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To solve the problem of finding the modulus of the complex number \((1 - \cos \theta + i \sin \theta)^{-2}\), let's break it down step by step.

### Step 1: Identify the complex number
The given complex number is:
\[
z = 1 - \cos \theta + i \sin \theta
\]

### Step 2: Calculate the modulus of \(z\)
The modulus of a complex number \(z = a + bi\) is given by:
\[
|z| = \sqrt{a^2 + b^2}
\]
For \(z = 1 - \cos \theta + i \sin \theta\), we have:
- Real part \(a = 1 - \cos \theta\)
- Imaginary part \(b = \sin \theta\)

Thus, the modulus of \(z\) is:
\[
|z| = \sqrt{(1 - \cos \theta)^2 + (\sin \theta)^2}
\]

### Step 3: Simplify the modulus expression
Expanding the expression under the square root:
\[
|z| = \sqrt{(1 - \cos \theta)^2 + \sin^2 \theta}
\]
\[
|z| = \sqrt{1 - 2\cos \theta + \cos^2 \theta + \sin^2 \theta}
\]
Using the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\):
\[
|z| = \sqrt{1 - 2\cos \theta + 1} = \sqrt{2(1 - \cos \theta)}
\]

### Step 4: Use the trigonometric identity
Recall that \(1 - \cos \theta = 2\sin^2 \left(\frac{\theta}{2}\right)\):
\[
|z| = \sqrt{2 \times 2\sin^2 \left(\frac{\theta}{2}\right)} = \sqrt{4\sin^2 \left(\frac{\theta}{2}\right)} = 2\sin \left(\frac{\theta}{2}\right)
\]

### Step 5: Calculate the modulus of the inverse squared
The modulus of the inverse squared is:
\[
\left| z^{-2} \right| = \frac{1}{|z|^2}
\]
So:
\[
\left| (1 - \cos \theta + i \sin \theta)^{-2} \right| = \frac{1}{(2\sin \left(\frac{\theta}{2}\right))^2}
\]
\[
= \frac{1}{4\sin^2 \left(\frac{\theta}{2}\right)}
\]

### Final Answer
The modulus of the complex number \((1 - \cos \theta + i \sin \theta)^{-2}\) is:
\[
\boxed{\frac{1}{4\sin^2 \left(\frac{\theta}{2}\right)}}
\]

Would you like to go into more detail on any part of this solution? Here are some related questions to consider:

1. What is the significance of the modulus in complex numbers?
2. How can we interpret the result geometrically on the complex plane?
3. How would the modulus change if the exponent were different?
4. Can you provide a real-world application where such a complex number modulus might be relevant?
5. How does the Pythagorean identity help simplify the modulus of complex numbers?

**Tip:** When simplifying trigonometric expressions, always consider using identities like the Pythagorean identity and the half-angle formulas to make the expressions more manageable.

The modulus of the complex number \((1 - \cos \theta + i \sin \theta)^{-2}\) is

Learn how to calculate the modulus of the complex number (1 - cos θ + i sin θ)^{-2} with a detailed, step-by-step solution. This problem involves complex numbers, trigonometry, and important identities like the Pythagorean theorem and half-angle identity. Suitable for Grades 11-12.

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