If -3-in- 2 n - n- the set of integers- then n is a multiple of

The correct option is D 12 (√(3) i)^n =(2(√(3)2 i2))^n =(2e^i π6)^n =2^ne^i nπ6 Hence for n=12 , we get 2^ne^ i2π =2^n ...

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if √3-in= 2...

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if $(\sqrt{3} - i)^{n} = 2^{n},$ $n ϵ Z$ the set of integers, then $n$ is a multiple of

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n

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1 + ω^{n} + ω^{2 n} =

2^{n} - (n - 1) 2^{n - 2} + \frac{(n - 2) (n - 3)}{⌊ 2} - \frac{(n - 3) (n - 4) (n - 5)}{⌊ 3} 2^{n - 6} + . . . . .;

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1 - (n - 1) + \frac{(n - 2) (n - 2)}{⌊ 2} - \frac{(n - 3) (n - 4) (n - 5)}{⌊ 3} + . . . . = (- 1)^{n}

(\sqrt{3} - i)^{n} = 2^{n}, n \in N

n

z = - 1 + i \sqrt{3}

n

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α^{2 n} + 2^{n} α^{n} + 2^{2 n}

State whether each of the following sets is a finite set or an infinite set:

(i) The set of multiples of 8.

(ii) The set of integers less than 10.

(iii) The set of whole numbers less than 12.

(iv) ${x : x = 3 n - 2, n \in W, n \leq 8}$ (v) ${x : x = 3 n - 2, n \in Z, n \leq 8}$ (vi) ${x : x = \frac{n - 2}{n + 1}, n \in W}$

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