A disc missing $p$ carries a holomorphic logarithm of $z-p$

Given: An open disc $D(c,\rho )$ with $\rho >0$ and a point $p\mathrm{\notin }D(c,\rho )$.

If $D(a,r)$ is an open disc with $r>0$ and $h:D(a,r)\to \mathbb{C}$ is holomorphic and nowhere zero, then there is a holomorphic $L:D(a,r)\to \mathbb{C}$ with $\exp (L(z))=h(z)$ for every $z\in D(a,r)$ (A nonvanishing holomorphic function on a disc has a holomorphic logarithm).

If $L$ and $h$ are holomorphic on an open $U$ with $\exp \circ L=h$, then $h$ is nowhere zero and $L^{\mathrm{\prime }}=h^{\mathrm{\prime }}\mathrm{/}h$; if $U$ misses $p$ and $h(z)=z-p$, then $L^{\mathrm{\prime }}(z)=1\mathrm{/}(z-p)$ (A holomorphic logarithm is a primitive of the logarithmic derivative).

$\ker (\exp )=2\pi i\mathbb{Z}$, and $\exp z=\exp w$ exactly when $z-w\in 2\pi i\mathbb{Z}$ ($\ker (\exp )=2\pi i\mathbb{Z}$, and $\exp z=\exp w$ exactly when $z-w\in 2\pi i\mathbb{Z}$).

The continuous image of a connected subset is a connected subset (A continuous image of a connected space is connected, and connectedness is a topological property).

A subset $E\subseteq \mathbb{R}$ is connected exactly when it is order-convex: $x,z\in E$ and $x\le w\le z$ imply $w\in E$ (The connected subspaces of $\mathbb{R}$ with its usual topology are exactly the order-convex subsets, the published characterisation transported by the identification of the two descriptions of "open in $\mathbb{R}$").

A path-connected subset of a topological space is a connected subset (Every path-connected space is connected, and every path component lies inside a component); a subset is path-connected when any two of its points are joined by a continuous map from $[0,1]$ with image inside it (Paths, path-connected spaces and path components).

A subset $U\subseteq {\mathbb{R}}^m$ is convex when $(1-t)x+ty\in U$ for all $x,y\in U$ and $t\in [0,1]$ (A convex subset of ${\mathbb{R}}^m$ contains every line segment between two of its points).

$B(x,r)=\{y:d(x,y)<r\}$ (Open ball, closed ball and sphere in a metric space).

Linear combinations, products and nonvanishing quotients of functions complex differentiable at a point are complex differentiable there; constants have derivative $0$ and the identity has derivative $1$ (Linearity, product, reciprocal, and quotient rules for complex derivatives).

A function complex differentiable at a point is continuous there (Complex differentiability at a point implies continuity there).

$\mathrm{\mid }zw\mathrm{\mid }=\mathrm{\mid }z\mathrm{\mid }\mathrm{\mid }w\mathrm{\mid }$ and $\mathrm{\mid }z+w\mathrm{\mid }\le \mathrm{\mid }z\mathrm{\mid }+\mathrm{\mid }w\mathrm{\mid }$ for complex $z,w$ (Conjugation is an involutive real-field automorphism, $z\bar{z}=\mathrm{\mid }z{\mathrm{\mid }}^2$, and modulus is definite, multiplicative, and subadditive).

For $z=a+bi$ with $a,b$ real, $\mathrm{Im}z=b$ and $\mathrm{\mid }z\mathrm{\mid }=\sqrt{a^2+b^2}$ (Real and imaginary parts, complex conjugation, and modulus).

The integers form an ordered commutative ring, and their canonical image in $\mathbb{R}$ is discrete; hence if $m<n$ then $m+\frac{1}{2}$ lies strictly between them and is not an integer (The integers form a commutative ring, The integers form a totally ordered ring, Integer part: for every real $x$ there is exactly one integer $m$ with $m\le x<m+1$).