Question

The principal argument of the complex number $\dfrac{{{\left( 1+i \right)}^{5}}{{\left( 1+\sqrt{3i} \right)}^{2}}}{-2i\left( -\sqrt{3}+i \right)}$ is
(a) $\dfrac{19\pi }{12}$
(b) $-\dfrac{7\pi }{12}$
(c) $-\dfrac{5\pi }{12}$
(d) $\dfrac{5\pi }{12}$

Answer 1

Hint: In this question, we will use properties of the argument of complex numbers to simplify the expression. Then apply the argument formula to find the principal argument of the given expression.

Complete step by step answer:

We know that, argument of a complex number $z=a+ib$ is given by,
$\arg z={{\tan }^{-1}}\dfrac{b}{a}\cdots \cdots \left( i \right)$,
Where $\arg z$ is the argument of $z$.
In the given question, we will also use some properties of argument of complex numbers which are given as below,
For two complex numbers ${{z}_{1}}$ and ${{z}_{2}}$, properties are,
$\arg {{z}_{1}}^{n}=n\arg {{z}_{1}}\cdots \cdots \left( ii \right)$,
$\arg \left( {{z}_{1}}\times {{z}_{2}} \right)=\arg {{z}_{1}}+\arg {{z}_{2}}\cdots \cdots \left( iii \right)$,
$\arg \left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)=\arg {{z}_{1}}-\arg {{z}_{2}}\cdots \cdots \left( iv \right)$.
Now, the given expression in the question is,
$\dfrac{{{\left( 1+i \right)}^{5}}{{\left( 1+\sqrt{3i} \right)}^{2}}}{-2i\left( -\sqrt{3}+i \right)}$.
So, we can write its argument as,
$\arg \left( \dfrac{{{\left( 1+i \right)}^{5}}{{\left( 1+\sqrt{3i} \right)}^{2}}}{-2i\left( -\sqrt{3}+i \right)} \right)$
Here, applying property $\left( iv \right)$, we get,
$=\arg \left( {{\left( 1+i \right)}^{5}}{{\left( 1+\sqrt{3i} \right)}^{2}} \right)-\arg \left( -2i\left( -\sqrt{3}+i \right) \right)$
Now, applying property $\left( iii \right)$ in both the argument terms, we get,
$\begin{aligned} & =\arg {{\left( 1+i \right)}^{5}}+\arg {{\left( 1+\sqrt{3i} \right)}^{2}}-\left[ \arg \left( -2i \right)+\arg \left( -\sqrt{3}+i \right) \right] \\\ & =\arg {{\left( 1+i \right)}^{5}}+\arg {{\left( 1+\sqrt{3i} \right)}^{2}}-\arg \left( -2i \right)-\arg \left( -\sqrt{3}+i \right)\cdots \cdots \left( v \right) \\\ \end{aligned}$
Now, applying property $\left( ii \right)$ in first two terms of above expression, we get,
$=5\arg \left( 1+i \right)+2\arg \left( 1+\sqrt{3i} \right)-\arg \left( -2i \right)-\arg \left( -\sqrt{3}+i \right)$
Now, using the equation $\left( i \right)$ in $\arg \left( 1+i \right)$, we get, $a=1$ and $b=1$. Therefore,
$\arg \left( 1+i \right)={{\tan }^{-1}}\left( \dfrac{1}{1} \right)={{\tan }^{-1}}1=\dfrac{\pi }{4}$
Now, using equation $\left( i \right)$ in $\arg \left( 1+\sqrt{3i} \right)$ , we get, $a=1$ and $b=\sqrt{3}$. Therefore,
$\arg \left( 1+\sqrt{3i} \right)={{\tan }^{-1}}\left( \dfrac{\sqrt{3}}{1} \right)={{\tan }^{-1}}\sqrt{3}=\dfrac{\pi }{3}$
Now, using the equation $\left( i \right)$ in $\arg \left( -2i \right)$, we get, $a=0$ and $b=-2$. Therefore,
$\arg \left( -2i \right)={{\tan }^{-1}}\left( \dfrac{-2}{0} \right)={{\tan }^{-1}}\left( -\infty \right)=-\dfrac{\pi }{2}$
Now, using equation $\left( i \right)$ in$\arg \left( -\sqrt{3}+i \right)$ , we get, $a=-\sqrt{3}$ and $b=1$. Therefore,
$\arg \left( -\sqrt{3}+i \right)={{\tan }^{-1}}\left( \dfrac{1}{-\sqrt{3}} \right)={{\tan }^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right)=\pi -\dfrac{\pi }{6}=\dfrac{5\pi }{6}$
Using the above calculated values in equation $\left( v \right)$, we get,
$\arg \left( \dfrac{{{\left( 1+i \right)}^{5}}{{\left( 1+\sqrt{3i} \right)}^{2}}}{-2i\left( -\sqrt{3}+i \right)} \right)=5\left( \dfrac{\pi }{4} \right)+2\left( \dfrac{\pi }{3} \right)-\left( -\dfrac{\pi }{2} \right)-\left( \dfrac{5\pi }{6} \right)$
$=\dfrac{5\pi }{4}+\dfrac{2\pi }{3}+\dfrac{\pi }{2}-\dfrac{5\pi }{6}$
Taking LCM of all the terms to add them, we get,
$\begin{aligned} & =\dfrac{5\pi \times 3}{4\times 3}+\dfrac{2\pi \times 4}{3\times 4}+\dfrac{\pi \times 6}{2\times 6}-\dfrac{5\pi \times 2}{6\times 2} \\\ & =\dfrac{15\pi }{12}+\dfrac{8\pi }{12}+\dfrac{6\pi }{12}-\dfrac{10\pi }{12} \\\ & =\dfrac{15\pi +8\pi +6\pi -10\pi }{12} \\\ & =\dfrac{19\pi }{12} \\\ \end{aligned}$
But, the principal argument lies in the interval $\left( -\pi ,\pi \right)$ and tangent function is a periodic function with period $2\pi$.
Also, we can write $\dfrac{19\pi }{12}$ as, $\dfrac{24\pi -5\pi }{12}=2\pi -\dfrac{5\pi }{12}$.
Therefore, principal argument will be,
$\dfrac{5\pi }{12}$.

Hence, the required principal argument is $\dfrac{5\pi }{12}$.

Note: You can do this question in an alternative way by first solving the whole expression to bring it to the form of $a+ib$, and then use the argument formula. But that way calculation will become very long.