac{1}{2 \pi i} \oint_{C_n} \frac{\Gamma'(z)}{\Gamma(z)} - \frac{d/dz (\Gamma(z) + \theta \cdot \Gamma(1-z))}{\Gamma(z) + \theta\cdot \Gamma(1-z)} \right|$$ $$= \left| \frac{1}{2 \pi i} \oint_{C_n} \frac{\theta \Gamma(1-z) (\psi(1-z) + \psi(z))} {\Gamma(z) + \theta \cdot \Gamma(1-z)} \right|$$ $$ \le \frac{8 |\theta| \cdot \log(n) \pi}{2 \pi \cdot \Gamma(n - 1/2)} \oint_{C_n} \frac{1} {|\Gamma(z) + \theta \cdot \Gamma(1-z)|}$$ $$ \le \frac{8 |\theta| \cdot \log(n) \pi}{2 \pi \cdot \Gamma(n - 1/2)} \cdot \frac{1}{1/2 \Gamma(n - 1/2) + 1} \ll 1,$$ where $\theta = \pm 1$ (or anything small) and $n \ge 15$, where the final inequality holds by a huuuge margin. It follows that $\Gamma(z) + \theta \cdot\Gamma(1-z)$ and $\Gamma(z)$ have the same number of zeros minus the number of poles in $C_n$. Since $\Gamma(z)$ has no zeros and poles in $C_n$, it follows that $\Gamma(z) + \theta\cdot\Gamma(1-z)$ has the same number of zeros and poles. It has exactly one pole, and thus exactly one zero. If $\theta = \pm 1$ (and so in particular is real), by the Schwarz reflection principle, this zero is forced to be real. By symmetry, the same argument applies in the region $z = s + i t$ with $|t| \le 4$ and $s \le -15$. Combined with the above argument, this reduces the claim to $z = s + i t$ with $|s| \le 15$ and $|t| \le 4$ where the claim can be checked directly.

Hence all the zeros are either in $\mathbf{R}$, or lie on the line $1/2 + i \mathbf{R}$.

This is a continuation of the argument above, which completes the argument.

Let $C_n$ denote the square with vertices $[n \pm 1/2, \pm 4 I]$ for a positive integer $n$. We have the following inequalities for $z \in C_n$ and $n \ge 15$: $$|\sin(\pi z)| \ge 1, \quad z \in C_n.$$ $$|\Gamma(z)| \ge \frac{1}{2} |\Gamma(n - 1/2)|,$$ $$|\Gamma(1-z)| \le \frac{\pi}{\Gamma(n - 1/2)} \le 1,$$ $$|\psi(1-z)|, |\psi(z)| \le 2 \log(n), $$

The first is easy, the second follows from Stirling's formula (this requires $n$ to be big enough, and also requires $z$ to have imaginary part at most $4$), the third follows from the previous two by the reflection formula for $\Gamma(z)$, the last follows by induction and by the formula $\psi(z+1) = \psi(z) + 1/z$. It follows that $$\left| \frac{1}{2 \pi i} \oint_{C_n} \frac{\Gamma'(z)}{\Gamma(z)} - \frac{d/dz (\Gamma(z) + \theta \cdot \Gamma(1-z))}{\Gamma(z) + \Gamma(1-z)} \right|$$ $$= \left| \frac{1}{2 \pi i} \oint_{C_n} \frac{\theta \Gamma(1-z) (\psi(1-z) + \psi(z))} {\Gamma(z) + \theta \cdot \Gamma(1-z)} \right|$$ $$ \le \frac{8 |\theta| \cdot \log(n) \pi}{2 \pi \cdot \Gamma(n - 1/2)} \oint_{C_n} \frac{1} {|\Gamma(z) + \theta \cdot \Gamma(1-z)|}$$ $$ \le \frac{8 |\theta| \cdot \log(n) \pi}{2 \pi \cdot \Gamma(n - 1/2)} \cdot \frac{1}{1/2 \Gamma(n - 1/2) + 1} \ll 1,$$ where $\theta = \pm 1$ (or anything small) and $n \ge 15$, where the final inequality holds by a huuuge margin. It follows that $\Gamma(z) + \theta \Gamma(1-z)$ and $\Gamma(z)$ have the same number of zeros minus the number of poles in $C_n$. Since $\Gamma(z)$ has no zeros and poles in $C_n$, it follows that $\Gamma(z) + \Gamma(1-z)$ has the same number of zeros and poles. It has exactly one pole, and thus exactly one zero. If $\theta = \pm 1$ (and so in particular is real), By the Schwarz reflection principle, this zero is forced to real. By symmetry, the same argument applies in the region $z = s + i t$ with $|t| \le 4$ and $s \le -15$. Combined with the above argument, this reduces the claim to $z = s + i t$ with $|s| \le 15$ and $|t| \le 4$ where the claim can be checked directly.

