S1:∣z∣=1,z+iz−i=z−iz+i⇒(z−i)(z−i)=(z+i)(z+i)∣z∣2−i(z+z)−1=∣z∣2+i(z+z)−1i(z+z)=0z+z=2cosθ=0⇒cosθ=0z=0+0i,∣z∣=1 S1:z+1z−1+z+1z−1=0(z−1)(z+1)+(z+1)(z−1)=0⇒∣z∣2+(z−z)−1+∣z∣2+(z−z)−1=0∣z∣2=1
Among the statements
(S1) : The set {z∈C−{−i}:∣z∣=1 and z+iz−i is purely real} contains exactly two elements, and (S2) : The set {z∈C−{−1}:∣z∣=1 and z+1z−1 is purely imaginary contains infinitely many elements.
Held on 7 Apr 2025 · Verified 6 Jul 2026.
both are incorrect
only (S1) is correct
only (S2) is correct
both are correct
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