If 1- - -2 be the three cube roots of unity- then 1- -n-12 n-11-2n-

Given (1+ω)∏_n=1^2n 1(1+ω^2^n) Expanding, we get (1+ω)(1+ω^2)(1+ω^4)(1+ω^8) ⋯(1+ω^2^2n 1) upto 2 n terms (1+ω)(1+ω^2)(1+ω)(1+ω^2)⋯⋯(1+ω^2) upto 2n terms Using p ...

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Standard XII

Mathematics

Cube Root of a Complex Number

If 1, ω, ω2 b...

Question

If $1, ω, ω^{2}$ be the three cube roots of unity, then
$(1 + ω) 2 n - 1 \prod n = 1 (1 + ω^{2^{n}}) =$

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Solution

Given $(1 + ω) 2 n - 1 \prod n = 1 (1 + ω^{2^{n}})$
Expanding, we get $(1 + ω) (1 + ω^{2}) (1 + ω^{4}) (1 + ω^{8}) \dots (1 + ω^{2^{2 n - 1}})$ to $2 n$ factors
$(1 + ω) (1 + ω^{2}) (1 + ω) (1 + ω^{2}) \dots \dots (1 + ω^{2})$ upto $2 n$ factors
Using property $1 + ω + ω^{2} = 0$
$= (- ω^{2}) (- ω) (- ω^{2}) (- ω) \dots 2 n terms$
$= 1 [∵ ω^{3} = 1]$

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Cube Root of a Complex Number

Standard XII Mathematics

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If 1,ω,ω2 be the three cube roots of unity, then (1+ω)2n−1∏n=1(1+ω2n)=

Given (1+ω)2n−1∏n=1(1+ω2n) Expanding, we get (1+ω)(1+ω2)(1+ω4)(1+ω8)⋯(1+ω22n−1) to 2n factors (1+ω)(1+ω2)(1+ω)(1+ω2)⋯⋯(1+ω2) upto 2n factors Using property 1+ω+ω2=0 =(−ω2)(−ω)(−ω2)(−ω)⋯2n terms =1 [∵ω3=1]

If $1, ω, ω^{2}$ be the three cube roots of unity, then
$(1 + ω) 2 n - 1 \prod n = 1 (1 + ω^{2^{n}}) =$