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`

Answer

To find the modulus of the complex number \( z = 1 + i \tan \alpha \), we follow these steps:

### Step 1: Identify the real and imaginary parts
The complex number \( z \) can be expressed in the form \( z = a + ib \), where:
- \( a = 1 \) (the real part)
- \( b = \tan \alpha \) (the imaginary part)

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

### Step 3: Simplify the expression
Calculating the squares:
\[ |z| = \sqrt{1 + \tan^2 \alpha} \]
Using the trigonometric identity \( 1 + \tan^2 \alpha = \sec^2 \alpha \):
\[ |z| = \sqrt{\sec^2 \alpha} \]

### Step 4: Final result
Since the square root of \( \sec^2 \alpha \) is \( |\sec \alpha| \) and given that \( \alpha \) is in the range \( \pi < \alpha < \frac{3\pi}{2} \), where \( \sec \alpha \) is negative, we have:
\[ |z| = -\sec \alpha \]

However, since modulus is always positive, we take the positive value:
\[ |z| = \sec \alpha \]

### Conclusion
Thus, the modulus of \( z \) is:
\[ |z| = \sec \alpha \]