Past exam of the mathematics course of the University of Cambridge / 2005 / ia / Paper 1 / 5C / 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 givesBoth values have $|z-1{|}^2=2-\sqrt{2}<1$ 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 yieldsthat 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 $y\ne 0$, then $x=1/9$, but the real equation would require $9y^2=-17/81$, impossible for real $y$. ThusIt 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.