Question

The amplitude of the complex number is:
\[\frac{(\sqrt{3}+i)(1-\sqrt{3}i)}{(-1+i)(-1-i)}\]

• A. $\dfrac{\pi}{2}$
• B. $\dfrac{\pi}{3}$
• C. $\dfrac{5\pi}{12}$
• D. $-\dfrac{\pi}{6}$

Hint:

To find the amplitude (argument) of a complex number $a+ib$, use $\tan\theta = \frac{b}{a}$ and choose the angle based on the quadrant of $(a,b)$.

Answer 1

The Correct Option is D

We are given:
\[ z = \frac{(\sqrt{3}+i)(1-\sqrt{3}i)}{(-1+i)(-1-i)} \] First, simplify the denominator:
\[ (-1+i)(-1-i) = (-1)^2 - i^2 = 1 - (-1) = 2 \] So,
\[ z = \frac{(\sqrt{3}+i)(1-\sqrt{3}i)}{2} \] Now expand the numerator:
\[ (\sqrt{3}+i)(1-\sqrt{3}i) = \sqrt{3}(1-\sqrt{3}i) + i(1-\sqrt{3}i) \] \[ = \sqrt{3} - 3i + i - i^2\sqrt{3} \] Since $i^2 = -1$,
\[ = \sqrt{3} - 3i + i + \sqrt{3} = 2\sqrt{3} - 2i \] Hence,
\[ z = \frac{2\sqrt{3} - 2i}{2} = \sqrt{3} - i \] Now find the amplitude (argument) of \( z = \sqrt{3} - i \).
\[ \text{tan} \, \theta = \frac{-1}{\sqrt{3}} = -\frac{1}{\sqrt{3}} \] So,
\[ \theta = -\frac{\pi}{6} \] \[ \boxed{\text{Amplitude} = -\frac{\pi}{6}} \]