Question

For any integer $n$, $\arg \left( {\dfrac{{{{\left( {\sqrt 3 + i} \right)}^{4n + 1}}}}{{{{\left( {1 - i\sqrt 3 } \right)}^{4n}}}}} \right)$ equals:
$\left( 1 \right)\dfrac{\pi }{3}$
$\left( 2 \right)\dfrac{\pi }{6}$
$\left( 3 \right)\dfrac{{2\pi }}{3}$
$\left( 4 \right)\dfrac{{5\pi }}{6}$

Answer 1

In order to solve this question, first of all we will simplify the complex number $\left( {\dfrac{{{{\left( {\sqrt 3 + i} \right)}^{4n + 1}}}}{{{{\left( {1 - i\sqrt 3 } \right)}^{4n}}}}} \right)$ into its simplest form that will be as $a + bi$. Then, we will calculate the value of $a$ and $b$. After that, we will use the formula ${\tan ^{ - 1}}\left( {\dfrac{b}{a}} \right)$ to get the answer to the question.

Complete step-by-step solution:
Since, the given expression is $\left( {\dfrac{{{{\left( {\sqrt 3 + i} \right)}^{4n + 1}}}}{{{{\left( {1 - i\sqrt 3 } \right)}^{4n}}}}} \right)$.
Now, we will simplify this equation. We will take term $2$ common from $\left( {\sqrt 3 + i} \right)$ and $\left( {1 - i\sqrt 3 } \right)$ as:
$\Rightarrow \left( {\dfrac{{{{\left( {\sqrt 3 + i} \right)}^{4n + 1}}}}{{{{\left( {1 - i\sqrt 3 } \right)}^{4n}}}}} \right) = \left( {\dfrac{{{2^{4n + 1}}{{\left( {\dfrac{{\sqrt 3 }}{2} + \dfrac{1}{2}i} \right)}^{4n + 1}}}}{{{2^{4n}}{{\left( {\dfrac{1}{2} - i\dfrac{{\sqrt 3 }}{2}} \right)}^{4n}}}}} \right)$
Then, we will convert the complex number in the form of $\left( {\cos \theta + i\sin \theta } \right)$ and $\left( {\cos \theta - i\sin \theta } \right)$ as,
$\Rightarrow \left( {\dfrac{{{{\left( {\sqrt 3 + i} \right)}^{4n + 1}}}}{{{{\left( {1 - i\sqrt 3 } \right)}^{4n}}}}} \right) = \left( {\dfrac{{{2^{4n + 1}}{{\left( {\cos \dfrac{\pi }{6} + i\sin \dfrac{\pi }{6}} \right)}^{4n + 1}}}}{{{2^{4n}}{{\left( {\cos \dfrac{\pi }{3} - i\sin \dfrac{\pi }{3}} \right)}^{4n}}}}} \right)$
As, we know that in a complex number $\left( {\cos \theta + i\sin \theta } \right)$ is denoted as ${e^{i\theta }}$ and $\left( {\cos \theta - i\sin \theta } \right)$ is denoted as ${e^{ - i\theta }}$. So, we will use this condition to simplify the above expression as,
$\Rightarrow \left( {\dfrac{{{{\left( {\sqrt 3 + i} \right)}^{4n + 1}}}}{{{{\left( {1 - i\sqrt 3 } \right)}^{4n}}}}} \right) = \left( {\dfrac{{{2^{4n + 1}}{e^{i\dfrac{\pi }{6}}}^{\left( {4n + 1} \right)}}}{{{2^{4n}}{e^{ - i\dfrac{\pi }{3}4n}}}}} \right)$
Here, we will use the rule of division of powers as:
$\Rightarrow \left( {\dfrac{{{{\left( {\sqrt 3 + i} \right)}^{4n + 1}}}}{{{{\left( {1 - i\sqrt 3 } \right)}^{4n}}}}} \right) = {2^{4n + 1 - 4n}}{e^{i\dfrac{\pi }{6}\left( {4n + 1} \right) - \left( { - 4ni\dfrac{\pi }{3}} \right)}}$
Now, we will cancel out the equal like term and will open the bracket to proceed further.
$\Rightarrow \left( {\dfrac{{{{\left( {\sqrt 3 + i} \right)}^{4n + 1}}}}{{{{\left( {1 - i\sqrt 3 } \right)}^{4n}}}}} \right) = {2^1}{e^{4ni\dfrac{\pi }{6} + i\dfrac{\pi }{6} + 4ni\dfrac{\pi }{3}}}$
After that, we will use the rule of subtraction of fraction to simplify the above expression.

