we are in C. (E): z^3 +(4−5i)z^2 +(8−20i)z−40i=0 1) Show that (E) has one imaginary pure root 2) solve (E)


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Question Number 120324 by mathocean1 last updated on 30/Oct/20

$we are in C .$ $(E) : z^{3} + (4 - 5 i) z^{2} + (8 - 20 i) z - 40 i = 0$ $1) Show that (E) has one imaginary pure root$ $2) solve (E)$
	Answered by Olaf last updated on 31/Oct/20
	$(1) Let z = iy z^3 +(4−5i)z^2 +(8−20i)z−40i = −iy^3 −(4−5i)y^2 +(20+8i)y−40i = −4y^2 +20y+i(−y^3 +5y^2 +8y−40) = −4y(y−5)−i(y^3 −5y^2 −8y+40) If y = 5 : −4y(y−5) = 0 and y^3 −5y^2 −8y+40 = 125−125−40+40 = 0 ⇒ 5i is a root of the polynome z^3 +(4−5i)z^2 +(8−20i)z−40i z^3 +(4−5i)z^2 +(8−20i)z−40i = (z−5i)(z^2 +wz+8) By identification : { ((w−5i = 4−5i)),((−5wi+8 = 8−20i)) :} ⇒ w = 4 z^3 +(4−5i)z^2 +(8−20i)z−40i = (z−5i)(z^2 +4z+8) = (z−5i)[(z+2)^2 +4] (z+2)^2 +4 = 0 ⇔ z = −1±i ⇒ S = {−1−i ; −1+i ; 5i }$
	$(1)$ $Let z = i y$ $z^{3} + (4 - 5 i) z^{2} + (8 - 20 i) z - 40 i$ $= - {i y}^{3} - (4 - 5 i) y^{2} + (20 + 8 i) y - 40 i$ $= - 4 y^{2} + 20 y + i (- y^{3} + 5 y^{2} + 8 y - 40)$ $= - 4 y (y - 5) - i (y^{3} - 5 y^{2} - 8 y + 40)$ $If y = 5 :$ $- 4 y (y - 5) = 0$ $and y^{3} - 5 y^{2} - 8 y + 40 = 125 - 125 - 40 + 40 = 0$ $\Rightarrow 5 i is a root of the polynome$ $z^{3} + (4 - 5 i) z^{2} + (8 - 20 i) z - 40 i$ $z^{3} + (4 - 5 i) z^{2} + (8 - 20 i) z - 40 i$ $= (z - 5 i) (z^{2} + w z + 8)$ $By identification :$ ${\begin{cases} w - 5 i = 4 - 5 i \\ - 5 w i + 8 = 8 - 20 i \end{cases}$ $\Rightarrow w = 4$ $z^{3} + (4 - 5 i) z^{2} + (8 - 20 i) z - 40 i$ $= (z - 5 i) (z^{2} + 4 z + 8)$ $= (z - 5 i) [{(z + 2)}^{2} + 4]$ ${(z + 2)}^{2} + 4 = 0 \Leftrightarrow z = - 1 \pm i$ $\Rightarrow S = {- 1 - i; - 1 + i; 5 i}$
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