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The zeroes of $\cosh z$ occur at $z=i\pi (n+\frac{1}{2})$, $n\in \mathbb{Z}$. They are simple because $\sinh z\ne 0$ there. Squaring the denominator therefore givesThere are no other finite singularities; this follows from the local zero-order criterion for a pole.

At $z=1$, numerator and denominator both have simple zeroes. Their derivatives give the removable valueAt each positive integer $n\ge 2$, the numerator is nonzero and the denominator has a simple zero, so these are simple poles. At zero the expansion on the principal sheet iswhich contains a fractional power. Thus zero is a branch point with pole-like leading growth, not an ordinary isolated pole. The negative real axis is the chosen branch cut. On local continuations across its nonzero points, negative integers also give simple poles because the continued square root is $\pm i\sqrt{|n|}$, never one. They lie outside the open principal branch domain; noninteger negative points are merely cut points, not intrinsic singularities.

The completed Dirichlet L-function isFor nonprincipal primitive Dirichlet characters, termwise Mellin transformation initially in $\Re s>1$ givesThe Dirichlet character theta function decays exponentially at infinity; its transformation makes it decay faster than any power at zero. Hence the integral is entire in $s$. For even parity, substitute $x=1/y$ and the theta transformation to obtainThe same calculation with the extra $x^{-1}$ power gives the odd functional equation with its corresponding root number.

The gamma function has no zeros and has simple poles at nonpositive integers. Thus the nontrivial zeros of $L$ and $\Lambda$ coincide with multiplicities. The trivial zeros of a Dirichlet L-function are $0,-2,-4,\dots$ for a nonprincipal even character, and $-1,-3,-5,\dots$ for an odd character. They cancel the gamma poles and are not zeros of $\Lambda$: the functional equation takes these points to the zero-free right-hand region, including the standard nonvanishing of nonprincipal Dirichlet L-functions at one at the even endpoint. The canceled zeros are simple.

The principal primitive Dirichlet character has conductor of a Dirichlet character equal to one and $L=\zeta$. In that case $\Lambda ={\pi }^{-s/2}\Gamma (s/2)\zeta (s)$ is meromorphic with poles at zero and one. Its canceled trivial zeros begin at $-2$, while $\zeta (0)=-1/2$ is not zero. Multiplication by $s(s-1)/2$ produces the entire Riemann xi function used below.

Let $q_0$ be the conductor of a Dirichlet character and ${\chi }_0$ the inducing primitive even character. Removing the Euler factors absent from $L(s,\chi )$ gives the imprimitive Dirichlet L-function Euler correctionThe primitive functional equation therefore givesEquivalently, replace the final primitive function by $L(1-s,\overline{\chi })/P_{\overline{\chi }}(1-s)$, interpreted as a meromorphic identity with removable values handled by continuation. It is the conductor of a Dirichlet character $q_0$, rather than the possibly inflated modulus $q$, that enters the gamma factor and root number. The complex conjugation bar in the original PDF is lost in the converted TeX.

The zeros are those of $L(s,{\chi }_0)$ together with the zeros of the finite Euler product, and multiplicities add. Since $|{\chi }_0(p)|=1$ at each extra prime, an extra factor vanishes at the imaginary points determined byEach extra prime creates infinitely many such points. Its nonzero-imaginary points are not zeros of the primitive function: the functional equation and nonvanishing of Dirichlet L-functions on the line one exclude them. Thus the zero sets are identical precisely when every prime dividing $q$ already divides $q_0$, making $P=1$. Increasing prime-power exponents alone can make a character imprimitive without changing its L-function. If $q_0=1$, the primitive function is zeta; the same Euler correction applies, with its pole at one retained.

For a nonprincipal Dirichlet character, complete periods sum to zero, so its partial sums are bounded by $q$. Partial summation at $X=q(1+t)$ yieldsThe head is bounded by $1+\log X$, hence $|L(1-it,\overline{\chi })|\ll \log (q+t)$. The completed functional equation givesThe stated gamma bounds make the ratio $O(t^{1/2})$: their exponential factors cancel, and their powers differ by $1/2$. Apply them directly for $t\ge 4$; the compact interval $2\le t\le 4$ is absorbed into the constant. ThereforeFor the conductor-one principal case, Euler summation for zeta at $1-it$, truncated at $X=t^2$, gives a harmonic-size head, a pole term of size $1/t$, and remainder $O((1+t)/X)$. It gives the same $O(\log t)$ bound before applying the zeta functional equation.

The Riemann xi function is the entire functionwith $\xi (s)=\xi (1-s)$. Its Hadamard factorization iswhere the Nontrivial zeros of the Riemann zeta function are repeated by multiplicity and the factors are canonical genus-one factors.

Here is the growth estimate needed for the Jensen zero-count bound. For $|s|\le 2T$, functional symmetry reduces to $\Re s\ge 1/2$. Euler summation truncated at $T^2$ bounds $(s-1)\zeta (s)$ by a fixed power of $T$, uniformly in that region; the multiplication cancels the pole at one. The logarithmic gamma estimate bounds $\log |\Gamma (s/2)|$ by $O(T\log T)$, including the bounded small-$s$ part separately. The remaining elementary factors obey the same bound. ThusFor a zero with $|\rho |\le T$, its contribution in Jensen's formula on radius $2T$ is at least $\log 2$. ConsequentlyIf a zero lies on the integration circle, use nearby radii and continuity of the zero-count estimate. Hence $n_{\xi }(T)=O(T\log T)$ for $T>2$. The growth also gives order at most one and justifies the stated Hadamard factorization; the zero-count bound gives convergence of its genus-one factors.

The printed logarithmic Stirling hint drops the term $-\frac{1}{2}\log z$. The correct expansion is $(z-\frac{1}{2})\log z-z+\frac{1}{2}\log (2\pi )+O(|z{|}^{-1})$ in a fixed sector. Its consequence $\log |\Gamma (z)|=O(|z|\log (2+|z|))$ is all that the argument needs.