Question Number 61651 by maxmathsup by imad last updated on 05/Jun/19
![let p(x) =(x+i(√3))^n +(x−i(√3))^n with x real 1) simlify p(x) 2) find the roots of P(x) 3)decompose inside C[x] p(x) 4) calculate ∫_0 ^1 p(x)dx](https://www.tinkutara.com/question/Q61651.png)
Commented by maxmathsup by imad last updated on 06/Jun/19
![1) we have P(x) =Σ_(k=0) ^n C_n ^k (i(√3))^k x^(n−k) +Σ_(k=0) ^n C_n ^k (−i(√3))^k x^(n−k) =Σ_(k=0) ^n C_n ^k { i^k +(−i)^k )3^(k/2) x^(n−k) =Σ_(k=2p) (...) +Σ_(k=2p+1) (...) =2Σ_(p=0) ^([(n/2)]) C_n ^(2p) (−1)^p 3^p x^(n−2p) = 2 Σ_(p=0) ^([(n/2)]) (−3)^p C_n ^(2p) x^(n−2p) another way with arctan we have P(x) = 2 Re((x+i(√3))^n ) but ∣x+i(√3)∣ =(√(x^2 +3)) ⇒ x+i(√3)=(√(x^2 +3))((x/( (√(x^2 +3)))) +i ((√3)/( (√(x^(2 ) +3))))) =r e^(iθ) ⇒r =(√(x^2 +3)) and tanθ =((√3)/x) ( we suppose x≠0) ⇒ θ arctan(((√3)/x)) ⇒x+i(√3)=(√(x^2 +3)) e^(iarctan(((√3)/x))) ⇒ (x+i(√3))^n =(x^2 +3)^(n/2) e^(in arctan(((√3)/x))) ⇒P(x) =2(x^2 +3)^(n/2) cos(narctan(((√3)/x)))](https://www.tinkutara.com/question/Q61702.png)
Commented by maxmathsup by imad last updated on 06/Jun/19
![2) P(x)=0 ⇔ (x−i(√3))^n =−(x+i(√3))^n ⇔(((x−i(√3))/(x+i(√3))))^n =−1 ⇔ Z^n =−1 with Z =((x−i(√3))/(x+i(√3))) ⇒Zx+i(√3)Z =x−i(√3) ⇒(Z−1)x =−i(√3)Z −i(√3) ⇒ (Z−1)x =−i(√3)(Z+1) ⇒x =i(√3)((1+Z)/(1−Z)) Z^n =−1 ⇒ Z^n =e^(i(2k+1)π) ⇒Z_k =e^(i(((2k+1)π)/n)) and k∈[[0,n−1]] ⇒the roots of P(x) are x_k =i(√3) ((1+e^((i(2k+1)π)/n) )/(1−e^((i(2k+1)π)/n) )) let skmplify x_k x_k =i(√3)((1+cos((2k+1)(π/n))+2i sin((2k+1)(π/(2n)))cos((2k+1)(π/(2n))))/(1−cos((2k+1)(π/n))−2i sin(2k+1)(π/(2n)))cos((2k+1)(π/(2n))))) =i(√3)((2cos^2 ((2k+1)(π/(2n))) +2i sin((2k+1)(π/(2n)))cos((2k+1)(π/(2n))))/(2sin^2 (l2k+1)(π/(2n))−2i sin(2k+1)(π/(2n))) cos(2k+1)(π/(2n)))) =i(√3)((cos(2k+1)(π/(2n))(e^(i(2k+1)(π/(2n))) ))/(−isin(2k+1)(π/(2n))( e^(i(2k+1)(π/(2n))) ))) =−(√3)cotan((2k+1)(π/(2n))) k∈[[0,n−1]] ×](https://www.tinkutara.com/question/Q61704.png)
Commented by maxmathsup by imad last updated on 06/Jun/19

Commented by maxmathsup by imad last updated on 06/Jun/19
![4) we have P(x) =2 Σ_(p=0) ^([(n/2)]) (−3)^p C_n ^(2p) x^(n−2p) ⇒ ∫_0 ^1 P(x)dx =2 Σ_(p=0) ^([(n/2)]) (−3)^p C_n ^(2p) [(1/(n−2p +1))x^(n−2p+1) ]_0 ^1 = 2 Σ_(p=0) ^([(n/2)]) (((−3)^p C_n ^(2p) )/(n−2p+1)) .](https://www.tinkutara.com/question/Q61712.png)