Question

The value of the following complex number ${{\left( 1+i \right)}^{5}}+{{\left( 1-i \right)}^{5}}$ is equal to:
(a) -8
(b) 8i
(c) 8
(d) 32

Answer 1

Hint: We are asked to find the value of this complex number ${{\left( 1+i \right)}^{5}}+{{\left( 1-i \right)}^{5}}$ which we are going to find by expanding each complex number using binomial expansion which is equal to ${{\left( 1+x \right)}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+......+{}^{n}{{C}_{n}}{{x}^{n}}$. After that we will use the different powers on iota as follows:
 $\begin{align}
  & {{i}^{4n}}=1 \\
 & {{i}^{4n+1}}=i \\
 & {{i}^{4n+2}}=-1 \\
 & {{i}^{4n+3}}=-i \\
\end{align}$

Complete step by step answer:
We have given the following complex number:
${{\left( 1+i \right)}^{5}}+{{\left( 1-i \right)}^{5}}$
We have to evaluate the above expression which we are going to do by expanding each complex numbers using the following binomial expansion:
${{\left( 1+x \right)}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+......+{}^{n}{{C}_{n}}{{x}^{n}}$
Let us expand ${{\left( 1+i \right)}^{5}}$ using binomial expansion we get,
${{\left( 1+i \right)}^{5}}={}^{5}{{C}_{0}}+{}^{5}{{C}_{1}}i+{}^{5}{{C}_{2}}{{i}^{2}}+{}^{5}{{C}_{3}}{{i}^{3}}+{}^{5}{{C}_{4}}{{i}^{4}}+{}^{5}{{C}_{5}}{{i}^{5}}$
Now, to evaluate the above we are going to use the following combinatorial formula and the different powers of iota.
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
The value of different powers of iota as follows:
$\begin{align}
  & {{i}^{4n}}=1 \\
 & {{i}^{4n+1}}=i \\
 & {{i}^{4n+2}}=-1 \\
 & {{i}^{4n+3}}=-i \\
\end{align}$
Using the above different powers of iota we are going to find the values of iota given in the binomial expansion:
$\begin{align}
  & {{i}^{^{2}}}=-1 \\
 & {{i}^{3}}=-i \\
 & {{i}^{4}}=1 \\
 & {{i}^{5}}=i \\
\end{align}$
$\begin{align}
  & {{\left( 1+i \right)}^{5}}=\dfrac{5!}{0!5!}+\dfrac{5!}{1!4!}i+\dfrac{5!}{2!3!}\left( -1 \right)+\dfrac{5!}{3!2!}\left( -i \right)+\dfrac{5!}{4!1!}\left( 1 \right)+\dfrac{5!}{5!0!}i \\
 & \Rightarrow {{\left( 1+i \right)}^{5}}=1+\dfrac{5.4!}{4!}i-\dfrac{5.4.3!}{2.1.3!}-i\dfrac{5.4.3!}{3!2.1}+\dfrac{5.4!}{4!1!}+i \\
 & \Rightarrow {{\left( 1+i \right)}^{5}}=1+5i-10-10i+5+i \\
 & \Rightarrow {{\left( 1+i \right)}^{5}}=-4-4i \\
\end{align}$
Similarly, we can find the value of ${{\left( 1-i \right)}^{5}}$.
\[{{\left( 1-i \right)}^{5}}={}^{5}{{C}_{0}}+{}^{5}{{C}_{1}}\left( -i \right)+{}^{5}{{C}_{2}}{{\left( -i \right)}^{2}}+{}^{5}{{C}_{3}}{{\left( -i \right)}^{3}}+{}^{5}{{C}_{4}}{{\left( -i \right)}^{4}}+{}^{5}{{C}_{5}}{{\left( -i \right)}^{5}}\]
\[\begin{align}
  & {{\left( 1-i \right)}^{5}}=\dfrac{5!}{0!5!}+\dfrac{5!}{1!4!}\left( -i \right)+\dfrac{5!}{2!3!}\left( \left( -1 \right) \right)+\dfrac{5!}{3!2!}\left( -\left( -i \right) \right)+\dfrac{5!}{4!1!}\left( 1 \right)+\dfrac{5!}{5!0!}\left( -i \right) \\
 & \Rightarrow {{\left( 1-i \right)}^{5}}=1-\dfrac{5.4!}{4!}i-\dfrac{5.4.3!}{2.1.3!}+i\dfrac{5.4.3!}{3!2.1}+\dfrac{5.4!}{4!1!}-i \\
 & \Rightarrow {{\left( 1-i \right)}^{5}}=1-5i-10+10i+5-i \\
 & \Rightarrow {{\left( 1-i \right)}^{5}}=-4+4i \\
\end{align}\]
Now, adding these two complex numbers we get,
$\begin{align}
  & {{\left( 1+i \right)}^{5}}+{{\left( 1-i \right)}^{5}} \\
 & =-4-4i-4+4i \\
 & =-8 \\
\end{align}$
Hence, we got the evaluation of the given complex number as -8.

