= Solution
Let $\pi:\mathbb C\to\mathbb C/\Lambda$ be the <universal covering map>. Since the <complex plane> is a <simply connected domain>, the <lifting criterion for a covering space> gives a <holomorphic lift between one-dimensional complex tori> $F:\mathbb C\to\mathbb C$ such that
$$
\pi\circ F=f\circ\pi.
$$
For every $\lambda\in\Lambda$, the difference $F(z+\lambda)-F(z)$ belongs to the discrete <period lattice> $\Lambda$. It depends continuously on $z$, so it is constant. Differentiating shows that the <derivative> $F'$ is $\Lambda$-periodic. It is bounded on the compact closure of a <fundamental parallelogram>, and periodicity makes it bounded on all of $\mathbb C$. The <Liouville theorem> therefore makes $F'$ constant, and hence
$$
F(z)=az+b.
$$
Thus every holomorphic map has an <affine lift of a holomorphic map between one-dimensional complex tori>. If “map of complex tori” means an identity-preserving map, choose $F(0)=0$; then $b=0$, so $F$ is the required <linear map>. Without that convention the statement must say affine, since a nonzero <translation in a group> of the torus lifts to $z\mapsto z+b$.
The <Weierstrass elliptic function> of $\Lambda$ is
$$
\wp_\Lambda(z)=\frac1{z^2}+
\sum_{\omega\in\Lambda\setminus\{0\}}
\left(\frac1{(z-\omega)^2}-\frac1{\omega^2}\right).
$$
The subtracted term gives the <Normal convergence of the Weierstrass elliptic-function series> away from $\Lambda$. Put
$$
G_{2r}=\sum_{\omega\in\Lambda\setminus\{0\}}\omega^{-2r}.
$$
The <Laurent coefficients of the Weierstrass elliptic function> at zero give
$$
\wp(z)=z^{-2}+3G_4z^2+5G_6z^4+O(z^6),
\qquad
\wp'(z)=-2z^{-3}+6G_4z+20G_6z^3+O(z^5).
$$
Set $g_2=60G_4$ and $g_3=140G_6$. Direct substitution shows that the principal part and constant term of
$$
H(z)=\wp'(z)^2-4\wp(z)^3+g_2\wp(z)+g_3
$$
vanish at zero. Since $H$ is an <elliptic function>, translation gives the same cancellation at every point of the <period lattice>. Every apparent <isolated singularity> is therefore a <removable singularity>, so $H$ is an <entire function>. It is periodic and hence bounded on the translates of a compact <fundamental parallelogram>. The <Liouville theorem> gives $H=0$, which proves the <Weierstrass elliptic differential equation>
$$
\boxed{\wp'(z)^2=4\wp(z)^3-g_2\wp(z)-g_3.}
$$
Finally suppose that $f$ is a biholomorphic <group homomorphism>. Its identity-preserving lift has the form $F(z)=\zeta z$ by the first part. Since the inverse map also lifts linearly,
$$
\zeta\Lambda=\Lambda.
$$
Choose a $\mathbb Z$-basis of $\Lambda$. Multiplication by $\zeta$ is then represented by a <unimodular matrix> $A\in\operatorname{GL}_2(\mathbb Z)$. As a real-linear transformation of $\mathbb C$, it has <determinant> $|\zeta|^2>0$; hence $\det A=1$ and $|\zeta|=1$. The <characteristic polynomial> of $A$ and the <Cayley-Hamilton theorem> give
$$
\zeta^2-\operatorname{tr}(A)\zeta+1=0.
$$
This already has the required form with $m=-\operatorname{tr}(A)\in\mathbb Z$ and $n=1$. Moreover $2\operatorname{Re}\zeta=\operatorname{tr}(A)$, so $|\zeta|=1$ forces $\operatorname{tr}(A)\in\{-2,-1,0,1,2\}$. Thus $\zeta$ is a <root of unity>, of order $1$, $2$, $3$, $4$, or $6$, completing the description of an <automorphism of a one-dimensional complex torus>.
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