The complex number 2 n 1-i 2n - 1-i 2n 2 n - n - Z- is equal to

2 ^ n (1+i) ^ 2n + (1+i) ^ 2n 2 ^ n , n ∈ I (1+p) ^ 2 =1 1+2i=2i 2 ^ n (2i) ^ n + (2i) ^ n 2 ^ n = 2 ^ n i^ n x 2 ^ n + i^ n x 2 ^ ...

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Addition of Vectors

The complex ...

Question

$T h e c o m p l e x n u m b e r \frac{2^{n}}{{(1 + i)}^{2 n}} + \frac{{(1 + i)}^{2 n}}{2^{n}}, n \in Z, i s e q u a l t o$

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Solution

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n \in I

{(1 + i)}^{2 n} + {(1 - i)}^{2 n} = - 2^{n + 1}

i = \sqrt{- 1}

n

\frac{2^{n}}{(1 + i)^{2 n}} + \frac{(1 + i)^{2 n}}{2^{n}}, n \in I

{(1 + i)}^{2 n} + {(1 - i)}^{2 n} = 2^{(n + 1)} c o s (n π / 2)

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

n ϵ Z

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