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Question Number 71564 by 20190927 last updated on 17/Oct/19

(z−i)^4 =−7+24i

(zi)4=7+24i

Commented by mathmax by abdo last updated on 18/Oct/19

we have ∣−7+24i∣=(√(7^2  +24^2 ))=(√(49+576))=(√(625))=25 ⇒  −7+24i =25 e^(−iarctan(((24)/7)))   and (z−i)^4  =25 e^(−iarctan(((24)/7)))  let   z−i=re^(iθ)  ⇒r^4 e^(i4θ) =25 e^(−iarctan(((24)/7)))  ⇒r=^4 (√(25))=(5^2 )^(1/4) =(√5)  4θ =−arctan(((24)/7))+2kπ ⇒θ_k =−((arctan(((24)/7)))/4) +((kπ)/2)  and k∈[[0,3]] so the roots ofthis equations are  Z_k =(√5)e^(i(((kπ)/2)−((arctan(((24)/7)))/4)))  +i   with 0≤k≤3

wehave7+24i∣=72+242=49+576=625=257+24i=25eiarctan(247)and(zi)4=25eiarctan(247)letzi=reiθr4ei4θ=25eiarctan(247)r=425=(52)14=54θ=arctan(247)+2kπθk=arctan(247)4+kπ2andk[[0,3]]sotherootsofthisequationsareZk=5ei(kπ2arctan(247)4)+iwith0k3

Commented by 20190927 last updated on 21/Oct/19

thank you

thankyou

Answered by MJS last updated on 17/Oct/19

x^4 =y ⇒  x_(1, 2) =±(y)^(1/4)   x_(3, 4) =±i(y)^(1/4)     ((−7+24i))^(1/4) =2+i  ⇒ z−i=±(2+i) ∨ z−i=±i(2+i)  ⇒ z=−2∨z=2+2i∨−1+3i∨1−i

x4=yx1,2=±y4x3,4=±iy47+24i4=2+izi=±(2+i)zi=±i(2+i)z=2z=2+2i1+3i1i

Commented by 20190927 last updated on 21/Oct/19

thank you

thankyou

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