Question

If \( z = 1 + i \tan \theta \), where \( \pi < \theta < \dfrac{3\pi}{2} \), then \( |z| \) is equal to

• A. \( 1+\tan\theta \)
• B. \( 2\tan\theta \)
• C. \( \sec\theta \)
• D. \( -\sec\theta \)
• E. \( \cosec\theta \)

Hint:

After using identities like \( 1+\tan^2\theta=\sec^2\theta \), always check the quadrant before removing the square root. The sign depends on whether \( \sec\theta \) is positive or negative in that interval.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The question asks for the modulus, \(|z|\), of a complex number \(z\). The modulus of a complex number \(z = x + iy\) is its distance from the origin on the complex plane and is always a non-negative real number.
Step 2: Key Formula or Approach:
The modulus of \(z = x + iy\) is calculated using the formula:
\[ |z| = \sqrt{x^2 + y^2} \] We will also use the fundamental trigonometric identity: \(1 + \tan^2\theta = \sec^2\theta\).
Step 3: Detailed Explanation:
The given complex number is \(z = 1 + i \tan \theta\).
The real part is \(x = 1\) and the imaginary part is \(y = \tan \theta\).
Apply the modulus formula:
\[ |z| = \sqrt{(1)^2 + (\tan \theta)^2} = \sqrt{1 + \tan^2\theta} \] Using the trigonometric identity, we replace \(1 + \tan^2\theta\) with \(\sec^2\theta\):
\[ |z| = \sqrt{\sec^2\theta} \] The square root of a squared term is the absolute value of that term:
\[ |z| = |\sec \theta| \] Now we must evaluate \(|\sec \theta|\) based on the given interval for \(\theta\).
The interval is \(\pi<\theta<\frac{3\pi}{2}\), which places \(\theta\) in the third quadrant of the unit circle.
In the third quadrant, \(\cos \theta\) is negative.
Since \(\sec \theta = \frac{1}{\cos \theta}\), \(\sec \theta\) is also negative in the third quadrant.
The absolute value of a negative number is its negation. Therefore:
\[ |\sec \theta| = -\sec \theta \quad (\text{for } \theta \text{ in quadrant III}) \] Step 4: Final Answer:
The modulus \(|z|\) is equal to \(-\sec \theta\). Therefore, option (D) is correct.