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: Write the complex number in standard form.
We are given \[ z=1+i\tan\theta \] A complex number in the form \[ z=a+ib \] has modulus \[ |z|=\sqrt{a^2+b^2} \] Here, \[ a=1 \quad \text{and} \quad b=\tan\theta \]

Step 2: Apply the modulus formula.
Using the modulus formula, \[ |z|=\sqrt{1^2+(\tan\theta)^2} \] \[ |z|=\sqrt{1+\tan^2\theta} \]

Step 3: Use the standard trigonometric identity.
We know the identity \[ 1+\tan^2\theta=\sec^2\theta \] Therefore, \[ |z|=\sqrt{\sec^2\theta} \] \[ |z|=|\sec\theta| \] This absolute value is important because modulus is always non-negative.

Step 4: Use the given interval for \( \theta \).
It is given that \[ \pi<\theta<\frac{3\pi}{2} \] This means \( \theta \) lies in the third quadrant.
In the third quadrant: \[ \sin\theta0 \] Since \[ \sec\theta=\frac{1}{\cos\theta} \] and \( \cos\theta<0 \), we get \[ \sec\theta<0 \]

Step 5: Simplify the absolute value.
Because \( \sec\theta \) is negative in the third quadrant, \[ |\sec\theta|=-\sec\theta \] Hence, \[ |z|=-\sec\theta \]

Step 6: Verify that the result is positive.
Since modulus must always be non-negative, let us check: \[ -\sec\theta > 0 \] because \( \sec\theta<0 \) in the third quadrant.
So the expression \( -\sec\theta \) is perfectly consistent with the value of a modulus.

Step 7: Final conclusion.
Thus, \[ |z|=\boxed{-\sec\theta} \] Therefore, the correct option is \[ \boxed{(4)\ -\sec\theta} \]