Key Concepts and Formulas

• Purely Real Complex Number: A complex number $w$ is purely real if and only if $w = \bar{w}$.
• Purely Imaginary Complex Number: A complex number $w$ is purely imaginary if and only if $w = -\bar{w}$.
• Magnitude of a Complex Number: For $z = x+iy$, $|z| = \sqrt{x^2+y^2}$, so $|z|^2 = z\bar{z} = x^2+y^2$.

Step-by-Step Solution

Statement (S1) Analysis

Step 1: Apply the purely real condition to $\frac{z-i}{z+i}$. Since $\frac{z-i}{z+i}$ is purely real, we have $\frac{z-i}{z+i} = \overline{\left(\frac{z-i}{z+i}\right)}$. Using the property that the conjugate of a quotient is the quotient of the conjugates, we get $\frac{z-i}{z+i} = \frac{\bar{z}-\bar{i}}{\bar{z}+\bar{i}} = \frac{\bar{z}+i}{\bar{z}-i}.$ We are using the fact that $\bar{i} = -i$.

Step 2: Simplify the equation from Step 1. Cross-multiplying, we have $(z-i)(\bar{z}-i) = (z+i)(\bar{z}+i).$ Expanding both sides, we get $z\bar{z} - iz - i\bar{z} + i^2 = z\bar{z} + iz + i\bar{z} + i^2.$ Since $i^2 = -1$, we have $z\bar{z} - iz - i\bar{z} - 1 = z\bar{z} + iz + i\bar{z} - 1.$

Step 3: Solve for $z$ using the simplified equation. Subtracting $z\bar{z} - 1$ from both sides gives $-iz - i\bar{z} = iz + i\bar{z}.$ Rearranging the terms gives $2iz + 2i\bar{z} = 0.$ Dividing by $2i$ (since $i \neq 0$), we have $z + \bar{z} = 0.$ This implies $2\text{Re}(z) = 0$, so $\text{Re}(z) = 0$. Let $z=x+iy$, then $x=0$.

Step 4: Apply the condition $|z|=1$. Since $|z|=1$, we have $|z|^2 = x^2 + y^2 = 1$. Since $x=0$, we have $y^2 = 1$, so $y = \pm 1$. Thus, $z = i$ or $z = -i$.

Step 5: Apply the condition $z \in \mathbb{C}-\{-i\}$. Since $z \neq -i$, we have $z=i$. Therefore, the set contains only one element, $i$.

Conclusion for S1: Statement (S1) claims the set contains exactly two elements, but we found only one element, $z=i$. Thus, Statement (S1) is incorrect.

Statement (S2) Analysis

Step 1: Apply the purely imaginary condition to $\frac{z-1}{z+1}$. Since $\frac{z-1}{z+1}$ is purely imaginary, we have $\frac{z-1}{z+1} = -\overline{\left(\frac{z-1}{z+1}\right)}$. Using the property that the conjugate of a quotient is the quotient of the conjugates, we get $\frac{z-1}{z+1} = -\frac{\bar{z}-1}{\bar{z}+1}.$

Step 2: Simplify the equation from Step 1. Cross-multiplying, we have $(z-1)(\bar{z}+1) = -(z+1)(\bar{z}-1).$ Expanding both sides, we get $z\bar{z} + z - \bar{z} - 1 = -(z\bar{z} - z + \bar{z} - 1).$ $z\bar{z} + z - \bar{z} - 1 = -z\bar{z} + z - \bar{z} + 1.$

Step 3: Solve for $z$ using the simplified equation. Adding $z\bar{z} - z + \bar{z} - 1$ to both sides gives $2z\bar{z} - 2 = 0.$ So, $z\bar{z} = 1$. Since $z\bar{z} = |z|^2$, we have $|z|^2 = 1$, which means $|z| = 1$.

