If `(z+1)/(z+i)` is a purely imaginary number (where`(i=sqrt(-1)`), then z lies on a

Options

A. straight line
B. circle
C. circle with radius = `1/sqrt(2)`
D. circle passing through the origin

Answer

To solve the problem, we need to analyze the expression \((z + 1)/(z + i)\) and determine the conditions under which it is purely imaginary. Let's denote \(z\) as \(z = x + iy\), where \(x\) and \(y\) are real numbers.

### Step-by-Step Solution:

1. **Substituting \(z\)**:
\[ z + 1 = (x + iy) + 1 = (x + 1) + iy \]
\[ z + i = (x + iy) + i = x + i(y + 1) \]

2. **Forming the Fraction**:
\[ \frac{z + 1}{z + i} = \frac{(x + 1) + iy}{x + i(y + 1)} \]

3. **Multiplying by the Conjugate**:
To simplify, we multiply the numerator and denominator by the conjugate of the denominator:
\[ \frac{((x + 1) + iy)(x - i(y + 1))}{(x + i(y + 1))(x - i(y + 1))} \]

4. **Calculating the Denominator**:
\[ (x + i(y + 1))(x - i(y + 1)) = x^2 + (y + 1)^2 \]

5. **Calculating the Numerator**:
\[ ((x + 1) + iy)(x - i(y + 1)) = (x + 1)x + (x + 1)(-i(y + 1)) + iyx + iy(-i(y + 1)) \]
\[ = x^2 + x + iyx - i(y + 1)(x + 1) - y(y + 1) \]
\[ = x^2 + x + y^2 + y + i(yx - (y + 1)(x + 1)) \]

6. **Separating Real and Imaginary Parts**:
The expression becomes:
\[ \frac{(x^2 + x + y^2 + y) + i(yx - (y + 1)(x + 1))}{x^2 + (y + 1)^2} \]

7. **Setting the Real Part to Zero**:
For the fraction to be purely imaginary, the real part must equal zero:
\[ x^2 + x + y^2 + y = 0 \]

8. **Rearranging the Equation**:
Rearranging gives:
\[ x^2 + x + y^2 + y = 0 \]

9. **Completing the Square**:
Completing the square for \(x\) and \(y\):
\[ (x + \frac{1}{2})^2 - \frac{1}{4} + (y + \frac{1}{2})^2 - \frac{1}{4} = 0 \]
\[ (x + \frac{1}{2})^2 + (y + \frac{1}{2})^2 = \frac{1}{2} \]

10. **Identifying the Geometric Shape**:
This is the equation of a circle centered at \((-1/2, -1/2)\) with a radius of \(\frac{1}{\sqrt{2}}\).

### Conclusion:
Thus, \(z\) lies on a circle centered at \((-1/2, -1/2)\) with radius \(\frac{1}{\sqrt{2}}\).