If `pi lt theta lt 2pi and z=1+costheta+isintheta`, then find the value of |z|.

Answer

To find the value of \(|z|\) where \(z = 1 + \cos\theta + i\sin\theta\) and \(\pi < \theta < 2\pi\), we will follow these steps:

### Step 1: Identify the components of \(z\)
We have:
\[ z = 1 + \cos\theta + i\sin\theta \]
Here, the real part \(a = 1 + \cos\theta\) and the imaginary part \(b = \sin\theta\).

### Step 2: Calculate the modulus of \(z\)
The modulus of a complex number \(a + ib\) is given by:
\[ |z| = \sqrt{a^2 + b^2} \]
Substituting the values of \(a\) and \(b\):
\[ |z| = \sqrt{(1 + \cos\theta)^2 + (\sin\theta)^2} \]

### Step 3: Expand the expression
Now, we will expand the expression:
\[ |z| = \sqrt{(1 + \cos\theta)^2 + \sin^2\theta} \]
Expanding \((1 + \cos\theta)^2\):
\[ = \sqrt{1 + 2\cos\theta + \cos^2\theta + \sin^2\theta} \]

### Step 4: 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 5: Factor out the common term
We can factor out the 2:
\[ |z| = \sqrt{2(1 + \cos\theta)} = \sqrt{2} \sqrt{1 + \cos\theta} \]

### Step 6: Use the double angle formula
Using the double angle formula for cosine, where \(1 + \cos\theta = 2\cos^2\left(\frac{\theta}{2}\right)\):
\[ |z| = \sqrt{2} \sqrt{2\cos^2\left(\frac{\theta}{2}\right)} = 2|\cos\left(\frac{\theta}{2}\right)| \]

### Step 7: Determine the sign of \(\cos\left(\frac{\theta}{2}\right)\)
Since \(\pi < \theta < 2\pi\), it follows that \(\frac{\pi}{2} < \frac{\theta}{2} < \pi\). In this range, \(\cos\left(\frac{\theta}{2}\right)\) is negative. Therefore, we take the modulus:
\[ |z| = 2|\cos\left(\frac{\theta}{2}\right)| \]

### Final Answer
Thus, the value of \(|z|\) is:
\[ |z| = 2(-\cos\left(\frac{\theta}{2}\right)) = -2\cos\left(\frac{\theta}{2}\right) \quad \text{(since it's negative)} \]