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.
Which of the following is correct?
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Step-by-Step Solution
Step 1: Analyze Statement S1
For a complex number to be purely real, it must be equal to its conjugate. We are given that z is a complex number such that ∣z∣=1 and z=−i. We set the expression equal to its conjugate to find the conditions for it to be purely real.
Step 2: Simplify the equation for S1
We expand both sides of the equation and simplify. Since i2=−1, these terms cancel out. We also know that zzˉ=∣z∣2, which also cancels out. This leaves us with an equation involving z and its conjugate zˉ. Dividing by 2i simplifies it further to z+zˉ=0.
Step 3: Solve for z in S1
The condition z+zˉ=0 implies that the real part of z is zero. Since we are also given that ∣z∣=1, we can write z=x+iy. Then x=0, so z=iy. Substituting this into ∣z∣=1, we get ∣iy∣=1, which means ∣i∣∣y∣=1, so 1⋅∣y∣=1, implying y=±1. Thus, z=i or z=−i. However, the problem states z=−i. Therefore, only z=i is a solution. This means statement S1 contains exactly one element, not two.
Step 4: Analyze Statement S2
For a complex number to be purely imaginary, it must be equal to the negative of its conjugate. We are given that z is a complex number such that ∣z∣=1 and z=−1. We set the expression equal to the negative of its conjugate to find the conditions for it to be purely imaginary.
Step 5: Simplify the equation for S2
We expand both sides of the equation and simplify. We use the property zzˉ=∣z∣2. Since ∣z∣=1, we substitute ∣z∣2=1 into the equation.
Step 6: Solve for z in S2
After substituting ∣z∣2=1, the equation simplifies to z−zˉ=z−zˉ. This is an identity, which means it is true for all z that satisfy the initial conditions. The initial conditions are ∣z∣=1 and z=−1. The set of all complex numbers z such that ∣z∣=1 (excluding z=−1) represents a circle in the complex plane (the unit circle with one point removed). This set contains infinitely many elements. Therefore, statement S2 is correct.