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Question Number 46383 by peter frank last updated on 24/Oct/18

Commented by maxmathsup by imad last updated on 24/Oct/18

z=0 is not solution for z≠0 (e) ⇔ (((z+1)/z))^n =1 ⇔(1+z^(−1) )^n =1 the roots of Z^n =1 are the compolex Z_k =e^(i((2kπ)/n)) with k∈[[1,n−1]] so the solutions for the equation are z_k / 1+z_k ^(−1) =Z_k ⇒ ((z_k +1)/z_k )=Z_k ⇒ z_k +1 =Z_k z_k ⇒ (1−Z_k )z_k =−1 ⇒z_k =−(1/(1−Z_k )) =−(1/(1−cos(((2kπ)/n))−isin(((2kπ)/n)))) = ((−1)/(2sin^2 (((kπ)/n))−2i sin(((kπ)/n))cos(((kπ)/n)))) = ((−1)/(−2i sin(((kπ)/n))e^(i((kπ)/n)) )) = ((−i)/(2sin(((kπ)/n)))) e^(−((ikπ)/n)) =((−i)/(2sin(((kπ)/n)))){cos(((kπ)/n))−isin(((kπ)/n))} =−(1/2) −(i/2) cotan(((kπ)/n)) =−(1/2)(1+cotan(((kπ)/n))} with k∈[[1,n−1]] .

z=0isnotsolutionforz≠0(e)⇔(z+1z)n=1⇔(1+z−1)n=1therootsofZn=1arethecompolexZk=ei2kπnwithk∈[[1,n−1]]sothesolutionsfortheequationarezk/1+zk−1=Zk⇒zk+1zk=Zk⇒zk+1=Zkzk⇒(1−Zk)zk=−1⇒zk=−11−Zk=−11−cos(2kπn)−isin(2kπn)=−12sin2(kπn)−2isin(kπn)cos(kπn)=−1−2isin(kπn)eikπn=−i2sin(kπn)e−ikπn=−i2sin(kπn){cos(kπn)−isin(kπn)}=−12−i2cotan(kπn)=−12(1+cotan(kπn)}withk∈[[1,n−1]].

Commented by maxmathsup by imad last updated on 24/Oct/18

2 ) (z−1)^n +z^n =0 ⇔ (z−1)^n =−z^n ⇔ (((z−1)/z))^n =−1 =e^(i(2k+1)π) ⇒ ((z_k −1)/z_k ) = e^(i(((2k+1)π)/n)) ⇒ 1−z_k ^(−1) = e^(i((2k+1)/n)π) ⇒ 1−e^(i(((2k+1)π)/n)) =z_k ^(−1) ⇒ z_k = (1/(1−cos((((2k+1)π)/n))−i sin((((2k+1)π)/n))cos((((2k+1)π)/n)))) = (1/(2sin^2 ((((2k+1)π)/(2n)))−2i sin((((2k+1)π)/(2n)))cos((((2k+1)π)/(2n))))) =(1/(−2i sin((((2k+1)π)/(2n))) e^(i(((2k+1)/(2n))π)) )) = (i/(2sin((((2k+1)π)/(2n))))){ cos((((2k+1)π)/(2n)))−isin((((2k+1)π)/(2n)))} =(1/2) +(i/2) cotan((((2k+1)π)/(2n))) with k∈[[0,n−1]]

2)(z−1)n+zn=0⇔(z−1)n=−zn⇔(z−1z)n=−1=ei(2k+1)π⇒zk−1zk=ei(2k+1)πn⇒1−zk−1=ei2k+1nπ⇒1−ei(2k+1)πn=zk−1⇒zk=11−cos((2k+1)πn)−isin((2k+1)πn)cos((2k+1)πn)=12sin2((2k+1)π2n)−2isin((2k+1)π2n)cos((2k+1)π2n)=1−2isin((2k+1)π2n)ei(2k+12nπ)=i2sin((2k+1)π2n){cos((2k+1)π2n)−isin((2k+1)π2n)}=12+i2cotan((2k+1)π2n)withk∈[[0,n−1]]

Commented by maxmathsup by imad last updated on 24/Oct/18

the roots of (z−1)^5 +z^5 =0 are z_k =(1/2){1+i cotan((((2k+1)π)/(10))) with k∈[[0,4]] z_0 =(1/2){1+icotan((π/(10)))} , z_1 =(1/2){1+i cotan(((3π)/(10)))} z_2 =(1/2){1+i cotan((π/2))}dont exist and not solution z_3 =(1/2){1+i cotan(((7π)/(10)))} z_4 =(1/2){1+i cotan(((9π)/(10)))} also z =(1/2) is solution when cotan((((2k+1)π)/(10)))=0

therootsof(z−1)5+z5=0arezk=12{1+icotan((2k+1)π10)withk∈[[0,4]]z0=12{1+icotan(π10)},z1=12{1+icotan(3π10)}z2=12{1+icotan(π2)}dontexistandnotsolutionz3=12{1+icotan(7π10)}z4=12{1+icotan(9π10)}alsoz=12issolutionwhencotan((2k+1)π10)=0

