The value of in+i-n, for i=√-1 and n ∈ I is:

Question

The value of in+i-n, for i=√-1 and n ∈ I is: (A) (2 n /(1- i )2 m )+((1+ i )2 n /2 n ) (B) ((1+ i )2 n /2 n )+((1- i )2 n /2 n ) (C) ((1+i)2 n/2 n )+(2 n /(1- i )2 n ) (D) (2 n /(1+i)2 n )+(2 n /(1-i)2 n )

Tardigrade Admin · Accepted Answer

Note that ((1+i)2 n /2 n )=(((1+ i )2) n /2 n )=((2 i ) n /2 n )= i n ⇒ (2 n /(1+ i )2 n )= i - n and ((1-i)2 n/2n)=i-n ⇒ (2n/(1-i)2 n)=in

Q. The value of $i^{n} + i^{- n}$ , for $i = - 1$ and $n \in I$ is:

$\frac{2 ^{n}}{( 1 - i ) ^{2 m}} + \frac{( 1 + i ) ^{2 n}}{2 ^{n}}$

$\frac{( 1 + i ) ^{2 n}}{2 ^{n}} + \frac{( 1 - i ) ^{2 n}}{2 ^{n}}$

$\frac{( 1 + i ) ^{2 n}}{2 ^{n}} + \frac{2 ^{n}}{( 1 - i ) ^{2 n}}$

$\frac{2 ^{n}}{( 1 + i ) ^{2 n}} + \frac{2 ^{n}}{( 1 - i ) ^{2 n}}$

Note that $\frac{( 1 + i ) ^{2 n}}{2 ^{n}} = \frac{( ( 1 + i ) ^{2} ) ^{n}}{2 ^{n}} = \frac{( 2 i ) ^{n}}{2 ^{n}} = i^{n} \Rightarrow \frac{2 ^{n}}{( 1 + i ) ^{2 n}} = i^{- n}$
and $\frac{( 1 - i ) ^{2 n}}{2 ^{n}} = i^{- n} \Rightarrow \frac{2 ^{n}}{( 1 - i ) ^{2 n}} = i^{n}$

Q. The value of in+i−n, for i=−1​ and n∈I is:

(1−i)2m2n​+2n(1+i)2n​

2n(1+i)2n​+2n(1−i)2n​

2n(1+i)2n​+(1−i)2n2n​

(1+i)2n2n​+(1−i)2n2n​