= Solution
Pulling back the factors of automorphy in the <Appell–Humbert theorem> by $z\mapsto cz$, where $c=a+bi$, pulls back both the <Hermitian form> and the <semicharacter of a complex lattice>. Thus \b[the explicit answer is]
$$
\boxed{H(z,w)=(a^2+b^2)z\overline w,\qquad \alpha(\lambda)=\alpha_1((a+bi)\lambda).}
$$
Writing $\lambda=m+ni$ makes the second formula completely explicit:
$$
\alpha(m+ni)=(-1)^{(am-bn)(bm+an)}=(-1)^{ab(m+n)+(a^2+b^2)mn}.
$$
The second equality is an equality of signs, obtained by reducing the exponents modulo $2$. The <Hermitian form> follows from $H_1(cz,cw)=|c|^2H_1(z,w)$. Its integer parameter equals the norm of the <Gaussian integer>, agreeing with the degree of this nonzero <isogeny of elliptic curves>. The formulas also include $c=0$, when the pulled-back <line bundle> is trivial.
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