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Question Number 120324 by mathocean1 last updated on 30/Oct/20
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)
weareinC.(E):z3+(4−5i)z2+(8−20i)z−40i=01)Showthat(E)hasoneimaginarypureroot2)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)Letz=iyz3+(4−5i)z2+(8−20i)z−40i=−iy3−(4−5i)y2+(20+8i)y−40i=−4y2+20y+i(−y3+5y2+8y−40)=−4y(y−5)−i(y3−5y2−8y+40)Ify=5:−4y(y−5)=0andy3−5y2−8y+40=125−125−40+40=0⇒5iisarootofthepolynomez3+(4−5i)z2+(8−20i)z−40iz3+(4−5i)z2+(8−20i)z−40i=(z−5i)(z2+wz+8)Byidentification:{w−5i=4−5i−5wi+8=8−20i⇒w=4z3+(4−5i)z2+(8−20i)z−40i=(z−5i)(z2+4z+8)=(z−5i)[(z+2)2+4](z+2)2+4=0⇔z=−1±i⇒S={−1−i;−1+i;5i}

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