= Solution
\b[The assertion requires $p+q>0$.] As printed, it also includes $(p,q)=(0,0)$, where the nonzero constant function $1$ is closed but cannot be the <exterior derivative> of a negative-degree form. The zero constant is harmless; below we prove the intended positive-degree result.
Set $k=p+q>0$, let $\rho_t(z)=tz$, and let
$$
R=\sum_j\left(z_j\frac{\partial}{\partial z_j}+\bar z_j\frac{\partial}{\partial\bar z_j}\right).
$$
The <polydisc> is preserved by these dilations. The <radial homotopy operator> is
$$
\boxed{H\alpha=\int_0^1\rho_t^*(\iota_R\alpha)\,\frac{dt}{t}.}
$$
Here $\iota_R$ is the <interior product of a differential form>. For a smooth $k$-form, its pulled-back contraction is $O(t^k)$, so the integral and its coefficient derivatives converge at zero. The identity for a <pullback of a differential form> along the radial flow and <Cartan's magic formula> give
$$
\frac{d}{dt}\rho_t^*\alpha=\frac1t\rho_t^*\mathcal L_R\alpha
=\frac1t\rho_t^*(d\iota_R\alpha+\iota_Rd\alpha).
$$
Integrating yields $dH\alpha+Hd\alpha=\alpha-\rho_0^*\alpha$. The final pullback vanishes in positive degree; since $d\alpha=0$, we have $dH\alpha=\alpha$.
Each $\rho_t$ is a <holomorphic map>, so its <pullback of a differential form> preserves bidegree. Contracting with the $(1,0)$ part of $R$ lowers $p$ by one, and contracting with its $(0,1)$ part lowers $q$ by one. Therefore the primitive has precisely the permitted types:
$$
\boxed{\beta=H\alpha\in\mathcal A^{p-1,q}(D)\oplus\mathcal A^{p,q-1}(D),\qquad d\beta=\alpha.}
$$
This is a type-preserving refinement of the <Poincaré lemma>; when one index is zero, the corresponding negative-degree summand is simply absent.
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