If `z=1+i tan alpha where pi lt alpha lt(3pi)/(2),` then what is `|z|` equal to ?

Options

A. `sec alpha`
B. `-sec alpha`
C. `sec^(2)alpha`
D. `-sec^(2) alpha`

Explanation

To find the modulus of the complex number \( z = 1 + i \tan \alpha \), where \( \pi < \alpha < \frac{3\pi}{2} \), we can follow these steps:

### Step 1: Identify the Real and Imaginary Parts
The complex number can be expressed as:
- Real part: \( \text{Re}(z) = 1 \)
- Imaginary part: \( \text{Im}(z) = \tan \alpha \)

### Step 2: Use the Modulus Formula
The modulus of a complex number \( z = a + bi \) is given by:
\[ |z| = \sqrt{a^2 + b^2} \]
In our case:
\[ |z| = \sqrt{(1)^2 + (\tan \alpha)^2} \]

### Step 3: Simplify the Expression
Now, we can simplify the expression:
\[ |z| = \sqrt{1 + \tan^2 \alpha} \]
Using the trigonometric identity \( 1 + \tan^2 \alpha = \sec^2 \alpha \), we have:
\[ |z| = \sqrt{\sec^2 \alpha} \]
This simplifies to:
\[ |z| = |\sec \alpha| \]

### Step 4: Determine the Sign of \(\sec \alpha\)
Since \( \alpha \) lies in the interval \( \pi < \alpha < \frac{3\pi}{2} \), it is in the third quadrant. In the third quadrant, the cosine function is negative, which means that:
\[ \sec \alpha = \frac{1}{\cos \alpha} < 0 \]
Thus, we have:
\[ |z| = -\sec \alpha \]

### Final Answer
Therefore, the modulus \( |z| \) is:
\[ |z| = -\sec \alpha \]

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