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Question Number 48272 by gunawan last updated on 21/Nov/18

z^5 =32 find all root z

z5=32findallrootz

Commented by maxmathsup by imad last updated on 21/Nov/18

let z=r e^(iθ) so z^5 =32 ⇔r^5 e^(5iθ) =2^5 e^(i2kπ) ⇔r=2 and 5θ =2kπ ⇒ θ_k =((2kπ)/5) so the roots are z_k =2 e^(i((2kπ)/5)) and k∈[[0,4]] z_0 =2 ,z_1 =2 e^(i((2π)/5)) ,z_2 =2 e^(i((4π)/5)) , z_3 =2 e^(i((6π)/5)) , z_(4 ) =2 e^(i((8π)/5)) . z_1 =2(cos(((2π)/5))+isin(((2π)/5))) but cos(((2π)/5))=2cos^2 ((π/5))−1 =2(((1+(√5))/4))^2 −1 =2((6+2(√5))/(16)) −1 =((12+4(√5)−16)/(16)) =((4(√5)−4)/(16)) =(((√5)−1)/4) sin(((2π)/5))=(√(1−((((√5)−1)/4))^2 ))=(√(1−((6−2(√5))/(16))))=(√((10+2(√5))/(16)))=((√(10+2(√5)))/4) ⇒ z_1 =(((√5)−1)/2) +i((√(10+2(√5)))/2) z_2 =2 e^(i(π−(π/5))) =−2 e^(−i(π/5)) =−2(cos((π/5))−isin((π/5)))but cos((π/5))=((1+(√5))/4) sin((π/5))=(√(1−(((1+(√5))/4))^2 ))=(√(1−((6+2(√5))/(16))))=(√((10−2(√5))/(16)))=((√(10−2(√5)))/4) ⇒ z_2 =−((1+(√5))/2) +i((√(10−2(√5)))/2) z_3 =2 e^(iπ) e^(i(π/5)) =−2(cos((π/5))+isin((π/5))) =−((1+(√5))/2) −i((√(10−2(√5)))/2) z_4 =2 e^(i(2π−((2π)/5))) =2 e^(−i((2π)/5)) =2(cos(((2π)/5))−isin(((2π)/5))) =(((√5)−1)/2) −i((√(10+(√5)))/2)

letz=reiθsoz5=32⇔r5e5iθ=25ei2kπ⇔r=2and5θ=2kπ⇒θk=2kπ5sotherootsarezk=2ei2kπ5andk∈[[0,4]]z0=2,z1=2ei2π5,z2=2ei4π5,z3=2ei6π5,z4=2ei8π5.z1=2(cos(2π5)+isin(2π5))butcos(2π5)=2cos2(π5)−1=2(1+54)2−1=26+2516−1=12+45−1616=45−416=5−14sin(2π5)=1−(5−14)2=1−6−2516=10+2516=10+254⇒z1=5−12+i10+252z2=2ei(π−π5)=−2e−iπ5=−2(cos(π5)−isin(π5))butcos(π5)=1+54sin(π5)=1−(1+54)2=1−6+2516=10−2516=10−254⇒z2=−1+52+i10−252z3=2eiπeiπ5=−2(cos(π5)+isin(π5))=−1+52−i10−252z4=2ei(2π−2π5)=2e−i2π5=2(cos(2π5)−isin(2π5))=5−12−i10+52

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