Question

Find the value of the following expressions: ${{\left( 1+i \right)}^{6}}+{{\left( 1-i \right)}^{3}}$.

Answer 1

Hint: Simplify each of the terms of the given expression using the Binomial theorem which states that for any two numbers ‘x’ and ‘y’, we have ${{\left( x+y \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}{{y}^{0}}+{}^{n}{{C}_{1}}{{x}^{n-1}}{{y}^{1}}+...+{}^{n}{{C}_{n}}{{x}^{0}}{{y}^{n}}$. Use the fact that $i=\sqrt{-1}$ is a square root of unity to calculate the higher powers of $i$ to further simplify the given expression.

Complete step-by-step solution -
We have to calculate the value of the expression ${{\left( 1+i \right)}^{6}}+{{\left( 1-i \right)}^{3}}$. We observe that this is an expression with complex numbers.
To simplify the given expression, we will use Binomial Theorem which states that for any two numbers ‘x’ and ‘y’, we have ${{\left( x+y \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}{{y}^{0}}+{}^{n}{{C}_{1}}{{x}^{n-1}}{{y}^{1}}+...+{}^{n}{{C}_{n}}{{x}^{0}}{{y}^{n}}$.
Substituting $x=1,y=i,n=6$ in the above expression, we have ${{\left( 1+i \right)}^{6}}={}^{6}{{C}_{0}}{{\left( 1 \right)}^{6}}{{\left( i \right)}^{0}}+{}^{6}{{C}_{1}}{{\left( 1 \right)}^{5}}{{\left( i \right)}^{1}}+{}^{6}{{C}_{2}}{{\left( 1 \right)}^{4}}{{\left( i \right)}^{2}}+{}^{6}{{C}_{3}}{{\left( 1 \right)}^{3}}{{\left( i \right)}^{3}}+{}^{6}{{C}_{4}}{{\left( 1 \right)}^{2}}{{\left( i \right)}^{4}}+{}^{6}{{C}_{5}}{{\left( 1 \right)}^{1}}{{\left( i \right)}^{5}}+{}^{6}{{C}_{6}}{{\left( 1 \right)}^{0}}{{\left( i \right)}^{6}}$.
We know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.
So, we have ${{\left( 1+i \right)}^{6}}=\dfrac{6!}{0!6!}{{\left( 1 \right)}^{6}}{{\left( i \right)}^{0}}+\dfrac{6!}{1!5!}{{\left( 1 \right)}^{5}}{{\left( i \right)}^{1}}+\dfrac{6!}{2!4!}{{\left( 1 \right)}^{4}}{{\left( i \right)}^{2}}+\dfrac{6!}{3!3!}{{\left( 1 \right)}^{3}}{{\left( i \right)}^{3}}+\dfrac{6!}{2!4!}{{\left( 1 \right)}^{2}}{{\left( i \right)}^{4}}+\dfrac{6!}{1!5!}{{\left( 1 \right)}^{1}}{{\left( i \right)}^{5}}+\dfrac{6!}{0!6!}{{\left( 1 \right)}^{0}}{{\left( i \right)}^{6}}$.
Thus, we can rewrite the above expression as ${{\left( 1+i \right)}^{6}}=1+6i+15{{i}^{2}}+20{{i}^{3}}+15{{i}^{4}}+6{{i}^{5}}+{{i}^{6}}$.
We know that $i=\sqrt{-1}$. Thus, we have ${{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}}=-1,{{i}^{3}}={{i}^{2}}\times i=-i,{{i}^{4}}={{\left( {{i}^{2}} \right)}^{2}}=1,{{i}^{5}}={{i}^{4}}\times i=i,{{i}^{6}}={{i}^{4}}\times {{i}^{2}}=-1$.
Substituting these values in ${{\left( 1+i \right)}^{6}}=1+6i+15{{i}^{2}}+20{{i}^{3}}+15{{i}^{4}}+6{{i}^{5}}+{{i}^{6}}$, we have ${{\left( 1+i \right)}^{6}}=1+6i+15\left( -1 \right)+20\left( -i \right)+15\left( 1 \right)+6\left( i \right)+\left( -1 \right)$.
Thus, we have ${{\left( 1+i \right)}^{6}}=1+6i-15-20i+15+6i-1=-8i.....\left( 1 \right)$.
We will now simplify the expression ${{\left( 1-i \right)}^{3}}$.
Substituting $x=1,y=-i,n=3$ in the formula for Binomial Expansion ${{\left( x+y \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}{{y}^{0}}+{}^{n}{{C}_{1}}{{x}^{n-1}}{{y}^{1}}+...+{}^{n}{{C}_{n}}{{x}^{0}}{{y}^{n}}$, we have ${{\left( 1-i \right)}^{3}}={}^{3}{{C}_{0}}{{\left( 1 \right)}^{3}}{{\left( -i \right)}^{0}}+{}^{3}{{C}_{1}}{{\left( 1 \right)}^{2}}{{\left( -i \right)}^{1}}+{}^{3}{{C}_{2}}{{\left( 1 \right)}^{1}}{{\left( -i \right)}^{2}}+{}^{3}{{C}_{3}}{{\left( 1 \right)}^{0}}{{\left( -i \right)}^{3}}$.
We know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.
So, we have ${{\left( 1-i \right)}^{3}}=\dfrac{3!}{0!3!}{{\left( 1 \right)}^{3}}{{\left( -i \right)}^{0}}+\dfrac{3!}{1!2!}{{\left( 1 \right)}^{2}}{{\left( -i \right)}^{1}}+\dfrac{3!}{2!1!}{{\left( 1 \right)}^{1}}{{\left( -i \right)}^{2}}+\dfrac{3!}{0!3!}{{\left( 1 \right)}^{0}}{{\left( -i \right)}^{3}}$.
Thus, we can rewrite the above expression as ${{\left( 1-i \right)}^{3}}=1+3\left( -i \right)+3{{\left( -i \right)}^{2}}+{{\left( -i \right)}^{3}}$.
We know that $i=\sqrt{-1}$. Thus, we have ${{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}}=-1,{{i}^{3}}={{i}^{2}}\times i=-i$.
Substituting these values in the above expression, we have ${{\left( 1-i \right)}^{3}}=1-3i-3+i$.
Thus, we have ${{\left( 1-i \right)}^{3}}=-2-2i.....\left( 2 \right)$.
Substituting equation (1) and (2) in the expression ${{\left( 1+i \right)}^{6}}+{{\left( 1-i \right)}^{3}}$, we have ${{\left( 1+i \right)}^{6}}+{{\left( 1-i \right)}^{3}}=-8i-2-2i$.
Simplifying the above expression, we have ${{\left( 1+i \right)}^{6}}+{{\left( 1-i \right)}^{3}}=-8i-2-2i=-2-10i=-2\left( 1+5i \right)$.
Hence, the value of the expression ${{\left( 1+i \right)}^{6}}+{{\left( 1-i \right)}^{3}}$ is $-2-10i=-2\left( 1+5i \right)$.

Note: We can’t solve this question without expanding the expression using the Binomial Theorem and then calculating higher powers of $i=\sqrt{-1}$. We can write any complex number in the form $a+ib$, where $ib$ is the imaginary part and $a$ is the real part.