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Question: For positive integers \({{n}_{1}},{{n}_{2}}\), the value of the expression \({{\left( 1+i \right)}^{...

For positive integers n1,n2{{n}_{1}},{{n}_{2}}, the value of the expression (1+i)n1+(1+i3)n1+(1+i5)n2+(1+i7)n2{{\left( 1+i \right)}^{{{n}_{1}}}}+{{\left( 1+{{i}^{3}} \right)}^{{{n}_{1}}}}+{{\left( 1+{{i}^{5}} \right)}^{{{n}_{2}}}}+{{\left( 1+{{i}^{7}} \right)}^{{{n}_{2}}}} is a real number if and only if
(a) n1=n2+1{{n}_{1}}={{n}_{2}}+1
(b) n1=n2−1{{n}_{1}}={{n}_{2}}-1
(c) n1=n2{{n}_{1}}={{n}_{2}}
(d) n1>0,n2>0{{n}_{1}}>0,{{n}_{2}}>0

Explanation

Solution

Hint: Simplify the given expression using the fact that i4n+1=i{{i}^{4n+1}}=i and i4n−1=−i{{i}^{4n-1}}=-i to simplify the terms of the given expression. Use the Binomial Theorem which states that for any two numbers ‘x’ and ‘y’, we have (x+y)n=nC0xny0+nC1xn−1y1+...+nCnx0yn{{\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}} to expand the terms of the given expression. Simplify the expression by cancelling out the like terms and solve the equation to find the condition for which the expression will be a real number.

Complete step-by-step solution -
We have to find the condition for which the expression (1+i)n1+(1+i3)n1+(1+i5)n2+(1+i7)n2{{\left( 1+i \right)}^{{{n}_{1}}}}+{{\left( 1+{{i}^{3}} \right)}^{{{n}_{1}}}}+{{\left( 1+{{i}^{5}} \right)}^{{{n}_{2}}}}+{{\left( 1+{{i}^{7}} \right)}^{{{n}_{2}}}} is a real number, where n1,n2{{n}_{1}},{{n}_{2}} are positive integers.
We know that i=−1i=\sqrt{-1} is a square root of unity. We also know that i4n+1=i{{i}^{4n+1}}=i and i4n−1=−i{{i}^{4n-1}}=-i.
Substituting n=1n=1 in the formula i4n−1=−i{{i}^{4n-1}}=-i, we have i4(1)−1=i3=−i{{i}^{4\left( 1 \right)-1}}={{i}^{3}}=-i.
Substituting n=1n=1 in the formula i4n+1=i{{i}^{4n+1}}=i, we have i4(1)+1=i5=i{{i}^{4\left( 1 \right)+1}}={{i}^{5}}=i.
Substituting n=2n=2 in the formula i4n−1=−i{{i}^{4n-1}}=-i, we have i4(2)−1=i7=−i{{i}^{4\left( 2 \right)-1}}={{i}^{7}}=-i.
Thus, we can rewrite the expression (1+i)n1+(1+i3)n1+(1+i5)n2+(1+i7)n2{{\left( 1+i \right)}^{{{n}_{1}}}}+{{\left( 1+{{i}^{3}} \right)}^{{{n}_{1}}}}+{{\left( 1+{{i}^{5}} \right)}^{{{n}_{2}}}}+{{\left( 1+{{i}^{7}} \right)}^{{{n}_{2}}}} as (1+i)n1+(1+i3)n1+(1+i5)n2+(1+i7)n2=(1+i)n1+(1−i)n1+(1+i)n2+(1−i)n2{{\left( 1+i \right)}^{{{n}_{1}}}}+{{\left( 1+{{i}^{3}} \right)}^{{{n}_{1}}}}+{{\left( 1+{{i}^{5}} \right)}^{{{n}_{2}}}}+{{\left( 1+{{i}^{7}} \right)}^{{{n}_{2}}}}={{\left( 1+i \right)}^{{{n}_{1}}}}+{{\left( 1-i \right)}^{{{n}_{1}}}}+{{\left( 1+i \right)}^{{{n}_{2}}}}+{{\left( 1-i \right)}^{{{n}_{2}}}}.
We will now expand the above expression using the Binomial Theorem.
We know that the Binomial Theorem which states that for any two numbers ‘x’ and ‘y’, we have (x+y)n=nC0xny0+nC1xn−1y1+...+nCnx0yn{{\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}} .