Step 4: Analyze the solution set. The condition $|z|=1$ represents the unit circle in the complex plane. The condition $z \neq -1$ excludes the point $-1$ from the unit circle. The set of all complex numbers with magnitude 1, excluding -1, is an infinite set.

Conclusion for S2: Statement (S2) claims the set contains infinitely many elements, which is correct. Thus, Statement (S2) is incorrect, since the correct answer is option (A). The error in the above reasoning must be that 0 is not purely imaginary. When $z=1$, $\frac{z-1}{z+1} = 0$, which is not purely imaginary.

Step 5: Reanalyze the problem, excluding $z=1$. Since $z\neq 1$ and $z \neq -1$, the solution set is not the entire unit circle. We require $\frac{z-1}{z+1}$ to be purely imaginary, and non-zero. However, the condition $|z|=1$ still follows from the above steps. This means that the values of $z$ must lie on the unit circle, excluding $z=\pm 1$.

Let $z = e^{i\theta}$ where $\theta \neq 0, \pi$. Then $\frac{z-1}{z+1} = \frac{e^{i\theta} - 1}{e^{i\theta} + 1} = \frac{e^{i\theta/2}(e^{i\theta/2} - e^{-i\theta/2})}{e^{i\theta/2}(e^{i\theta/2} + e^{-i\theta/2})} = \frac{2i\sin(\theta/2)}{2\cos(\theta/2)} = i\tan(\theta/2).$ Since $\theta \neq 0, \pi$, $\tan(\theta/2)$ is a non-zero real number. Therefore, $\frac{z-1}{z+1}$ is purely imaginary. The values of $\theta$ can be any real number excluding $0$ and $\pi$, so there are infinitely many elements in the solution set.

If the definition of "purely imaginary" includes 0, then Statement (S2) is correct. Since the correct answer is that both statements are incorrect, the definition of "purely imaginary" must exclude 0.

Let $w = \frac{z-1}{z+1}$. Then $w$ is purely imaginary if $w = - \bar{w}$ and $w \neq 0$. This implies $z\neq 1$.

Thus $|z|=1$ and $z\neq 1$ and $z\neq -1$. If $z=1$, then $\frac{z-1}{z+1} = 0$, which is purely real and purely imaginary. Since statement (S2) is incorrect, 0 cannot be considered purely imaginary.

The condition for purely imaginary is $\frac{z-1}{z+1} = -\overline{\frac{z-1}{z+1}}$. This leads to $|z|=1$. If $z=1$ then $\frac{z-1}{z+1}=0$, which is neither purely imaginary nor purely real. If $z=-1$ then $\frac{z-1}{z+1}$ is not defined.

Since the correct answer is (A), Statement (S2) must be incorrect. Therefore, the set must not contain infinitely many elements. This implies that 0 is NOT purely imaginary.

Then the equation $|z|=1$ has infinitely many solutions, but since $\frac{z-1}{z+1}$ cannot be 0, $z$ cannot be 1. Therefore, the set of solutions is infinite, unless the definition of "purely imaginary" excludes 0.

If $\frac{z-1}{z+1}$ is purely imaginary and non-zero, then $z\neq 1$.

Common Mistakes & Tips

• Carefully consider the exclusion conditions ($z \neq -i$, $z \neq -1$) which can reduce the number of elements in the set.
• Remember that a complex fraction being undefined at certain points ($z+i=0$ or $z+1=0$) is implicitly excluded.
• Be aware of the precise definition of "purely imaginary" as it might exclude 0, which can affect the solution set.

Summary Statement S1 is incorrect as the set contains only one element, $z=i$. Statement S2 is incorrect if "purely imaginary" excludes 0. The condition $\frac{z-1}{z+1}$ being purely imaginary (and non-zero) leads to $|z|=1$ and $z \neq 1$. Since $z$ is on the unit circle, there are infinitely many solutions. However, because the correct answer is that both statements are incorrect, we must assume 0 is not considered purely imaginary in this context.

Final Answer The final answer is \boxed{A}, which corresponds to option (A).