= Solution
For the unheaded continuation, set $w=\log z$ in the food equation. The two square roots are $w=\pm i\pi/4$, so exponentiating gives
$$
\boxed{z=e^{i\pi/4},\ e^{-i\pi/4}=\frac{1\pm i}{\sqrt2}\quad\text{(food)}.}
$$
Both values have $|z-1|^2=2-\sqrt2<1$ and hence lie strictly inside the disk. Both also have the indicated principal logarithms. Allowing the multivalued <complex logarithm> produces no additional points: the only permitted values of $w$ are still the two square roots already used.
For drink, the PDF squares only the denominator. Put $z=x+iy$. Its numerator is $(3x+iy)/2$ and its unsquared denominator is $(x+3iy)/2$, which is nonzero unless $z=0$. Multiplying the equation by this denominator squared and equating the <real part> and <imaginary part> yields
$$
\frac32x=\frac34x^2-\frac{27}4y^2,\qquad\frac12y=\frac92xy,
$$
that is $x^2-9y^2=2x$ and $y(1-9x)=0$. If $y=0$, the real equation gives $x=0$ or $x=2$, and zero is excluded. If $y\ne0$, then $x=1/9$, but the real equation would require $9y^2=-17/81$, impossible for real $y$. Thus
$$
\boxed{z=2\quad\text{is the only drink point}.}
$$
It lies on the allowed disk boundary, and direct substitution gives $3/1^2=3$. The denominator-only square matters: squaring the entire quotient would define a different locus.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/ia/paper-1-complex-locations.png]
{title=Food points inside the allowed complex disk and the unique drink point on its boundary}
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