Question

The value of ${\left( {1 + i} \right)^8} + {\left( {1 - i} \right)^8}$ is:
A. $16$
B. $ - 16$
C. $32$
D. $ - 32$

Answer 1

Hint: In the given problem, we need to evaluate the value of the given expression involving complex numbers. The given question requires knowledge of the concepts of complex numbers and its representation in different forms. We must know the conversion of a complex number into polar form and Euler form. After converting the complex number into Euler form, we do the operations on them as given in the question and get to the final answer.

Complete step by step answer:
In the question, we need to evaluate the value of ${\left( {1 + i} \right)^8} + {\left( {1 - i} \right)^8}$. So, we have two complex numbers ${Z_1} = \left( {1 + i} \right)$ and ${Z_2} = \left( {1 - i} \right)$.Now, we will find the modulus and argument of both the complex numbers.
So, we have, ${Z_1} = \left( {1 + i} \right)$.
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} + {{(1)}^2}} = \sqrt 2 $
So, the modulus of the given complex number ${Z_1} = \left( {1 + i} \right)$ is $\left( {\sqrt 2 } \right)$.
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:
$\left| {{Z_2}} \right| = \sqrt {{{(1)}^2} + {{( - 1)}^2}} = \sqrt 2 $
So, the modulus of the given complex number ${Z_2} = \left( {1 - i} \right)$ is $\left( {\sqrt 2 } \right)$.

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,
${Z_1} = \left( {1 + i} \right) = \sqrt 2 \left( {\cos \theta + \sin \theta } \right)$
Comparing both sides, we get,
$ \Rightarrow \cos \theta = \dfrac{1}{{\sqrt 2 }}$ and $\sin \theta = \dfrac{1}{{\sqrt 2 }}$.
Since both sine and cosine are positive. So, the argument of complex numbers is in the first quadrant. Also, we know that values of $\sin \left( {\dfrac{\pi }{4}} \right)$ and $\cos \left( {\dfrac{\pi }{4}} \right)$ is $\dfrac{1}{{\sqrt 2 }}$.
Hence, the argument of complex number ${Z_1} = \left( {1 + i} \right)$ is $\left( {\dfrac{\pi }{4}} \right)$.

Also, ${Z_2} = \left( {1 - i} \right) = \sqrt 2 \left( {\cos \theta + \sin \theta } \right)$
Comparing both sides, we get,
$ \Rightarrow \cos \theta = \dfrac{1}{{\sqrt 2 }}$ and $\sin \theta = - \dfrac{1}{{\sqrt 2 }}$.
Since both sine is negative and cosine is positive. So, the argument of complex numbers is in the fourth quadrant. Also, we know that values of $\sin \left( {\dfrac{\pi }{4}} \right)$ and $\cos \left( {\dfrac{\pi }{4}} \right)$ is $\dfrac{1}{{\sqrt 2 }}$.
Hence, the argument of complex number ${Z_2} = \left( {1 - i} \right)$ is $\left( {2\pi - \dfrac{\pi }{4}} \right) = \left( {\dfrac{{7\pi }}{4}} \right)$.
So, we get the complex numbers as ${Z_1} = \left( {1 + i} \right) = \sqrt 2 \left( {\cos \left( {\dfrac{\pi }{4}} \right) + \sin \left( {\dfrac{\pi }{4}} \right)} \right)$ and ${Z_2} = \left( {1 - i} \right) = \sqrt 2 \left( {\cos \left( {\dfrac{{7\pi }}{4}} \right) + \sin \left( {\dfrac{{7\pi }}{4}} \right)} \right)$ .

We can also write the complex numbers IN Euler form as $r{e^{i\theta }}$.
So, we get, ${Z_1} = \sqrt 2 {e^{i\left( {\dfrac{\pi }{4}} \right)}}$ and ${Z_2} = \sqrt 2 {e^{i\left( {\dfrac{{7\pi }}{4}} \right)}}$.
Now, we do the operations over the Euler’s form of both the complex numbers as required in the question. So, we get,
 ${\left( {1 + i} \right)^8} + {\left( {1 - i} \right)^8}$
$ \Rightarrow {\left( {\sqrt 2 {e^{i\left( {\dfrac{\pi }{4}} \right)}}} \right)^8} + {\left( {\sqrt 2 {e^{i\left( {\dfrac{{7\pi }}{4}} \right)}}} \right)^8}$
Now, using the law of exponents ${\left( {{a^x}} \right)^y} = {a^{xy}}$, we get,
\[ \Rightarrow {\left( {\sqrt 2 } \right)^8}{e^{i\left( {2\pi } \right)}} + {\left( {\sqrt 2 } \right)^8}{e^{i\left( {14\pi } \right)}}\]

Now, evaluating the powers of $\sqrt 2 $, we get,
\[ \Rightarrow 16{e^{i\left( {2\pi } \right)}} + 16{e^{i\left( {14\pi } \right)}}\]
Now, we know that the expansion of ${e^{i\theta }}$ is the same as $\left( {\cos \theta + i\sin \theta } \right)$. So, we get,
\[ \Rightarrow 16\left[ {\cos \left( {2\pi } \right) + i\sin \left( {2\pi } \right)} \right] + 16\left[ {\cos \left( {14\pi } \right) + i\sin \left( {14\pi } \right)} \right]\]
We know that the value of sine for any integral multiple of $\pi $ is zero. So, we get,
\[ \Rightarrow 16\left[ {\cos \left( {2\pi } \right) + 0} \right] + 16\left[ {\cos \left( {14\pi } \right) + 0} \right]\]
Substituting in the values of $\cos 2\pi $ and $\cos 14\pi $,
\[ \Rightarrow 16\left[ 1 \right] + 16\left[ 1 \right]\]
\[ \therefore 32\]
So, the value of ${\left( {1 + i} \right)^8} + {\left( {1 - i} \right)^8}$ is $32$.

Hence, option C 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 $\left( {\cos \theta + i\sin \theta } \right)$.