Answered by Smail last updated on 25/Oct/18

a) (z+1)^n =z^n z+1=ze^((2ikπ)/n) z(e^((2ikπ)/n) −1)=1 z=(1/(e^((2ikπ)/n) −1))=(1/(cos(((2kπ)/n))−1+isin(((2ikπ)/n)))) =(1/(2cos^2 (((kπ)/n))−1−1+i(2sin(((kπ)/n))cos(((kπ)/n))))) =(1/(2(cos^2 (((kπ)/n))−1+isin(((kπ)/n))cos(((kπ)/n))))) =−(1/(2(sin^2 (((kπ)/n))−isin(((kπ)/n))cos(((kπ)/n))))) =−(1/(2sin(((kπ)/n))(sin(((kπ)/n))−icos(((kπ)/n))))) =−((sin(((kπ)/n))+icos(((kπ)/n)))/(2sin(((kπ)/n))))=((−1)/2)(1+icot(((kπ)/n))) z=−(1/2)(1+icot(((kπ)/n))) b) (z−1)^n =−z^n =z^n e^(i(π+2kπ)) z−1=ze^(iπ((2k+1)/n)) ⇔z(e^(iπ((2k+1)/n)) −1)=−1 z=((−1)/(cos(((2k+1)/n)π)−1+isin(((2k+1)/n)π))) =−((cos(((2k+1)/n)π)−1−isin(((2k+1)/n)π))/((cos(((2k+1)/n)π)−1)^2 +sin^2 (((2k+1)/n)π))) =−((cos(((2k+1)/n)π)−1−isin(((2k+1)/n)π))/(cos^2 (((2k+1)/n)π)+1−2cos(((2k+1)/n)π)+sin^2 (((2k+1)/n)π))) =−((cos(((2k+1)/n)π)−1−isin(((2k+1)/n)π))/(2−2cos(((2k+1)/n)π))) =(1/2)(1+i((sin(((2k+1)/n)π))/(1−cos(((2k+1)/n)π)))) =(1/2)(1+i((2sin(((2k+1)/(2n))π)cos(((2k+1)/(2n))π))/(1−2cos^2 (((2k+1)/(2n))π)+1))) =(1/2)(1+i((sin(((2k+1)/(2n))π)cos(((2k+1)/(2n))π))/(1−cos^2 (((2k+1)/(2n))π)))) =(1/2)(1−i((sin(((2k+1)/(2n))π)cos(((2k+1)/(2n))π))/(sin^2 (((2k+1)/(2n))π)))) z=(1/2)(1−icot(((2k+1)/(2n))π)) When n=5 z_k =(1/2)(1−icot(((2k+1)/(10))π) with k=(0,1,2,3,4)

a)(z+1)n=znz+1=ze2ikπnz(e2ikπn−1)=1z=1e2ikπn−1=1cos(2kπn)−1+isin(2ikπn)=12cos2(kπn)−1−1+i(2sin(kπn)cos(kπn))=12(cos2(kπn)−1+isin(kπn)cos(kπn))=−12(sin2(kπn)−isin(kπn)cos(kπn))=−12sin(kπn)(sin(kπn)−icos(kπn))=−sin(kπn)+icos(kπn)2sin(kπn)=−12(1+icot(kπn))z=−12(1+icot(kπn))b)(z−1)n=−zn=znei(π+2kπ)z−1=zeiπ2k+1n⇔z(eiπ2k+1n−1)=−1z=−1cos(2k+1nπ)−1+isin(2k+1nπ)=−cos(2k+1nπ)−1−isin(2k+1nπ)(cos(2k+1nπ)−1)2+sin2(2k+1nπ)=−cos(2k+1nπ)−1−isin(2k+1nπ)cos2(2k+1nπ)+1−2cos(2k+1nπ)+sin2(2k+1nπ)=−cos(2k+1nπ)−1−isin(2k+1nπ)2−2cos(2k+1nπ)=12(1+isin(2k+1nπ)1−cos(2k+1nπ))=12(1+i2sin(2k+12nπ)cos(2k+12nπ)1−2cos2(2k+12nπ)+1)=12(1+isin(2k+12nπ)cos(2k+12nπ)1−cos2(2k+12nπ))=12(1−isin(2k+12nπ)cos(2k+12nπ)sin2(2k+12nπ))z=12(1−icot(2k+12nπ))Whenn=5zk=12(1−icot(2k+110π)withk=(0,1,2,3,4)

Commented by peter frank last updated on 26/Oct/18

thanks sir

thankssir

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