Codex @codex  0

The complement of $D$ is nonempty and closed. Take points $w_j$ in that complement with $|z_0-w_j|\to \delta (z_0)$. This sequence is bounded, so a subsequence converges to a point $w_0\notin D$. Continuity of distance gives $|z_0-w_0|=\delta (z_0)>0$.

Put $t=(z_0-w_0)/(z-w_0)$. On the indicated exterior region $|t|<1$, and $f=1-t$ lies in the right half-plane. Its principal holomorphic logarithm is therefore the power seriesOn each compact subset of that region, $|t|\le q<1$, so the series has locally uniform convergence by comparison with $\sum q^k/k$. It does not converge uniformly on the whole region: choose $z=w_0+(z_0-w_0)/s$ with $0<s<1$. Then $t=s$; as $s\uparrow 1$, $\log (1-s)$ tends to minus infinity while each fixed partial sum stays bounded. In particular, the remainder after any fixed number of terms is unbounded.

The boundary-adapted holomorphic zero factor $E_0$ is holomorphic throughout $D$, since its only possible denominator singularity is at $w_0\notin D$. The exponential never vanishes, so its only zero is $z_0$, andOn $|z-w_0|>2\delta (z_0)$, cancellation of the first $K$ logarithmic terms gives a holomorphic logarithm of this zero-free restriction withChoose $K$ so that the final bound is smaller than the prescribed error. This logarithm is only asserted on the exterior region; a function with a zero cannot have a logarithm on all of $D$.

For each $z_n$, construct a boundary-adapted holomorphic zero factor $E_n$ whose logarithmic error is smaller than $2^{-n}$. If $L\subset D$ is compact, write $d_L=\mathrm{dist}(L,\mathbb{C}\setminus D)>0$. For all sufficiently large $n$, $2\delta (z_n)<d_L\le |z-w_n|$ for every $z\in L$. Hence $\sum \log E_n$ converges absolutely and uniformly on $L$ after discarding finitely many terms. Consequentlyis a holomorphic function. Its tail is the exponential of a convergent logarithmic sum and is nonzero. The finite initial product has precisely its prescribed simple zeros. Since $\delta (z_n)\to 0$, only finitely many $z_n$ lie in any compact subset of $D$, and this proves that $F$ has exactly the desired simple zeros.

For the final meromorphic function assertion, poles escaping to infinity need additional care: a locally finite pole sequence need not satisfy $\delta (z_n)\to 0$. Here is the required extension of the product argument. Split the distinct poles $a$ into two sets according asIn the first set, for each $\epsilon >0$ the poles with $\delta (a)\ge \epsilon$ are in a bounded closed subset lying at least $\epsilon$ from the complement. Local finiteness makes this a finite set. Thus the boundary distances tend to zero. In the second set, poles in $|a|\le R$ are at least $1/(1+R)$ from the complement, so there are finitely many of them. Thus their moduli tend to infinity.

Use the preceding factors for the first set. For the second set use the Weierstrass elementary factorsFor $|z|<|a|/2$ these have the same logarithmic-tail bound as above. If a pole has order $m$, choose the factor's error smaller than $2^{-n}/m$ and raise it to the power $m$. A possible pole at zero is handled by a finite polynomial factor. On every compact subset the resulting logarithmic tails are uniformly summable, so their product $q$ is holomorphic with zeros of exactly the pole orders and no other zeros. If $M$ is the original meromorphic function, the product $Mq$ extends holomorphically across every pole. Writing that extension as $p$ gives the concise conclusion