All the points in the set `S={(alpha+i)/(alpha-i):alpha in R } (i= sqrt(-1))lie` on a

Options

A. cicle whose radius is `sqrt(2)`
B. straight line whose slope is -1
C. circle whose radius is 1.
D. straight line whose slope is 1.

Explanation

To solve the problem, we need to analyze the set \( S = \left\{ \frac{\alpha + i}{\alpha - i} : \alpha \in \mathbb{R} \right\} \) and determine where all the points in this set lie.

### Step 1: Rationalize the expression
We start with the expression:
\[ z = \frac{\alpha + i}{\alpha - i} \]
To simplify this, we multiply the numerator and denominator by the conjugate of the denominator:
\[ z = \frac{(\alpha + i)(\alpha + i)}{(\alpha - i)(\alpha + i)} = \frac{(\alpha + i)^2}{\alpha^2 + 1} \]

### Step 2: Expand the numerator
Now, we expand the numerator:
\[ (\alpha + i)^2 = \alpha^2 + 2\alpha i - 1 \]
Thus, we have:
\[ z = \frac{\alpha^2 - 1 + 2\alpha i}{\alpha^2 + 1} \]

### Step 3: Separate real and imaginary parts
Now we can separate the real and imaginary parts:
\[ z = \frac{\alpha^2 - 1}{\alpha^2 + 1} + i \frac{2\alpha}{\alpha^2 + 1} \]
Let \( x = \frac{\alpha^2 - 1}{\alpha^2 + 1} \) and \( y = \frac{2\alpha}{\alpha^2 + 1} \).

### Step 4: Find the relationship between \( x \) and \( y \)
Now, we will find \( x^2 + y^2 \):
\[ x^2 = \left(\frac{\alpha^2 - 1}{\alpha^2 + 1}\right)^2 = \frac{(\alpha^2 - 1)^2}{(\alpha^2 + 1)^2} \]
\[ y^2 = \left(\frac{2\alpha}{\alpha^2 + 1}\right)^2 = \frac{4\alpha^2}{(\alpha^2 + 1)^2} \]
Adding these:
\[ x^2 + y^2 = \frac{(\alpha^2 - 1)^2 + 4\alpha^2}{(\alpha^2 + 1)^2} \]
Expanding the numerator:
\[ (\alpha^2 - 1)^2 + 4\alpha^2 = \alpha^4 - 2\alpha^2 + 1 + 4\alpha^2 = \alpha^4 + 2\alpha^2 + 1 = (\alpha^2 + 1)^2 \]
Thus:
\[ x^2 + y^2 = \frac{(\alpha^2 + 1)^2}{(\alpha^2 + 1)^2} = 1 \]

### Conclusion
The equation \( x^2 + y^2 = 1 \) represents a circle with a radius of 1 centered at the origin in the complex plane. Therefore, all points in the set \( S \) lie on the unit circle.

### Final Answer
The points in the set \( S \) lie on the unit circle.

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