Question

The argument of $\dfrac{{\left( {1 - i\sqrt 3 } \right)}}{{\left( {1 + i\sqrt 3 } \right)}}$ is
A.$60^\circ$
B.$120^\circ$
C.$210^\circ$
D.$240^\circ$

Answer 1

First we will first rationalize the given expression by multiplying numerator and denominator by $1 + i\sqrt 3$. Then use the property, ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$ in the denominator and ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$in the numerator of the obtained equation to simplify it. Then we will use the trigonometric values to find the argument.

Complete step-by-step answer:
We are given $\dfrac{{\left( {1 - i\sqrt 3 } \right)}}{{\left( {1 + i\sqrt 3 } \right)}}$.
Let us assume that $z = \dfrac{{\left( {1 - i\sqrt 3 } \right)}}{{\left( {1 + i\sqrt 3 } \right)}}$.
Rationalizing the given expression by multiplying numerator and denominator by $1 + i\sqrt 3$, we get

$$\Rightarrow z = \dfrac{{\left( {1 - i\sqrt 3 } \right)}}{{\left( {1 + i\sqrt 3 } \right)}} \times \dfrac{{\left( {1 - i\sqrt 3 } \right)}}{{\left( {1 - i\sqrt 3 } \right)}} \\\ \Rightarrow z = \dfrac{{{{\left( {1 - i\sqrt 3 } \right)}^2}}}{{\left( {1 + i\sqrt 3 } \right)\left( {1 - i\sqrt 3 } \right)}} \\\$$

Using the property, ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$ in the denominator and ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$in the numerator of the above equation, we get

$$\Rightarrow z = \dfrac{{{1^2} - 2 \times 1 \times i\sqrt 3 + {{\left( {i\sqrt 3 } \right)}^2}}}{{{1^2} - {{\left( {i\sqrt 3 } \right)}^2}}} \\\ \Rightarrow z = \dfrac{{1 - 2i\sqrt 3 + 3{i^2}}}{{1 - 3{i^2}}} \\\$$

Using the property of complex number ${i^2} = - 1$ in the above equation, we get

$$\Rightarrow z = \dfrac{{1 - 2i\sqrt 3 + 3\left( { - 1} \right)}}{{1 - 3\left( { - 1} \right)}} \\\ \Rightarrow z = \dfrac{{1 - 2i\sqrt 3 - 3}}{{1 + 3}} \\\ \Rightarrow z = \dfrac{{ - 2i\sqrt 3 - 2}}{4} \\\$$

Taking 2 common from the numerator in the above equation, we get

$$\Rightarrow z = \dfrac{{2\left( { - i\sqrt 3 - 1} \right)}}{4} \\\ \Rightarrow z = \dfrac{{ - i\sqrt 3 - 1}}{2} \\\ \Rightarrow z = - \dfrac{{i\sqrt 3 }}{2} - \dfrac{1}{2} \\\$$

We know that the standard equation for the complex number $z$ is $\cos \theta + i\sin \theta$.
Using the standard equation and the trigonometric values, $\sin 240^\circ = - \dfrac{{\sqrt 3 }}{2}$ and $\cos 240^\circ = - \dfrac{1}{2}$ in the above expression, we get
$\Rightarrow \cos 240^\circ + i\sin 240^\circ$
Thus, the argument of the given expression is $240^\circ$.
Hence, option D is correct.

Note: In solving these types of questions, students should know the basic properties and values of trigonometric functions. Since $- \dfrac{{\sqrt 3 }}{2}$ and $- \dfrac{1}{2}$ are both negative values, it lies in the fourth quadrant.