Question

The principal argument of the complex number $z = \frac{1 + \sin \frac{\pi}{3} + i \cos \frac{\pi}{3}}{1 + \sin \frac{\pi}{3} - i \cos \frac{\pi}{3}}$ is:

• A. $\frac{\pi}{3}$
• B. $\frac{\pi}{6}$
• C. $\frac{2\pi}{3}$
• D. $\frac{\pi}{2}$
• E. $\frac{\pi}{4}$

Hint:

Any complex number of the form $\frac{1 + \bar{w}}{1 + w}$ where $|w|=1$ will have an argument related to the argument of $w$. Specifically, if $w = \cos \alpha - i \sin \alpha$, the result simplifies beautifully into a single exponential term.

Answer 1

The Correct Option is B

Concept: For a complex number $z = \frac{1 + \sin \theta + i \cos \theta}{1 + \sin \theta - i \cos \theta}$, we can simplify the expression using trigonometric identities to find the argument. A useful identity is $1 + \sin \theta = 1 + \cos(\pi/2 - \theta)$ and $\cos \theta = \sin(\pi/2 - \theta)$.

Step 1: Convert to standard form using $\alpha = \pi/2 - \pi/3 = \pi/6$.
Let $\theta = \pi/3$. Then $\sin \theta = \cos(\pi/6)$ and $\cos \theta = \sin(\pi/6)$.
The expression becomes: \[ z = \frac{1 + \cos \frac{\pi}{6} + i \sin \frac{\pi}{6}}{1 + \cos \frac{\pi}{6} - i \sin \frac{\pi}{6}} \]

Step 2: Apply half-angle formulas.
Recall $1 + \cos 2\phi = 2 \cos^2 \phi$ and $\sin 2\phi = 2 \sin \phi \cos \phi$. Let $2\phi = \pi/6$, so $\phi = \pi/12$. \[ z = \frac{2 \cos^2 \frac{\pi}{12} + i (2 \sin \frac{\pi}{12} \cos \frac{\pi}{12})}{2 \cos^2 \frac{\pi}{12} - i (2 \sin \frac{\pi}{12} \cos \frac{\pi}{12})} \] Factor out $2 \cos \frac{\pi}{12}$: \[ z = \frac{\cos \frac{\pi}{12} + i \sin \frac{\pi}{12}}{\cos \frac{\pi}{12} - i \sin \frac{\pi}{12}} = \frac{e^{i\pi/12}}{e^{-i\pi/12}} = e^{i(\pi/12 + \pi/12)} = e^{i\pi/6} \]

Step 3: Identify the principal argument.
The resulting form is $e^{i\theta}$, where $\theta = \pi/6$. The principal argument is $\pi/6$.