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

Hint: 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.