So, we have (1+i)n1=n1C0(1)n1(i)0+n1C1(1)n1−1(i)1+n1C2(1)n1−2(i)2+...+n1Cn1(1)0(i)n1=n1C0+n1C1i−n1C2−n1C3i...+n1Cn1(i)n1{{\left( 1+i \right)}^{{{n}_{1}}}}={}^{{{n}_{1}}}{{C}_{0}}{{\left( 1 \right)}^{{{n}_{1}}}}{{\left( i \right)}^{0}}+{}^{{{n}_{1}}}{{C}_{1}}{{\left( 1 \right)}^{{{n}_{1}}-1}}{{\left( i \right)}^{1}}+{}^{{{n}_{1}}}{{C}_{2}}{{\left( 1 \right)}^{{{n}_{1}}-2}}{{\left( i \right)}^{2}}+...+{}^{{{n}_{1}}}{{C}_{{{n}_{1}}}}{{\left( 1 \right)}^{0}}{{\left( i \right)}^{{{n}_{1}}}}={}^{{{n}_{1}}}{{C}_{0}}+{}^{{{n}_{1}}}{{C}_{1}}i-{}^{{{n}_{1}}}{{C}_{2}}-{}^{{{n}_{1}}}{{C}_{3}}i...+{}^{{{n}_{1}}}{{C}_{{{n}_{1}}}}{{\left( i \right)}^{{{n}_{1}}}}.
Similarly, we have (1+i)n1=n1C0(1)n1(−i)0+n1C1(1)n1−1(−i)1+n1C2(1)n1−2(−i)2+...+n1Cn1(1)0(−i)n1=n1C0−n1C1i−n1C2+n1C3i...+n1Cn1(−i)n1{{\left( 1+i \right)}^{{{n}_{1}}}}={}^{{{n}_{1}}}{{C}_{0}}{{\left( 1 \right)}^{{{n}_{1}}}}{{\left( -i \right)}^{0}}+{}^{{{n}_{1}}}{{C}_{1}}{{\left( 1 \right)}^{{{n}_{1}}-1}}{{\left( -i \right)}^{1}}+{}^{{{n}_{1}}}{{C}_{2}}{{\left( 1 \right)}^{{{n}_{1}}-2}}{{\left( -i \right)}^{2}}+...+{}^{{{n}_{1}}}{{C}_{{{n}_{1}}}}{{\left( 1 \right)}^{0}}{{\left( -i \right)}^{{{n}_{1}}}}={}^{{{n}_{1}}}{{C}_{0}}-{}^{{{n}_{1}}}{{C}_{1}}i-{}^{{{n}_{1}}}{{C}_{2}}+{}^{{{n}_{1}}}{{C}_{3}}i...+{}^{{{n}_{1}}}{{C}_{{{n}_{1}}}}{{\left( -i \right)}^{{{n}_{1}}}}.
We also have (1+i)n2=n2C0(1)n2(i)0+n2C1(1)n2−1(i)1+n2C2(1)n2−2(i)2+...+n2Cn2(1)0(i)n2=n2C0+n2C1i−n2C2−n2C3i...+n2Cn2(i)n2{{\left( 1+i \right)}^{{{n}_{2}}}}={}^{{{n}_{2}}}{{C}_{0}}{{\left( 1 \right)}^{{{n}_{2}}}}{{\left( i \right)}^{0}}+{}^{{{n}_{2}}}{{C}_{1}}{{\left( 1 \right)}^{{{n}_{2}}-1}}{{\left( i \right)}^{1}}+{}^{{{n}_{2}}}{{C}_{2}}{{\left( 1 \right)}^{{{n}_{2}}-2}}{{\left( i \right)}^{2}}+...+{}^{{{n}_{2}}}{{C}_{{{n}_{2}}}}{{\left( 1 \right)}^{0}}{{\left( i \right)}^{{{n}_{2}}}}={}^{{{n}_{2}}}{{C}_{0}}+{}^{{{n}_{2}}}{{C}_{1}}i-{}^{{{n}_{2}}}{{C}_{2}}-{}^{{{n}_{2}}}{{C}_{3}}i...+{}^{{{n}_{2}}}{{C}_{{{n}_{2}}}}{{\left( i \right)}^{{{n}_{2}}}} and (1+i)n2=n2C0(1)n2(−i)0+n2C1(1)n2−1(−i)1+n2C2(1)n2−2(−i)2+...+n2Cn2(1)0(−i)n2=n2C0−n2C1i−n2C2+n2C3i...+n2Cn2(−i)n2{{\left( 1+i \right)}^{{{n}_{2}}}}={}^{{{n}_{2}}}{{C}_{0}}{{\left( 1 \right)}^{{{n}_{2}}}}{{\left( -i \right)}^{0}}+{}^{{{n}_{2}}}{{C}_{1}}{{\left( 1 \right)}^{{{n}_{2}}-1}}{{\left( -i \right)}^{1}}+{}^{{{n}_{2}}}{{C}_{2}}{{\left( 1 \right)}^{{{n}_{2}}-2}}{{\left( -i \right)}^{2}}+...+{}^{{{n}_{2}}}{{C}_{{{n}_{2}}}}{{\left( 1 \right)}^{0}}{{\left( -i \right)}^{{{n}_{2}}}}={}^{{{n}_{2}}}{{C}_{0}}-{}^{{{n}_{2}}}{{C}_{1}}i-{}^{{{n}_{2}}}{{C}_{2}}+{}^{{{n}_{2}}}{{C}_{3}}i...+{}^{{{n}_{2}}}{{C}_{{{n}_{2}}}}{{\left( -i \right)}^{{{n}_{2}}}}.
