Angle subtended by chord of a circle at the centre is twice the angle subtended at the circumference . If OP is rotated through an angle `phi` in anti - clockwise direction to become OQ then ` OQ = OP e^(iphi)`
If ` | z - 3| = 3 " then " ( z - 6)/( z)` is equal to

Options

A. i tan (arg z)
B. i cot (arg z)
C. cot (arg z)
D. tan [arg (z - 3)]

Answer

To solve the problem, we need to analyze the given equation and the conditions provided. We are given the equation \( | z - 3| = 3 \). This describes a circle in the complex plane centered at \( z = 3 \) with a radius of 3.

### Step-by-Step Solution:

1. **Understanding the Circle**:
The equation \( | z - 3| = 3 \) means that the complex number \( z \) lies on a circle centered at \( 3 + 0i \) (or simply \( 3 \) on the real axis) with a radius of \( 3 \).

2. **Parametrizing the Circle**:
We can express \( z \) in terms of a parameter \( \theta \):
\[ z = 3 + 3e^{i\theta} \]
where \( e^{i\theta} = \cos(\theta) + i\sin(\theta) \).

3. **Finding \( z - 6 \)**:
We need to compute \( z - 6 \):
\[ z - 6 = (3 + 3e^{i\theta}) - 6 = 3e^{i\theta} - 3 = 3(e^{i\theta} - 1) \]

4. **Finding \( \frac{z - 6}{z} \)**:
Now we compute \( \frac{z - 6}{z} \):
\[ \frac{z - 6}{z} = \frac{3(e^{i\theta} - 1)}{3 + 3e^{i\theta}} = \frac{e^{i\theta} - 1}{1 + e^{i\theta}} \]

5. **Simplifying the Expression**:
To simplify \( \frac{e^{i\theta} - 1}{1 + e^{i\theta}} \), we can multiply the numerator and the denominator by the conjugate of the denominator:
\[ \frac{(e^{i\theta} - 1)(1 + e^{-i\theta})}{(1 + e^{i\theta})(1 + e^{-i\theta})} \]
The denominator simplifies to:
\[ 1 + e^{i\theta} + e^{-i\theta} + 1 = 2 + 2\cos(\theta) = 2(1 + \cos(\theta)) \]
The numerator simplifies to:
\[ e^{i\theta} + 1 - e^{-i\theta} - 1 = e^{i\theta} - e^{-i\theta} = 2i\sin(\theta) \]
Thus, we have:
\[ \frac{2i\sin(\theta)}{2(1 + \cos(\theta))} = \frac{i\sin(\theta)}{1 + \cos(\theta)} \]

6. **Final Expression**:
Therefore, the final expression for \( \frac{z - 6}{z} \) is:
\[ \frac{z - 6}{z} = \frac{i\sin(\theta)}{1 + \cos(\theta)} \]