If z = 1 -cos `theta+ isin theta ` then |z| is equal to

Options

A. `2 sin"" (theta)/(2)`
B. `2 cos ""(theta)/(2)`
C. `2| sin""(theta)/(2)|`
D. `2 |cos ""(theta)/(2)|`

Answer

To find the modulus of the complex number \( z = 1 - \cos \theta + i \sin \theta \), we will follow these steps:

### Step 1: Write down the expression for \( z \)
Given:
\[ z = 1 - \cos \theta + i \sin \theta \]

### Step 2: Use the formula for the modulus of a complex number
The modulus of a complex number \( z = a + bi \) is given by:
\[ |z| = \sqrt{a^2 + b^2} \]
In our case, \( a = 1 - \cos \theta \) and \( b = \sin \theta \).

### Step 3: Substitute \( a \) and \( b \) into the modulus formula
Now, we can calculate the modulus:
\[ |z| = \sqrt{(1 - \cos \theta)^2 + (\sin \theta)^2} \]

### Step 4: Expand the expression
Expanding \( (1 - \cos \theta)^2 \):
\[ (1 - \cos \theta)^2 = 1 - 2\cos \theta + \cos^2 \theta \]
Thus,
\[ |z| = \sqrt{1 - 2\cos \theta + \cos^2 \theta + \sin^2 \theta} \]

### Step 5: Use the Pythagorean identity
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ |z| = \sqrt{1 - 2\cos \theta + 1} = \sqrt{2 - 2\cos \theta} \]

### Step 6: Factor out the common term
We can factor out the 2:
\[ |z| = \sqrt{2(1 - \cos \theta)} \]

### Step 7: Use the double angle formula
Recall that \( 1 - \cos \theta = 2\sin^2\left(\frac{\theta}{2}\right) \):
\[ |z| = \sqrt{2 \cdot 2\sin^2\left(\frac{\theta}{2}\right)} = \sqrt{4\sin^2\left(\frac{\theta}{2}\right)} \]

### Step 8: Simplify the expression
Taking the square root:
\[ |z| = 2|\sin\left(\frac{\theta}{2}\right)| \]

### Final Answer
Thus, the modulus of \( z \) is:
\[ |z| = 2|\sin\left(\frac{\theta}{2}\right)| \]

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