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Question Number 54991 by Tawa1 last updated on 15/Feb/19

Find all the roots of: z^4 + 16i = 0

Findalltherootsof:z4+16i=0

Commented by maxmathsup by imad last updated on 15/Feb/19

z^4 +16i =0 ⇔ z^4 =−16i let z =r e^(iθ) ⇒r^4 e^(i4θ) =16(−i)=16 e^(−((iπ)/2)) ⇒ r^4 =16 and 4θ =−(π/2) +2kπ ⇒r =2 and θ_k =−(π/8) +((kπ)/2) and k ∈[[0,3]] so the roots of this equation are z_k =2 e^(i(−(π/8)+((kπ)/2))) ⇒z_0 =2 e^(−((iπ)/8)) z_1 = 2 e^(i(((3π)/8))) , z_2 =2 e^(i( ((7π)/8))) , z_3 =2 e^(i(((11π)/8))) also we have z^4 +16i =(z−z_0 )(z−z_1 )(z−z_2 )(z−z_3 ).

z4+16i=0⇔z4=−16iletz=reiθ⇒r4ei4θ=16(−i)=16e−iπ2⇒r4=16and4θ=−π2+2kπ⇒r=2andθk=−π8+kπ2andk∈[[0,3]]sotherootsofthisequationarezk=2ei(−π8+kπ2)⇒z0=2e−iπ8z1=2ei(3π8),z2=2ei(7π8),z3=2ei(11π8)alsowehavez4+16i=(z−z0)(z−z1)(z−z2)(z−z3).

Commented by Tawa1 last updated on 16/Feb/19

God bless you sir.

Godblessyousir.

Commented by maxmathsup by imad last updated on 16/Feb/19

you are welcome sir .

youarewelcomesir.

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