\Rightarrow \left( {\dfrac{{{{\left( {\sqrt 3 + i} \right)}^{4n + 1}}}}{{{{\left( {1 - i\sqrt 3 } \right)}^{4n}}}}} \right) = {2^1}{e^{4ni\left( {\dfrac{\pi }{6} + \dfrac{\pi }{3}} \right) + i\dfrac{\pi }{6}}} \\\ \Rightarrow \left( {\dfrac{{{{\left( {\sqrt 3 + i} \right)}^{4n + 1}}}}{{{{\left( {1 - i\sqrt 3 } \right)}^{4n}}}}} \right) = {2^1}{e^{4ni\left( {\dfrac{{\pi + 2\pi }}{6}} \right) + i\dfrac{\pi }{6}}} \\\ \end{aligned} $$ Now, we will use the addition to simplify the above step as: $$ \Rightarrow \left( {\dfrac{{{{\left( {\sqrt 3 + i} \right)}^{4n + 1}}}}{{{{\left( {1 - i\sqrt 3 } \right)}^{4n}}}}} \right) = 2{e^{4ni \times \dfrac{{3\pi }}{6} + i\dfrac{\pi }{6}}}$$ Here, we will find the product for the above step as: $$ \Rightarrow \left( {\dfrac{{{{\left( {\sqrt 3 + i} \right)}^{4n + 1}}}}{{{{\left( {1 - i\sqrt 3 } \right)}^{4n}}}}} \right) = 2{e^{12ni\dfrac{\pi }{6} + i\dfrac{\pi }{6}}}$$ Then, we will use division to proceed further as: $$ \Rightarrow \left( {\dfrac{{{{\left( {\sqrt 3 + i} \right)}^{4n + 1}}}}{{{{\left( {1 - i\sqrt 3 } \right)}^{4n}}}}} \right) = 2{e^{2ni\pi + i\dfrac{\pi }{6}}}$$ We can take term $$i$$ from the above step and can write the above term as: $$ \Rightarrow \left( {\dfrac{{{{\left( {\sqrt 3 + i} \right)}^{4n + 1}}}}{{{{\left( {1 - i\sqrt 3 } \right)}^{4n}}}}} \right) = 2{e^{i\left( {2n\pi + \dfrac{\pi }{6}} \right)}}$$ Now, we will write the above expression as $\left( {\cos \theta + i\sin \theta } \right)$. $$ \Rightarrow \left( {\dfrac{{{{\left( {\sqrt 3 + i} \right)}^{4n + 1}}}}{{{{\left( {1 - i\sqrt 3 } \right)}^{4n}}}}} \right) = 2\left( {\cos \left( {2n\pi + \dfrac{\pi }{6}} \right) + i\sin \left( {2n\pi + \dfrac{\pi }{6}} \right)} \right)$$ Here, we can write the above expression as: $$ \Rightarrow \left( {\dfrac{{{{\left( {\sqrt 3 + i} \right)}^{4n + 1}}}}{{{{\left( {1 - i\sqrt 3 } \right)}^{4n}}}}} \right) = 2\left( {\cos \dfrac{\pi }{6} + i\sin \dfrac{\pi }{6}} \right)$$ After that, we will substitute the corresponding value cosine and sine to simplify it as: $$ \Rightarrow \left( {\dfrac{{{{\left( {\sqrt 3 + i} \right)}^{4n + 1}}}}{{{{\left( {1 - i\sqrt 3 } \right)}^{4n}}}}} \right) = 2\left( {\dfrac{{\sqrt 3 }}{2} + i\dfrac{1}{2}} \right)$$ Then, open the bracket and cancel out the equal like term to simplify the obtained expression. $$ \Rightarrow \left( {\dfrac{{{{\left( {\sqrt 3 + i} \right)}^{4n + 1}}}}{{{{\left( {1 - i\sqrt 3 } \right)}^{4n}}}}} \right) = \sqrt 3 + i$$ Now, we will compare the obtained equation with $$a + bi$$ and will find the value of $a$ and $b$. $ \Rightarrow a + bi = \sqrt 3 + i $ $ \Rightarrow a = \sqrt 3 $ $ \Rightarrow b = 1 $ Here, we will use the formula of argument of complex number ${\tan ^{ - 1}}\left( {\dfrac{b}{a}} \right)$ to solve $\arg \left( {\dfrac{{{{\left( {\sqrt 3 + i} \right)}^{4n + 1}}}}{{{{\left( {1 - i\sqrt 3 } \right)}^{4n}}}}} \right)$. $ \Rightarrow \arg \left( {\dfrac{{{{\left( {\sqrt 3 + i} \right)}^{4n + 1}}}}{{{{\left( {1 - i\sqrt 3 } \right)}^{4n}}}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{1}{{\sqrt 3 }}} \right)$ As we know, $\tan \dfrac{\pi }{6} = \dfrac{1}{{\sqrt 3 }}$ . We will use this in the above expression to solve it. $\Rightarrow \arg \left( {\dfrac{{{{\left( {\sqrt 3 + i} \right)}^{4n + 1}}}}{{{{\left( {1 - i\sqrt 3 } \right)}^{4n}}}}} \right) = {\tan ^{ - 1}}\left( {\tan \dfrac{\pi }{6}} \right) $ $\Rightarrow \arg \left( {\dfrac{{{{\left( {\sqrt 3 + i} \right)}^{4n + 1}}}}{{{{\left( {1 - i\sqrt 3 } \right)}^{4n}}}}} \right) = \dfrac{\pi }{6} $ **Hence, option $2$ is the right answer.** **Note:** Argument of a complex number is angle use in the complex number. If a complex number is $\left( {\cos \theta + i\sin \theta } \right)$, the argument of the function is $\theta $ . Now, if the complex number is represented as $$a + bi$$, the argument of the complex number can be calculated by using the formula ${\tan ^{ - 1}}\left( {\dfrac{b}{a}} \right)$.