Hence all the zeros are either in $\mathbf{R}$, or lie on the line $1/2 + i \mathbf{R}$.

This is a continuation of the argument above, which completes the argument.

Let $C_n$ denote the square with vertices $[n \pm 1/2, \pm 4 I]$ for a positive integer $n$. We have the following inequalities for $z \in C_n$ and $n \ge 15$: $$|\sin(\pi z)| \ge 1, \quad z \in C_n.$$ $$|\Gamma(z)| \ge \frac{1}{2} \Gamma(n - 1/2),$$ $$|\Gamma(1-z)| \le \frac{\pi}{\Gamma(n - 1/2)} \le 1,$$ $$|\psi(1-z)|, |\psi(z)| \le 2 \log(n), $$

The first is easy, the second follows from Stirling's formula (this requires $n$ to be big enough, and also requires $z$ to have imaginary part at most $4$), the third follows from the previous two by the reflection formula for $\Gamma(z)$, the last follows by induction and by the formula $\psi(z+1) = \psi(z) + 1/z$. It follows that $$\left| \frac{1}{2 \pi i} \oint_{C_n} \frac{\Gamma'(z)}{\Gamma(z)} - \frac{d/dz (\Gamma(z) + \theta \cdot \Gamma(1-z))}{\Gamma(z) + \theta\cdot \Gamma(1-z)} \right|$$ $$= \left| \frac{1}{2 \pi i} \oint_{C_n} \frac{\theta \Gamma(1-z) (\psi(1-z) + \psi(z))} {\Gamma(z) + \theta \cdot \Gamma(1-z)} \right|$$ $$ \le \frac{8 |\theta| \cdot \log(n) \pi}{2 \pi \cdot \Gamma(n - 1/2)} \oint_{C_n} \frac{1} {|\Gamma(z) + \theta \cdot \Gamma(1-z)|}$$ $$ \le \frac{8 |\theta| \cdot \log(n) \pi}{2 \pi \cdot \Gamma(n - 1/2)} \cdot \frac{1}{1/2 \Gamma(n - 1/2) + 1} \ll 1,$$ where $\theta = \pm 1$ (or anything small) and $n \ge 15$, where the final inequality holds by a huuuge margin. It follows that $\Gamma(z) + \theta \Gamma(1-z)$ and $\Gamma(z)$ have the same number of zeros minus the number of poles in $C_n$. Since $\Gamma(z)$ has no zeros and poles in $C_n$, it follows that $\Gamma(z) + \Gamma(1-z)$ has the same number of zeros and poles. It has exactly one pole, and thus exactly one zero. If $\theta = \pm 1$ (and so in particular is real), by the Schwarz reflection principle, this zero is forced to real. By symmetry, the same argument applies in the region $z = s + i t$ with $|t| \le 4$ and $s \le -15$. Combined with the above argument, this reduces the claim to $z = s + i t$ with $|s| \le 15$ and $|t| \le 4$ where the claim can be checked directly.

Hence all the zeros are either in $\mathbf{R}$, or lie on the line $1/2 + i \mathbf{R}$.