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Question Number 70914 by 20190927 last updated on 09/Oct/19

1+(z+2i)+(z+2i)^2 +(z+2i)^3 +(z+2i)^4 =0  find z , z∈C

1+(z+2i)+(z+2i)2+(z+2i)3+(z+2i)4=0findz,zC

Commented by mathmax by abdo last updated on 09/Oct/19

let x=z+2i  (e) ⇔1+x+x^2 +x^3 +x^4 =0 ⇒((1−x^5 )/(1−x)) =0 and x≠1 ⇒  x^5 =1 and x≠1  butx^5 =1 ⇒x^5 =e^(i2kπ)  ⇒x_k =e^((i2kπ)/5)   and k∈[[1,4]] so the roots of (e) are z_k =x_k −2i  z_k =e^((i2kπ)/5) −2i  =cos(((2kπ)/5))+i(sin(((2kπ)/5))−2)=r_k e^(iθ_k )   with r_k =(√(cos(((2kπ)/5))^2  +(sin(((2kπ)/5))−2)^2 ))  and θ_k =arctan{((sin(((2kπ)/5))−2)/(cos(((2kπ)/5))))}   1≤k≤4

letx=z+2i(e)1+x+x2+x3+x4=01x51x=0andx1x5=1andx1butx5=1x5=ei2kπxk=ei2kπ5andk[[1,4]]sotherootsof(e)arezk=xk2izk=ei2kπ52i=cos(2kπ5)+i(sin(2kπ5)2)=rkeiθkwithrk=cos(2kπ5)2+(sin(2kπ5)2)2andθk=arctan{sin(2kπ5)2cos(2kπ5)}1k4

Commented by 20190927 last updated on 10/Oct/19

thank you

thankyou

Commented by mathmax by abdo last updated on 10/Oct/19

you are welcome.

youarewelcome.

Answered by mind is power last updated on 09/Oct/19

⇒(z+2i −1).(1+(z+2i)+(z+2i)^2 +(z+2i)^3 +(z+2i)^4 )=(z+2i)^5 −1=0  ⇒(z+2i)^5 =e^(2ikπ)   ⇒z=e^((2ikπ)/5) −2i  k∈{0,....4}  k≠0 for our solution

(z+2i1).(1+(z+2i)+(z+2i)2+(z+2i)3+(z+2i)4)=(z+2i)51=0(z+2i)5=e2ikπz=e2ikπ52ik{0,....4}k0foroursolution

Commented by 20190927 last updated on 10/Oct/19

thank you

thankyou

Commented by mind is power last updated on 10/Oct/19

y′re welcom

yrewelcom

Answered by MJS last updated on 09/Oct/19

z=x−2i  x^4 +x^3 +x^2 +x+1=0  (x−1)(x^4 +x^3 +x^2 +x+1)=0  x^5 −1=0  ⇒ x_0 =1 not valid because we added it  x_1 =−(1/4)+((√5)/4)+((√(10+2(√5)))/4)i  x_2 =−(1/4)−((√5)/4)+((√(10−2(√5)))/4)i  x_3 =−(1/4)−((√5)/4)−((√(10−2(√5)))/4)i  x_4 =−(1/4)+((√5)/4)−((√(10+2(√5)))/4)i  z_i =x_i −2i

z=x2ix4+x3+x2+x+1=0(x1)(x4+x3+x2+x+1)=0x51=0x0=1notvalidbecauseweaddeditx1=14+54+10+254ix2=1454+10254ix3=145410254ix4=14+5410+254izi=xi2i

Commented by 20190927 last updated on 10/Oct/19

thank you

thankyou

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