Among the statements:
(S1): The set {z∈C−{−i}:∣z∣=1,z+iz−i is purely real} contains exactly two elements.
(S2): The set {z∈C−{−1}:∣z∣=1,z+1z−1 is purely imaginary} contains infinitely many elements.
Solutions · 1
Official solution
Statement 1: z+iz−i purely real means it equals its own conjugate:
\[\frac{z-i}{z+i}=\frac{\bar z+i}{\bar z-i} \implies (z-i)(\bar z-i)=(z+i)(\bar z+i) \implies z\bar z - iz - i\bar z-1 = z\bar z+iz+i\bar z-1\]
\[\implies z+\bar z=0 \implies x=0 \text{ (the imaginary axis)}.\]
Combined with ∣z∣=1, this gives z=±i, but z=−i, so only z=i works --- a single element, not two. So (S1) is false.
Statement 2: z+1z−1 purely imaginary means:
\[\frac{z-1}{z+1}+\frac{\bar z-1}{\bar z+1}=0 \implies z\bar z-1=0 \implies |z|^2=1,\]
which is true for every point on the unit circle (except z=−1). So (S2) contains infinitely many elements --- true.
Only (S2) is correct.