Thus, we have (1+i)n1+(1−i)n1+(1+i)n2+(1−i)n2=n1C0+n1C1i−n1C2−n1C3i...+n1Cn1(i)n1+n1C0−n1C1i−n1C2+n1C3i...+n1Cn1(−i)n1+n2C0+n2C1i−n2C2−n2C3i...+n2Cn2(i)n2+n2C0−n2C1i−n2C2+n2C3i...+n2Cn2(−i)n2{{\left( 1+i \right)}^{{{n}_{1}}}}+{{\left( 1-i \right)}^{{{n}_{1}}}}+{{\left( 1+i \right)}^{{{n}_{2}}}}+{{\left( 1-i \right)}^{{{n}_{2}}}}={}^{{{n}_{1}}}{{C}_{0}}+{}^{{{n}_{1}}}{{C}_{1}}i-{}^{{{n}_{1}}}{{C}_{2}}-{}^{{{n}_{1}}}{{C}_{3}}i...+{}^{{{n}_{1}}}{{C}_{{{n}_{1}}}}{{\left( i \right)}^{{{n}_{1}}}}+{}^{{{n}_{1}}}{{C}_{0}}-{}^{{{n}_{1}}}{{C}_{1}}i-{}^{{{n}_{1}}}{{C}_{2}}+{}^{{{n}_{1}}}{{C}_{3}}i...+{}^{{{n}_{1}}}{{C}_{{{n}_{1}}}}{{\left( -i \right)}^{{{n}_{1}}}}+{}^{{{n}_{2}}}{{C}_{0}}+{}^{{{n}_{2}}}{{C}_{1}}i-{}^{{{n}_{2}}}{{C}_{2}}-{}^{{{n}_{2}}}{{C}_{3}}i...+{}^{{{n}_{2}}}{{C}_{{{n}_{2}}}}{{\left( i \right)}^{{{n}_{2}}}}+{}^{{{n}_{2}}}{{C}_{0}}-{}^{{{n}_{2}}}{{C}_{1}}i-{}^{{{n}_{2}}}{{C}_{2}}+{}^{{{n}_{2}}}{{C}_{3}}i...+{}^{{{n}_{2}}}{{C}_{{{n}_{2}}}}{{\left( -i \right)}^{{{n}_{2}}}}.
Simplifying the above expression, we have (1+i)n1+(1−i)n1+(1+i)n2+(1−i)n2=2n1C0−2n1C2...+n1Cn1(i)n1+n1Cn1(−i)n1+2n2C0−2n2C2+...+n2Cn2(i)n2+n2Cn2(−i)n2{{\left( 1+i \right)}^{{{n}_{1}}}}+{{\left( 1-i \right)}^{{{n}_{1}}}}+{{\left( 1+i \right)}^{{{n}_{2}}}}+{{\left( 1-i \right)}^{{{n}_{2}}}}=2{}^{{{n}_{1}}}{{C}_{0}}-2{}^{{{n}_{1}}}{{C}_{2}}...+{}^{{{n}_{1}}}{{C}_{{{n}_{1}}}}{{\left( i \right)}^{{{n}_{1}}}}+{}^{{{n}_{1}}}{{C}_{{{n}_{1}}}}{{\left( -i \right)}^{{{n}_{1}}}}+2{}^{{{n}_{2}}}{{C}_{0}}-2{}^{{{n}_{2}}}{{C}_{2}}+...+{}^{{{n}_{2}}}{{C}_{{{n}_{2}}}}{{\left( i \right)}^{{{n}_{2}}}}+{}^{{{n}_{2}}}{{C}_{{{n}_{2}}}}{{\left( -i \right)}^{{{n}_{2}}}}.
