Question

Find the value of ${{\left( 1+i \right)}^{5}}+{{\left( 1-i \right)}^{5}}$.
A. -8
B. 8i
C. 8
D. 32

Answer 1

Hint: We have been given an equation of complex identity. We use binomial theorem to form the sum of two numbers with power n as ${{\left( 1+x \right)}^{n}}+{{\left( 1-x \right)}^{n}}=2\left[ 1+{}^{n}{{C}_{2}}{{x}^{2}}+{}^{n}{{C}_{4}}{{x}^{4}}+{}^{n}{{C}_{6}}{{x}^{6}}+.. \right]$. We replace the values of x and n. then we use the indices values of $i=\sqrt{-1}$ to get the solution of the problem.

Complete step-by-step solution
We have been given a complex equation to solve ${{\left( 1+i \right)}^{5}}+{{\left( 1-i \right)}^{5}}$ where $i=\sqrt{-1}$.
We also have the identities of ${{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1$.
From the theorem we get \[{{\left( 1+x \right)}^{n}}=1+{}^{n}{{C}_{1}}{{x}^{1}}+{}^{n}{{C}_{2}}{{x}^{2}}+{}^{n}{{C}_{3}}{{x}^{3}}+.....+{}^{n}{{C}_{n}}{{x}^{n}}\] and for
\[{{\left( 1-x \right)}^{n}}=1-{}^{n}{{C}_{1}}{{x}^{1}}+{}^{n}{{C}_{2}}{{x}^{2}}-{}^{n}{{C}_{3}}{{x}^{3}}+.....+{{\left( -1 \right)}^{n}}{}^{n}{{C}_{n}}{{x}^{n}}\]. Adding that we get the sum of binomial of two numbers tell that ${{\left( 1+x \right)}^{n}}+{{\left( 1-x \right)}^{n}}=2\left[ 1+{}^{n}{{C}_{2}}{{x}^{2}}+{}^{n}{{C}_{4}}{{x}^{4}}+{}^{n}{{C}_{6}}{{x}^{6}}+.. \right]$.
So, the even positioned terms in the expansion get doubled and the oddly positioned terms get eliminated.
We replace the values $x=i,n=5$. We get ${{\left( 1+i \right)}^{5}}+{{\left( 1-i \right)}^{5}}=2\left[ 1+{}^{5}{{C}_{2}}{{i}^{2}}+{}^{5}{{C}_{4}}{{i}^{4}} \right]$.
Now we replace the values
$\begin{align}
  & {{\left( 1+i \right)}^{5}}+{{\left( 1-i \right)}^{5}} \\
 & =2\left[ 1-\dfrac{5!}{2!3!}+\dfrac{5!}{4!1!} \right] \\
 & =2\left[ 1-10+5 \right] \\
 & =-8 \\
\end{align}$
Therefore, the value of ${{\left( 1+i \right)}^{5}}+{{\left( 1-i \right)}^{5}}$ is -8. The correct option is A.

Note: In equation of ${{\left( 1+x \right)}^{n}}+{{\left( 1-x \right)}^{n}}=2\left[ 1+{}^{n}{{C}_{2}}{{x}^{2}}+{}^{n}{{C}_{4}}{{x}^{4}}+{}^{n}{{C}_{6}}{{x}^{6}}+.. \right]$, the end is finalised depending on the value of n. The term changes depending on if n is odd or even. In every binomial the number of terms is one greater than the power value of the binomial.