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

Options

A. straight line whose slope is 1
B. circle whose radius is1
C. circle whose radius is `sqrt(2)`
D. straight line whose slope is – 1

Answer

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

### Step 1: Rewrite the expression
Let \( z = \frac{\alpha + i}{\alpha - i} \).

### Step 2: Multiply numerator and denominator by the conjugate of the denominator
To simplify \( z \), we multiply the numerator and denominator by the conjugate of the denominator, which is \( \alpha + i \):
\[ z = \frac{(\alpha + i)(\alpha + i)}{(\alpha - i)(\alpha + i)} = \frac{\alpha^2 + 2i\alpha - 1}{\alpha^2 + 1} \]

### Step 3: Separate real and imaginary parts
Now, we can express \( z \) in terms of its 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 \( y \)
We need to eliminate \( \alpha \) to find a relationship between \( x \) and \( y \). From the equation for \( y \):
\[ \alpha = \frac{y(\alpha^2 + 1)}{2} \]
Substituting \( \alpha \) back into the equation for \( x \):
\[ x = \frac{\left(\frac{y(\alpha^2 + 1)}{2}\right)^2 - 1}{\left(\frac{y(\alpha^2 + 1)}{2}\right)^2 + 1} \]
This leads to a more complex expression, but we can also derive the relationship using the modulus.

### Step 5: Use the modulus property
Since \( z = \frac{z_1}{\overline{z_1}} \) where \( z_1 = \alpha + i \), we know that:
\[ |z| = \frac{|z_1|}{|\overline{z_1}|} = \frac{|z_1|}{|z_1|} = 1 \]
This means that all points \( z \) lie on the unit circle in the complex plane.

### Step 6: Conclusion
The equation \( |z| = 1 \) represents a circle with radius 1 centered at the origin. Therefore, the points in the set \( S \) lie on the unit circle.

### Final Answer
The correct option is that all the points in the set \( S \) lie on the unit circle.

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