We must have that the above expression is a real number. Thus, we have n1Cn1(i)n1+n1Cn1(−i)n1+n2Cn2(i)n2+n2Cn2(−i)n2=0{}^{{{n}_{1}}}{{C}_{{{n}_{1}}}}{{\left( i \right)}^{{{n}_{1}}}}+{}^{{{n}_{1}}}{{C}_{{{n}_{1}}}}{{\left( -i \right)}^{{{n}_{1}}}}+{}^{{{n}_{2}}}{{C}_{{{n}_{2}}}}{{\left( i \right)}^{{{n}_{2}}}}+{}^{{{n}_{2}}}{{C}_{{{n}_{2}}}}{{\left( -i \right)}^{{{n}_{2}}}}=0.
So, we have n1Cn1[(i)n1+(−i)n1]+n2Cn2[(i)n2+(−i)n2]=0⇒(i)n1+(−i)n1+(i)n2+(−i)n2=0{}^{{{n}_{1}}}{{C}_{{{n}_{1}}}}\left[ {{\left( i \right)}^{{{n}_{1}}}}+{{\left( -i \right)}^{{{n}_{1}}}} \right]+{}^{{{n}_{2}}}{{C}_{{{n}_{2}}}}\left[ {{\left( i \right)}^{{{n}_{2}}}}+{{\left( -i \right)}^{{{n}_{2}}}} \right]=0\Rightarrow {{\left( i \right)}^{{{n}_{1}}}}+{{\left( -i \right)}^{{{n}_{1}}}}+{{\left( i \right)}^{{{n}_{2}}}}+{{\left( -i \right)}^{{{n}_{2}}}}=0 as we know that nCn=n!0!n!=1{}^{n}{{C}_{n}}=\dfrac{n!}{0!n!}=1.
Thus, we have (i)n1+(−1)n1(i)n1+(i)n2+(−1)n2(i)n2=0⇒(i)n1(1+(−1)n1)+(i)n2(1+(−1)n2)=0{{\left( i \right)}^{{{n}_{1}}}}+{{\left( -1 \right)}^{{{n}_{1}}}}{{\left( i \right)}^{{{n}_{1}}}}+{{\left( i \right)}^{{{n}_{2}}}}+{{\left( -1 \right)}^{{{n}_{2}}}}{{\left( i \right)}^{{{n}_{2}}}}=0\Rightarrow {{\left( i \right)}^{{{n}_{1}}}}\left( 1+{{\left( -1 \right)}^{{{n}_{1}}}} \right)+{{\left( i \right)}^{{{n}_{2}}}}\left( 1+{{\left( -1 \right)}^{{{n}_{2}}}} \right)=0.
So, we must have (−1)n1=−1{{\left( -1 \right)}^{{{n}_{1}}}}=-1 and (−1)n2=−1{{\left( -1 \right)}^{{{n}_{2}}}}=-1. Thus, n1{{n}_{1}} and n2{{n}_{2}} must be both odd numbers. Considering all the options, the given condition holds if and only if n1=n2{{n}_{1}}={{n}_{2}}.
Hence, the condition required for (1+i)n1+(1+i3)n1+(1+i5)n2+(1+i7)n2{{\left( 1+i \right)}^{{{n}_{1}}}}+{{\left( 1+{{i}^{3}} \right)}^{{{n}_{1}}}}+{{\left( 1+{{i}^{5}} \right)}^{{{n}_{2}}}}+{{\left( 1+{{i}^{7}} \right)}^{{{n}_{2}}}} to be a real number is n1=n2{{n}_{1}}={{n}_{2}}, which is option (c).

Note: We can’t solve this question without calculating the higher values of i=−1i=\sqrt{-1}. We also observe that n1{{n}_{1}} and n2{{n}_{2}} can take any positive odd integer as its value. However, based on the given options, we must have n1=n2{{n}_{1}}={{n}_{2}}.