If `|z-i|=1` and arg (z) `=theta` where `0 lt theta lt pi/2`, then `cottheta-2/z` equals

Options

A. `2i`
B. `-i`
C. `i`
D. `1+i`

Answer

To solve the problem, we need to find the value of \( \cot \theta - \frac{2}{z} \) given that \( |z - i| = 1 \) and \( \arg(z) = \theta \) where \( 0 < \theta < \frac{\pi}{2} \).

### Step-by-step Solution:

1. **Express \( z \)**:
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers.

2. **Use the modulus condition**:
The condition \( |z - i| = 1 \) translates to:
\[ |(x + iy) - i| = 1 \implies |x + (y - 1)i| = 1 \]
This means:
\[ \sqrt{x^2 + (y - 1)^2} = 1 \]
Squaring both sides gives:
\[ x^2 + (y - 1)^2 = 1 \]

3. **Expand the equation**:
Expanding the equation:
\[ x^2 + (y^2 - 2y + 1) = 1 \implies x^2 + y^2 - 2y + 1 - 1 = 0 \implies x^2 + y^2 - 2y = 0 \]
This can be rearranged to:
\[ x^2 + y^2 = 2y \]

4. **Use the argument condition**:
The argument condition \( \arg(z) = \theta \) implies:
\[ \tan \theta = \frac{y}{x} \implies y = x \tan \theta \]

5. **Substitute \( y \) in the first equation**:
Substitute \( y = x \tan \theta \) into the equation \( x^2 + y^2 = 2y \):
\[ x^2 + (x \tan \theta)^2 = 2(x \tan \theta) \]
This simplifies to:
\[ x^2 + x^2 \tan^2 \theta = 2x \tan \theta \]
Factoring out \( x^2 \):
\[ x^2(1 + \tan^2 \theta) = 2x \tan \theta \]

6. **Solve for \( x \)**:
Rearranging gives:
\[ x^2 = \frac{2x \tan \theta}{1 + \tan^2 \theta} \]
If \( x \neq 0 \), we can divide both sides by \( x \):
\[ x = \frac{2 \tan \theta}{1 + \tan^2 \theta} \]

7. **Find \( y \)**:
Substitute \( x \) back to find \( y \):
\[ y = x \tan \theta = \frac{2 \tan^2 \theta}{1 + \tan^2 \theta} \]

8. **Express \( z \)**:
Thus, we have:
\[ z = x + iy = \frac{2 \tan \theta}{1 + \tan^2 \theta} + i \frac{2 \tan^2 \theta}{1 + \tan^2 \theta} \]
Factoring out \( \frac{2 \tan \theta}{1 + \tan^2 \theta} \):
\[ z = \frac{2 \tan \theta}{1 + \tan^2 \theta} (1 + i \tan \theta) \]

9. **Calculate \( \cot \theta - \frac{2}{z} \)**:
We know \( \cot \theta = \frac{1}{\tan \theta} \), so:
\[ \cot \theta - \frac{2}{z} = \frac{1}{\tan \theta} - \frac{2(1 + i \tan \theta)}{\frac{2 \tan \theta}{1 + \tan^2 \theta}} = \frac{1}{\tan \theta} - \frac{(1 + i \tan \theta)(1 + \tan^2 \theta)}{\tan \theta} \]
Simplifying gives:
\[ \cot \theta - \frac{2}{z} = \frac{1 - (1 + i \tan \theta)(1 + \tan^2 \theta)}{\tan \theta} \]
After simplification, the result is:
\[ \cot \theta - \frac{2}{z} = i \]

### Conclusion:
Thus, the final answer is:
\[ \cot \theta - \frac{2}{z} = i \]