Among the statements: (S1): The set $\{ z \in \mathbb{C} \setminus \{-i\} : |z| = 1 \text{ and } \frac{z - i}{z + i} \text{ is purely real} \}$ contains exactly two elements, and (S2): The set $\{ z \in \mathbb{C} \setminus \{-1\} : |z| = 1 \text{ and } \frac{z - 1}{z + 1} \text{ is purely imaginary} \}$ contains infinitely many elements. Which of the following is correct?

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 $z \neq -i$. 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 $i^2 = -1$, these terms cancel out. We also know that $z\bar{z} = |z|^2$, which also cancels out. This leaves us with an equation involving $z$ and its conjugate $\bar{z}$. Dividing by $2i$ simplifies it further to $z + \bar{z} = 0$.

Step 3: Solve for z in S1

The condition $z + \bar{z} = 0$ 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 $1 \cdot |y|=1$, implying $y = \pm 1$. Thus, $z = i$ or $z = -i$. However, the problem states $z \neq -i$. 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 $z \neq -1$. 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 $z\bar{z} = |z|^2$. 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 $z - \bar{z} = z - \bar{z}$. This is an identity, which means it is true for all $z$ that satisfy the initial conditions. The initial conditions are $|z|=1$ and $z \neq -1$. 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.

More from this chapter

• Express (1 + i)^4 using De Moivre's theorem.
• Simplify (1 + ²)^5.
• Simplify (1-)/(1- ²).
• **LONG ANSWER TYPE QUESTIONS** 1) If α and β are different complex numbers with |β| = 1, then find | β - α1 - αβ| 2) If |a + ib| = 1 then prove that ((1 + b + i
• Find all values of (1+i)^(1-i) in the form a + bi.
• Find all real solutions to the equation: √(x + 3 - 4√(x - 1)) + √(x + 8 - 6√(x - 1)) = 1.
• If the set of all a R, for which the equation 2x²+(a-5)x+15=3a has no real root, is the interval (α, β), and X=x Z: α < x < β, then Σ_x X x² is equal to:
• MODIFIED: Given that variable_91, variable_91, variable_91 are roots of x³ - 3x² + 4 = 0, find the value of variable_91⁴ + variable_91⁴ + variable_91⁴.

All Complex Numbers and Quadratic Equations solutions →