Question

The argument of

$\frac{1+i\sqrt{3}}{\sqrt{3}+i}$

where $i = \sqrt{-1}$, is

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

Answer 1

Answer: (B)

$\frac{\pi}{6}$

Answer 2

To find the argument of the complex number $Z = \frac{1+i\sqrt{3}}{\sqrt{3}+i}$, we can use the property of arguments: $\text{arg}\left(\frac{z_1}{z_2}\right) = \text{arg}(z_1) - \text{arg}(z_2)$.

Let $z_1 = 1+i\sqrt{3}$ and $z_2 = \sqrt{3}+i$.

1. Find the argument of $z_1 = 1+i\sqrt{3}$:

   The complex number $z_1$ is in the first quadrant (real part is positive, imaginary part is positive). Let $\theta_1 = \text{arg}(z_1)$. $\tan \theta_1 = \frac{\text{Imaginary part}}{\text{Real part}} = \frac{\sqrt{3}}{1} = \sqrt{3}$. Since $0 < \theta_1 < \frac{\pi}{2}$, we have $\theta_1 = \frac{\pi}{3}$.

2. Find the argument of $z_2 = \sqrt{3}+i$:

   The complex number $z_2$ is also in the first quadrant. Let $\theta_2 = \text{arg}(z_2)$. $\tan \theta_2 = \frac{\text{Imaginary part}}{\text{Real part}} = \frac{1}{\sqrt{3}}$. Since $0 < \theta_2 < \frac{\pi}{2}$, we have $\theta_2 = \frac{\pi}{6}$.

3. Calculate the argument of $Z = \frac{z_1}{z_2}$:

   $\text{arg}(Z) = \text{arg}(z_1) - \text{arg}(z_2)$ $\text{arg}(Z) = \frac{\pi}{3} - \frac{\pi}{6}$ To subtract, find a common denominator: $\text{arg}(Z) = \frac{2\pi}{6} - \frac{\pi}{6} = \frac{2\pi - \pi}{6} = \frac{\pi}{6}$.

Alternatively, we can first simplify the complex number by multiplying the numerator and denominator by the conjugate of the denominator: $Z = \frac{1+i\sqrt{3}}{\sqrt{3}+i} \times \frac{\sqrt{3}-i}{\sqrt{3}-i}$ $Z = \frac{(1)(\sqrt{3}) + (1)(-i) + (i\sqrt{3})(\sqrt{3}) + (i\sqrt{3})(-i)}{(\sqrt{3})^2 - (i)^2}$ $Z = \frac{\sqrt{3} - i + 3i - i^2\sqrt{3}}{3 - (-1)}$ $Z = \frac{\sqrt{3} + 2i + \sqrt{3}}{4}$ $Z = \frac{2\sqrt{3} + 2i}{4}$ $Z = \frac{\sqrt{3}}{2} + i\frac{1}{2}$

Now, find the argument of $Z = \frac{\sqrt{3}}{2} + i\frac{1}{2}$: Let $\phi = \text{arg}(Z)$. Since $Z$ is in the first quadrant. $\tan \phi = \frac{\text{Imaginary part}}{\text{Real part}} = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}}$. Therefore, $\phi = \frac{\pi}{6}$.