= Solution
{c}
Propagate from one <beam waist> to the other, using the <optical lens> <ray transfer matrix>:
$$
M=P(v)L(f)P(u)
=\begin{pmatrix}1-v/f&u+v-uv/f\\-1/f&1-u/f\end{pmatrix}.
$$
<Optical mode matching> requires $q_A=iz_A$ to transform to $q_B=iz_B$. Multiplying the denominator across gives
$$
iz_B(1-u/f-iz_A/f)=(1-v/f)iz_A+u+v-uv/f.
$$
Equating real and imaginary parts and multiplying by $f$ yields
$$
\boxed{v(u-f)+z_Az_B=uf,\qquad
vz_A-z_B(u-f)=z_Af.}
$$
Set $\delta=u-f$. The second equation gives $v=f+(z_B/z_A)\delta$. Substituting this into the first cancels the terms $f\delta$:
$$
\frac{z_B}{z_A}\delta^2+z_Az_B=f^2.
$$
Consequently the two algebraic branches are
$$
\boxed{\delta=\pm\sqrt{\frac{z_A}{z_B}f^2-z_A^2},\qquad
u=f\pm z_A\sqrt{\frac{f^2}{z_Az_B}-1},\qquad
v=f\pm z_B\sqrt{\frac{f^2}{z_Az_B}-1}.}
$$
Equivalently, $u=f+\delta$. The signs in $u$ and $v$ are the same. Real solutions require $\boxed{f^2\geq z_Az_B}$; a converging <optical lens> therefore needs $f\geq\sqrt{z_Az_B}$. At equality $u=v=f$. For the stated arrangement, additionally require $u,v\geq0$ and that the chosen <optical lens> position is accessible outside both <optical resonators>. A negative $v$ is an algebraic virtual-waist solution, not a downstream <beam waist>.
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