Question

For positive integers $n_1,n_2$ , the value of the expression $(1+i{)}^{n_1}+(1+i^3{)}^{n_1}+(1+i^5{)}^{n_2}+(1+i^7{)}^{n_2}$ is a real number if and only if

• A. $n_1=n_2+1$
• B. $n_1=n_2-1$
• C. $n_1=n_2$
• D. $n_1$ and $n_2$ can take any positive integral values

Answer 1

The correct option is D $n_1$ and $n_2$ can take any positive integral values

Given:- $n_1$ and $n_2$ are positive integers.

Value of ${(1+i)}^{n_1}+{(1+i^3)}^{n_1}+{(1+i^5)}^{n_2}+{(1+i^7)}^{n_2}$ is a real no.

To find the condition for which the value of given them is a real no.

We know that,

Expansion of ${(1+i)}^n={}^n{C}_0+{}^n{C}_{1}i+{}^n{C}_2{i}^2+{}^n{C}_3{i}^3+\cdots +{}^n{C}_n{i}^n$ ...... $\left(1\right)$

Expansion of ${(1-i)}^n={}^n{C}_0-{}^n{C}_{1}i+{}^n{C}_2{i}^2-{}^n{C}_3{i}^3+\cdots +(-1{)}^n{}^n{C}_n{i}^n$ ...... $\left(2\right)$

By adding equations $\left(1\right)$ and $\left(2\right)$, the odd terms get cancelled out and we get,

${(1+i)}^n+{(1-i)}^n=2{}^n{C}_0+2{}^n{C}_2{i}^2+2{}^n{C}_4{i}^4+\cdots$

We know that, $i=\sqrt{-1}$ $\Rightarrow i^2=-1$

$\therefore {(1+i)}^n+{(1-i)}^n=2{}^n{C}_0-2{}^n{C}_2+2{}^n{C}_4\cdots$

Therefore, the value of ${(1+i)}^n+{(1+i)}^n$

is always real

${(1+i)}^{n_1}+{(1+i^3)}^{n_1}+{(1+i^5)}^{n_2}+{(1+i^7)}^{n_2}$

$={(1+i)}^{n_1}+{[1+i^2(i\left)\right]}^{n_1}+{(1+i^5)}^{n_2}+{[1+\left(i^5\right)(i^2\left)\right]}^{n_2}$

$={(1+i)}^{n_1}+{(1-i)}^{n_1}+{(1+i^5)}^{n_2}+{(1-i^5)}^{n_2}$

always real

$=\text{real no.}+{[1+\left(i^2\right)(i^2\left)i\right]}^{n_2}+{[1-\left(i^5\right)(i^2\left)\right]}^{n_2}$

$=\text{real no.}+{(1+i)}^{n_2}+{(1-i)}^{n_2}$

always reals

Hence, any positive integral value of $n_1$ and $n_2$ can give the real value

Hence, $n_1$ and $n_2$ can take any positiveintegral value.