If `|z-1| =1` and arg`(z)=theta`, where `z ne 0` and `theta` is acute, then `(1-2/z)` is equal to

Options

A. `tantheta`
B. `I tantheta`
C. `tan theta/2`
D. `I tantheta/2`

Explanation

To solve the problem step by step, we will analyze the given conditions and derive the required expression.

### Given:
1. \( |z - 1| = 1 \)
2. \( \arg(z) = \theta \) (where \( \theta \) is acute)

### Step 1: Understand the geometric representation
The equation \( |z - 1| = 1 \) represents a circle in the complex plane with center at \( (1, 0) \) and radius \( 1 \). This means that any point \( z \) on this circle can be expressed as:
\[ z = 1 + e^{i\phi} \]
where \( \phi \) is the angle parameterizing the circle.

### Step 2: Express \( z \) in terms of \( \theta \)
Since \( \arg(z) = \theta \), we can express \( z \) in terms of \( \theta \):
\[ z = r e^{i\theta} \]
where \( r \) is the modulus of \( z \).

### Step 3: Find the relationship between \( r \) and \( \theta \)
From the circle equation, we know:
\[ |z - 1| = 1 \implies |r e^{i\theta} - 1| = 1 \]
This can be expanded as:
\[ \sqrt{(r \cos \theta - 1)^2 + (r \sin \theta)^2} = 1 \]
Squaring both sides gives:
\[ (r \cos \theta - 1)^2 + (r \sin \theta)^2 = 1 \]
Expanding this:
\[ (r^2 \cos^2 \theta - 2r \cos \theta + 1) + (r^2 \sin^2 \theta) = 1 \]
Combining terms:
\[ r^2 (\cos^2 \theta + \sin^2 \theta) - 2r \cos \theta + 1 = 1 \]
Using \( \cos^2 \theta + \sin^2 \theta = 1 \):
\[ r^2 - 2r \cos \theta = 0 \]
Factoring gives:
\[ r(r - 2 \cos \theta) = 0 \]
Since \( z \neq 0 \), we have:
\[ r = 2 \cos \theta \]

### Step 4: Substitute \( z \) back into the expression
Now we can express \( z \):
\[ z = 2 \cos \theta e^{i\theta} \]

### Step 5: Calculate \( \frac{2}{z} \)
We need to find \( \frac{2}{z} \):
\[ \frac{2}{z} = \frac{2}{2 \cos \theta e^{i\theta}} = \frac{1}{\cos \theta e^{i\theta}} = \frac{1}{\cos \theta} e^{-i\theta} \]

### Step 6: Calculate \( 1 - \frac{2}{z} \)
Now, substituting into the expression \( 1 - \frac{2}{z} \):
\[ 1 - \frac{2}{z} = 1 - \frac{1}{\cos \theta} e^{-i\theta} \]
To combine these, we need a common denominator:
\[ = \frac{\cos \theta - e^{-i\theta}}{\cos \theta} \]

### Step 7: Simplify the expression
Now, we can express \( e^{-i\theta} \) as \( \cos \theta - i \sin \theta \):
\[ = \frac{\cos \theta - (\cos \theta - i \sin \theta)}{\cos \theta} = \frac{i \sin \theta}{\cos \theta} = i \tan \theta \]

### Final Result
Thus, we find that:
\[ 1 - \frac{2}{z} = i \tan \theta \]