Question

If $ z = \sqrt{3} + i $, then the argument of $ z^2 e^{-i} $ is equal to

• A. \( \frac{e}{3} \)
• B. \( \frac{\pi}{3} \)
• C. \( \frac{\pi}{6} \)
• D. \( \frac{e}{6} \)

Hint:

When working with complex numbers, the argument of the product or quotient can be found by adding or subtracting the arguments of the individual terms.

Answer 1

The Correct Option is B

To solve for the argument of \( z^2 e^{-i} \), where \( z = \sqrt{3} + i \), follow these steps:

Step 1: Compute the argument of \( z \)

Express \( z \) in polar form. The modulus \(|z|\) is \(\sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{4} = 2\). The argument \( \theta \) is given by \(\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}\).

Step 2: Compute \( z^2 \)

Square the complex number \( z \):

\( z^2 = (\sqrt{3} + i)^2 = 3 + 2\sqrt{3}i - 1 = 2 + 2\sqrt{3}i \).

The modulus of \( z^2 \) is \(\sqrt{2^2 + (2\sqrt{3})^2} = \sqrt{4 + 12} = 4\).

The tangent \(\tan^{-1}\left(\frac{2\sqrt{3}}{2}\right) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}\).

Step 3: Compute \( z^2 e^{-i} \)

The argument of \( e^{-i} \) is \(-1\).

Add the arguments: \(\frac{\pi}{3} - 1\).

Step 4: Find the principal argument

Simplifying: \(\frac{\pi}{3} - 1 = \frac{\pi}{3}\) since its angle lies in the principal argument range between \(-\pi\) and \(\pi\).

The argument of \( z^2 e^{-i} \) is thus \(\boxed{\frac{\pi}{3}}\).

Answer 2

To solve the problem of finding the argument of \( z^2 e^{-i} \) where \( z = \sqrt{3} + i \), follow these steps:

Step 1: Determine the Argument of \( z \)
The number \( z = \sqrt{3} + i \) can be expressed in polar form. The argument \( \arg(z) \) is given by \( \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) \), which simplifies to \( \frac{\pi}{6} \).

Step 2: Compute \( z^2 \)
By squaring \( z \), we have \( z^2 = (\sqrt{3} + i)^2 = 3 + 2\sqrt{3}i - 1 = 2 + 2\sqrt{3}i \).
The argument \( \theta \) of \( z^2 \) is given by \( \tan^{-1}\left(\frac{2\sqrt{3}}{2}\right) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \).

Step 3: Apply Argument of \( e^{-i} \)
The argument of \( e^{-i} \) is \( -1 \cdot \arg(i) = -1 \cdot \frac{\pi}{2} = -\frac{\pi}{2} \).
Thus, the argument of \( z^2 e^{-i} = \arg(z^2) + \arg(e^{-i}) = \frac{\pi}{3} - \frac{\pi}{2} = \frac{\pi}{3} - \frac{3\pi}{6} + \frac{2\pi}{6} = \frac{\pi}{3} \).

The argument of \( z^2 e^{-i} \) simplifies to \( \frac{\pi}{3} \).