So, the correct answer is “Option A”.

Note: The other way to solve the above problem is to convert the complex numbers written in the brackets into Euler form.
The expression given above is equal to:
${{\left( 1+i \right)}^{5}}+{{\left( 1-i \right)}^{5}}$
We can write 1 + i in the form of Euler form as follows:
The complex number written in the brackets is in the form of:
$\cos \theta +i\sin \theta $
And Euler form of the above complex number is equal to:
$r{{e}^{i\theta }}$
Writing 1 + i in the Euler form as follows:
Multiplying and dividing $\sqrt{2}$ to 1 + i we get,
$\begin{align}
  & \dfrac{\sqrt{2}}{\sqrt{2}}\left( 1+i \right) \\
 & =\sqrt{2}\left( \dfrac{1}{\sqrt{2}}+\dfrac{i}{\sqrt{2}} \right) \\
\end{align}$
Rewriting the above expression as follows:
$\sqrt{2}\left( \cos \dfrac{\pi }{4}+i\sin \dfrac{\pi }{4} \right)$
Writing in the Euler form we get,
$\sqrt{2}{{e}^{i\dfrac{\pi }{4}}}$
Similarly we can write the Euler form for 1 – i as follows:
$\begin{align}
  & \dfrac{\sqrt{2}}{\sqrt{2}}\left( 1-i \right) \\
 & =\sqrt{2}\left( \dfrac{1}{\sqrt{2}}-\dfrac{i}{\sqrt{2}} \right) \\
\end{align}$
The above complex number occurs in the fourth quadrant so the angle in the Euler form changes from $\dfrac{\pi }{4}$ to $-\dfrac{\pi }{4}$.
$\sqrt{2}{{e}^{-i\dfrac{\pi }{4}}}$
Now, substituting these complex numbers of the Euler form in the given expression we get,
$\begin{align}
  & {{\left( \sqrt{2}{{e}^{i\dfrac{\pi }{4}}} \right)}^{5}}+{{\left( \sqrt{2}{{e}^{-i\dfrac{\pi }{4}}} \right)}^{5}} \\
 & ={{\left( \sqrt{2} \right)}^{5}}{{e}^{\dfrac{i5\pi }{4}}}+{{\left( \sqrt{2} \right)}^{5}}{{e}^{-\dfrac{i5\pi }{4}}} \\
\end{align}$
Converting the above Euler form into cosine and sine angle forms we get,
$\begin{align}
  & {{\left( \sqrt{2} \right)}^{5}}\left( \cos \dfrac{5\pi }{4}+i\sin \dfrac{5\pi }{4}+\cos \left( -\dfrac{5\pi }{4} \right)+i\sin \left( -\dfrac{5\pi }{4} \right) \right) \\
 & ={{\left( \sqrt{2} \right)}^{5}}\left( \cos \dfrac{5\pi }{4}+i\sin \dfrac{5\pi }{4}+\cos \left( -\dfrac{5\pi }{4} \right)-i\sin \left( \dfrac{5\pi }{4} \right) \right) \\
 & ={{\left( \sqrt{2} \right)}^{5}}\left( \cos \dfrac{5\pi }{4}+\cos \left( -\dfrac{5\pi }{4} \right) \right) \\
\end{align}$
We know that $\cos \left( -\theta \right)=\cos \theta $ so using this relation in the above expression we get,
$\begin{align}
  & {{\left( \sqrt{2} \right)}^{5}}\left( \cos \dfrac{5\pi }{4}+\cos \left( \dfrac{5\pi }{4} \right) \right) \\
 & ={{\left( \sqrt{2} \right)}^{5}}\left( -\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}} \right) \\
\end{align}$
$\begin{align}
  & ={{\left( \sqrt{2} \right)}^{5}}\left( -\dfrac{2}{\sqrt{2}} \right) \\
 & =-{{\left( \sqrt{2} \right)}^{4}}\left( 2 \right) \\
 & =-8 \\
\end{align}$
Hence, we are getting the same value as in the above solution.