Question Number 120324 by mathocean1 last updated on 30/Oct/20
$$\mathrm{we}\:\mathrm{are}\:\mathrm{in}\:\mathbb{C}. \\ $$$$\left(\mathrm{E}\right):\:\mathrm{z}^{\mathrm{3}} +\left(\mathrm{4}−\mathrm{5i}\right)\mathrm{z}^{\mathrm{2}} +\left(\mathrm{8}−\mathrm{20i}\right)\mathrm{z}−\mathrm{40i}=\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:\mathrm{Show}\:\mathrm{that}\:\left(\mathrm{E}\right)\:\mathrm{has}\:\mathrm{one}\:\mathrm{imaginary}\:\mathrm{pure}\:\mathrm{root} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{solve}\:\left(\mathrm{E}\right) \\ $$
Answered by Olaf last updated on 31/Oct/20
![(1) Let z = iy z^3 +(4−5i)z^2 +(8−20i)z−40i = −iy^3 −(4−5i)y^2 +(20+8i)y−40i = −4y^2 +20y+i(−y^3 +5y^2 +8y−40) = −4y(y−5)−i(y^3 −5y^2 −8y+40) If y = 5 : −4y(y−5) = 0 and y^3 −5y^2 −8y+40 = 125−125−40+40 = 0 ⇒ 5i is a root of the polynome z^3 +(4−5i)z^2 +(8−20i)z−40i z^3 +(4−5i)z^2 +(8−20i)z−40i = (z−5i)(z^2 +wz+8) By identification : { ((w−5i = 4−5i)),((−5wi+8 = 8−20i)) :} ⇒ w = 4 z^3 +(4−5i)z^2 +(8−20i)z−40i = (z−5i)(z^2 +4z+8) = (z−5i)[(z+2)^2 +4] (z+2)^2 +4 = 0 ⇔ z = −1±i ⇒ S = {−1−i ; −1+i ; 5i }](https://www.tinkutara.com/question/Q120412.png)
$$\left(\mathrm{1}\right) \\ $$$$\mathrm{Let}\:{z}\:=\:{iy} \\ $$$${z}^{\mathrm{3}} +\left(\mathrm{4}−\mathrm{5}{i}\right){z}^{\mathrm{2}} +\left(\mathrm{8}−\mathrm{20}{i}\right){z}−\mathrm{40}{i} \\ $$$$=\:−{iy}^{\mathrm{3}} −\left(\mathrm{4}−\mathrm{5}{i}\right){y}^{\mathrm{2}} +\left(\mathrm{20}+\mathrm{8}{i}\right){y}−\mathrm{40}{i} \\ $$$$=\:−\mathrm{4}{y}^{\mathrm{2}} +\mathrm{20}{y}+{i}\left(−{y}^{\mathrm{3}} +\mathrm{5}{y}^{\mathrm{2}} +\mathrm{8}{y}−\mathrm{40}\right) \\ $$$$=\:−\mathrm{4}{y}\left({y}−\mathrm{5}\right)−{i}\left({y}^{\mathrm{3}} −\mathrm{5}{y}^{\mathrm{2}} −\mathrm{8}{y}+\mathrm{40}\right) \\ $$$$ \\ $$$$\mathrm{If}\:{y}\:=\:\mathrm{5}\:: \\ $$$$−\mathrm{4}{y}\left({y}−\mathrm{5}\right)\:=\:\mathrm{0} \\ $$$$\mathrm{and}\:{y}^{\mathrm{3}} −\mathrm{5}{y}^{\mathrm{2}} −\mathrm{8}{y}+\mathrm{40}\:=\:\mathrm{125}−\mathrm{125}−\mathrm{40}+\mathrm{40}\:=\:\mathrm{0} \\ $$$$ \\ $$$$\Rightarrow\:\mathrm{5}{i}\:\mathrm{is}\:\mathrm{a}\:\mathrm{root}\:\mathrm{of}\:\mathrm{the}\:\mathrm{polynome} \\ $$$${z}^{\mathrm{3}} +\left(\mathrm{4}−\mathrm{5}{i}\right){z}^{\mathrm{2}} +\left(\mathrm{8}−\mathrm{20}{i}\right){z}−\mathrm{40}{i} \\ $$$$ \\ $$$${z}^{\mathrm{3}} +\left(\mathrm{4}−\mathrm{5}{i}\right){z}^{\mathrm{2}} +\left(\mathrm{8}−\mathrm{20}{i}\right){z}−\mathrm{40}{i} \\ $$$$=\:\left({z}−\mathrm{5}{i}\right)\left({z}^{\mathrm{2}} +{wz}+\mathrm{8}\right) \\ $$$$ \\ $$$$\mathrm{By}\:\mathrm{identification}\:: \\ $$$$\begin{cases}{{w}−\mathrm{5}{i}\:=\:\mathrm{4}−\mathrm{5}{i}}\\{−\mathrm{5}{wi}+\mathrm{8}\:=\:\mathrm{8}−\mathrm{20}{i}}\end{cases} \\ $$$$\Rightarrow\:{w}\:=\:\mathrm{4} \\ $$$$ \\ $$$${z}^{\mathrm{3}} +\left(\mathrm{4}−\mathrm{5}{i}\right){z}^{\mathrm{2}} +\left(\mathrm{8}−\mathrm{20}{i}\right){z}−\mathrm{40}{i} \\ $$$$=\:\left({z}−\mathrm{5}{i}\right)\left({z}^{\mathrm{2}} +\mathrm{4}{z}+\mathrm{8}\right) \\ $$$$=\:\left({z}−\mathrm{5}{i}\right)\left[\left({z}+\mathrm{2}\right)^{\mathrm{2}} +\mathrm{4}\right] \\ $$$$\left({z}+\mathrm{2}\right)^{\mathrm{2}} +\mathrm{4}\:=\:\mathrm{0}\:\Leftrightarrow\:{z}\:=\:−\mathrm{1}\pm{i} \\ $$$$ \\ $$$$\Rightarrow\:\mathcal{S}\:=\:\left\{−\mathrm{1}−{i}\:;\:−\mathrm{1}+{i}\:;\:\mathrm{5}{i}\:\right\} \\ $$