Question Number 101921 by Farruxjano last updated on 05/Jul/20

Please, help  i  cannot solve problems  such as: x^4 =1, x^5 =1, or x^n =1 n∈N...

Commented byprakash jain last updated on 05/Jul/20

Search internet for roots of unity.  You will find lot of information.  to start with i=(√(−1))  e^(iθ) =cos θ+isin θ  1=cos 2πk+isin 2πk  1^(1/n) =e^((2πj)/n)   (j=0,....n−1)  as you can see (e^((2πj)/n) )^n =e^(2jπ) =1

Answered by 1549442205 last updated on 06/Jul/20

Applying the Mauvra′s formular   (cosϕ+isinϕ)^n =cosnϕ+isinnϕ   Since 1=cos2kπ+isin2kπ ,so  i)For the eqs.x^4 =1 we get x=1^(1/4)   ⇔x=(cos2kπ+isin2kπ)^(1/4) =cos((2kπ)/4)+isin((2kπ)/4)(k=0,1,2,3)  we get four different roots   ii)For the eqs.x^5 =1⇔x=1^(1/5) .We get  x=(cos2kπ+isin2kπ)^(1/5) ⇔x=cos((2kπ)/5)+isin((2kπ)/5)(k∈{0,1,2,3,4})  weget five different  roots  iii)For eqs.x^n =1⇔x=1^(1/n) .We get  x=(cos2kπ+isin2kπ)^(1/n) =cos((2kπ)/n)+isin((2kπ)/n)(n∈{0,1,2,...n−1})  weget n different roots  for k=n,n+1,n+2,... we get the repeated roots