Question

The modulus and amplitude of ${\left( {1 + i\sqrt 3 } \right)^8}$ are:
(A) $256$ and $\left( {\dfrac{\pi }{3}} \right)$
(B) $256$ and $\left( {\dfrac{{2\pi }}{3}} \right)$
(C) $2$ and $\left( {\dfrac{{2\pi }}{3}} \right)$
(D) $256$ and $\left( {\dfrac{{8\pi }}{3}} \right)$

Answer 1

In the given problem, we need to evaluate the modulus and argument of the given complex number. The given question requires knowledge of the concepts of complex numbers and its evaluation of its different parameters. We must know the formula to calculate the modulus of a complex number and the procedure to calculate the argument of the same.

Complete step by step answer:
In the question, we need to evaluate the modulus and amplitude of ${\left( {1 + i\sqrt 3 } \right)^8}$.
So, let us consider ${Z_1} = \left( {1 + i\sqrt 3 } \right)$.
The absolute value of a complex number is given by $\left| Z \right|$ and it is calculated as: $\left| Z \right| = \sqrt {{x^2} + {y^2}}$.
Thus, putting in the values of x and y, we get the absolute value of given complex number as:
$\left| {{Z_1}} \right| = \sqrt {{{(1)}^2} + {{(\sqrt 3 )}^2}} = \sqrt {1 + 3} = \sqrt 4 = 2$
So, the modulus of the complex number ${Z_1} = \left( {1 + i\sqrt 3 } \right)$ is $\left( 2 \right)$.
Also, ${Z_2} = \left( {1 - i} \right)$.
Putting in the values of x and y, we get the absolute value of given complex number as:
Now, we know that a complex number can be represented in polar form as $r\left( {\cos \theta + i\sin \theta } \right)$, where r is the modulus of complex numbers and $\theta$ is the argument. So, we have,
${Z_1} = \left( {1 + i\sqrt 3 } \right) = 2\left( {\cos \theta + i\sin \theta } \right)$
Comparing both sides, we get,
$\Rightarrow \cos \theta = \dfrac{1}{2}$ and $\sin \theta = \dfrac{{\sqrt 3 }}{2}$.
Since both sine and cosine are positive. So, the argument of complex numbers is in the first quadrant. Also, we know that values of $\sin \left( {\dfrac{\pi }{3}} \right)$ is $\dfrac{{\sqrt 3 }}{2}$ and $\cos \left( {\dfrac{\pi }{3}} \right)$ is $\dfrac{1}{2}$.
Hence, the argument of complex number ${Z_1} = \left( {1 + i\sqrt 3 } \right)$ is $\left( {\dfrac{\pi }{3}} \right)$.
Now, we know that the modulus and argument of the complex number ${Z_1} = \left( {1 + i\sqrt 3 } \right)$ is $2$ and $\left( {\dfrac{\pi }{3}} \right)$ respectively. Now, we calculate for the complex number ${\left( {1 + i\sqrt 3 } \right)^8}$.
For this, we write the complex number in the Euler form of complex number.
We can write the complex numbers in Euler form as $r{e^{i\theta }}$.
So, we get, ${\left( {1 + i\sqrt 3 } \right)^8}$ as ${\left( {2{e^{i\left( {\dfrac{\pi }{3}} \right)}}} \right)^8}$.
Now, we simplify the expression using the law of exponents. So, using the law of exponents ${\left( {{a^x}} \right)^y} = {a^{xy}}$, we get,
$\Rightarrow {2^8}{e^{i\left( {8 \times \dfrac{\pi }{3}} \right)}}$
We know that ${2^8} = 256$. So, we get,
$\Rightarrow 256{e^{i\left( {\dfrac{{8\pi }}{3}} \right)}}$
Converting the complex number back to the polar form, we get,
$\Rightarrow 256\left( {\cos \left( {\dfrac{{8\pi }}{3}} \right) + i\sin \left( {\dfrac{{8\pi }}{3}} \right)} \right)$
Now, we know that the principal argument of a complex number lies in the range of $( - \pi ,\pi ]$. We also know that the periodicity of the sine and cosine functions is $2\pi$. So, we get,
$\Rightarrow 256\left( {\cos \left( {\dfrac{{8\pi }}{3} - 2\pi } \right) + i\sin \left( {\dfrac{{8\pi }}{3} - 2\pi } \right)} \right)$
$\Rightarrow 256\left( {\cos \left( {\dfrac{{2\pi }}{3}} \right) + i\sin \left( {\dfrac{{2\pi }}{3}} \right)} \right)$
Therefore, the modulus of the complex number ${\left( {1 + i\sqrt 3 } \right)^8}$ is $256$ and amplitude is $\left( {\dfrac{{2\pi }}{3}} \right)$. Hence, option (B) is the correct answer.

Note:
We must know the method for representing the complex number in the Euler’s form. We should know the process of finding the argument and modulus of a given complex number. We must take care while doing the calculations so as to be sure of the final answer. One must know that the expansion of ${e^{i\theta }}$ is the same as $\left( {\cos \theta + i\sin \theta } \right)$.