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

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

Explanation

Solution

We have been given an equation of complex identity. We use binomial theorem to form the sum of two numbers with power n as (1+x)n+(1−x)n=2[1+nC2x2+nC4x4+nC6x6+..]{{\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=−1i=\sqrt{-1} to get the solution of the problem.

Complete step-by-step solution
We have been given a complex equation to solve (1+i)5+(1−i)5{{\left( 1+i \right)}^{5}}+{{\left( 1-i \right)}^{5}} where i=−1i=\sqrt{-1}.
We also have the identities of i2=−1,i3=−i,i4=1{{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1.
From the theorem we get (1+x)n=1+nC1x1+nC2x2+nC3x3+.....+nCnxn{{\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
(1−x)n=1−nC1x1+nC2x2−nC3x3+.....+(−1)nnCnxn{{\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 (1+x)n+(1−x)n=2[1+nC2x2+nC4x4+nC6x6+..]{{\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=5x=i,n=5. We get (1+i)5+(1−i)5=2[1+5C2i2+5C4i4]{{\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
(1+i)5+(1−i)5 =2[1−5!2!3!+5!4!1!] =2[1−10+5] =−8 \begin{aligned} & {{\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{aligned}
Therefore, the value of (1+i)5+(1−i)5{{\left( 1+i \right)}^{5}}+{{\left( 1-i \right)}^{5}} is -8. The correct option is A.

Note: In equation of (1+x)n+(1−x)n=2[1+nC2x2+nC4x4+nC6x6+..]{{\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.