Equation (5): Weierstrass zeta addition identity

For every ordered pair $L=(\omega_1,\omega_2)$ of complex numbers that are linearly independent over $\mathbb R$, let $\Lambda=\{m\omega_1+n\omega_2:m,n\in\mathbb Z\}$. For $w\in\mathbb C$, define $Z_L(w)=\frac1w+\sum'_{\ell\in\Lambda\setminus\{0\}}\left(\frac1{w-\ell}+\frac1\ell+\frac{w}{\ell^2}\right)$, $P_L(w)=\sum'_{\ell\in\Lambda}\left(\frac1{(w-\ell)^2}-\frac1{\ell^2}\right)$, and $D_L(w)=-\sum'_{\ell\in\Lambda}\frac2{(w-\ell)^3}$; the omitted zero-lattice summand in $Z_L$ is explicitly defined to be zero. For all $z,v\in\mathbb C$ such that $z\notin\Lambda$, $v\notin\Lambda$, and $z+v\notin\Lambda$, the equality $2\bigl(P_L(v)-P_L(z)\bigr)Z_L(z+v)=2\bigl(Z_L(z)+Z_L(v)\bigr)\bigl(P_L(v)-P_L(z)\bigr)+D_L(v)-D_L(z)$ holds. Here every primed sum is the unordered infinite sum over the indicated lattice points, with value zero if the summand family is not summable, and complex division is total with $a/0=0$ for every $a\in\mathbb C$; in particular, the $1/\ell^2$ term at $\ell=0$ in $P_L$ is zero. The three exclusions ensure that $z$, $v$, and $z+v$ are nonzero and that their differences from lattice points are nonzero. There is no hypothesis that $z\ne v$ or that $P_L(z)\ne P_L(v)$: the case $z=v$ is included whenever $z,2z\notin\Lambda$, and when $P_L(z)=P_L(v)$ the asserted equality reduces to $D_L(v)=D_L(z)$.