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Question Number 112746 by bemath last updated on 09/Sep/20
If z = −i is the root of the equation z^3 +k.z^2 +(8+2(√2) i)z+8i =0 find the other roots
$$\mathrm{If}\:\mathrm{z}\:=\:−\mathrm{i}\:\mathrm{is}\:\mathrm{the}\:\mathrm{root}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{z}^{\mathrm{3}} +\mathrm{k}.\mathrm{z}^{\mathrm{2}} +\left(\mathrm{8}+\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{i}\right)\mathrm{z}+\mathrm{8i}\:=\mathrm{0} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{other}\:\mathrm{roots} \\ $$
Commented by malwan last updated on 09/Sep/20
the equation should be z^3 + kz^2 + (8 + 2(√2) i )z + 8i = 0 you posted it again as image
$${the}\:{equation}\:{should}\:{be} \\ $$$${z}^{\mathrm{3}} \:+\:{kz}^{\mathrm{2}} \:+\:\left(\mathrm{8}\:+\:\mathrm{2}\sqrt{\mathrm{2}}\:{i}\:\right){z}\:+\:\mathrm{8}{i}\:=\:\mathrm{0} \\ $$$$\:{you}\:{posted}\:{it}\:{again}\:{as}\:{image}\: \\ $$
Commented by bemath last updated on 09/Sep/20
oo yes
$$\mathrm{oo}\:\mathrm{yes} \\ $$
Answered by bobhans last updated on 09/Sep/20
z=−i ⇒ (−i)^3 +k(−i)^2 +(8+2(√2) i)(−i)+8i=0 ⇒i −k −8i+2(√2) +8i = 0 ⇒ k = 2(√2) +i By Horner′s rule z^3 +(2(√2) +i)z^2 +(8+2(√2) i)z +8i= 0 (z+i)(z^2 +2(√2) z+8) = 0 z_1 =−i, z_(2,3) = ((−2(√2) ±(√(8−32)))/2) z_2 = ((−2(√2)+2(√6) i)/2) = −(√2) +(√6) i ∣z_2 ∣ = (√8) = 2(√2) →z_2 = 2(√2)(−(1/2)+((√3)/2) i) z_2 = 2(√2) e^(((2π)/3)i) z_3 =((−2(√2)−2(√6) i)/2) = −(√2)−(√6) i z_3 = 2(√2) e^(−((2π)/3)i )
$$\mathrm{z}=−\mathrm{i}\:\Rightarrow\:\left(−\mathrm{i}\right)^{\mathrm{3}} +\mathrm{k}\left(−\mathrm{i}\right)^{\mathrm{2}} +\left(\mathrm{8}+\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{i}\right)\left(−\mathrm{i}\right)+\mathrm{8i}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{i}\:−\mathrm{k}\:−\mathrm{8i}+\mathrm{2}\sqrt{\mathrm{2}}\:+\mathrm{8i}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{k}\:=\:\mathrm{2}\sqrt{\mathrm{2}}\:+\mathrm{i} \\ $$$$\mathrm{By}\:\mathrm{Horner}'\mathrm{s}\:\mathrm{rule}\: \\ $$$$\:\mathrm{z}^{\mathrm{3}} +\left(\mathrm{2}\sqrt{\mathrm{2}}\:+\mathrm{i}\right)\mathrm{z}^{\mathrm{2}} +\left(\mathrm{8}+\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{i}\right)\mathrm{z}\:+\mathrm{8i}=\:\mathrm{0} \\ $$$$\left(\mathrm{z}+\mathrm{i}\right)\left(\mathrm{z}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{z}+\mathrm{8}\right)\:=\:\mathrm{0} \\ $$$$\mathrm{z}_{\mathrm{1}} =−\mathrm{i},\:\mathrm{z}_{\mathrm{2},\mathrm{3}} \:=\:\frac{−\mathrm{2}\sqrt{\mathrm{2}}\:\pm\sqrt{\mathrm{8}−\mathrm{32}}}{\mathrm{2}} \\ $$$$\mathrm{z}_{\mathrm{2}} \:=\:\frac{−\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{2}\sqrt{\mathrm{6}}\:\mathrm{i}}{\mathrm{2}}\:=\:−\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{6}}\:\mathrm{i} \\ $$$$\mid\mathrm{z}_{\mathrm{2}} \mid\:=\:\sqrt{\mathrm{8}}\:=\:\mathrm{2}\sqrt{\mathrm{2}}\:\rightarrow\mathrm{z}_{\mathrm{2}} \:=\:\mathrm{2}\sqrt{\mathrm{2}}\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{i}\right) \\ $$$$\mathrm{z}_{\mathrm{2}} \:=\:\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{e}^{\frac{\mathrm{2}\pi}{\mathrm{3}}\mathrm{i}} \\ $$$$\mathrm{z}_{\mathrm{3}} =\frac{−\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{2}\sqrt{\mathrm{6}}\:\mathrm{i}}{\mathrm{2}}\:=\:−\sqrt{\mathrm{2}}−\sqrt{\mathrm{6}}\:\mathrm{i} \\ $$$$\mathrm{z}_{\mathrm{3}} =\:\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{e}^{−\frac{\mathrm{2}\pi}{\mathrm{3}}\mathrm{i}\:} \\ $$
Commented by malwan last updated on 09/Sep/20
z_3 =−(√2)−(√6)i=[2(√(2 )); ((4π)/3)]=2(√2)e^(((4π)/3)i) or z_3 =−((√2)+(√6)i)=−[2(√2);(π/3)] =[2(√2); (π/3) + π]=[2(√2) ; ((4π)/3)]=2(√2)e^(((4π)/3)i)
$${z}_{\mathrm{3}} =−\sqrt{\mathrm{2}}−\sqrt{\mathrm{6}}{i}=\left[\mathrm{2}\sqrt{\mathrm{2}\:};\:\frac{\mathrm{4}\pi}{\mathrm{3}}\right]=\mathrm{2}\sqrt{\mathrm{2}}{e}^{\frac{\mathrm{4}\pi}{\mathrm{3}}{i}} \\ $$$${or}\:{z}_{\mathrm{3}} =−\left(\sqrt{\mathrm{2}}+\sqrt{\mathrm{6}}{i}\right)=−\left[\mathrm{2}\sqrt{\mathrm{2}};\frac{\pi}{\mathrm{3}}\right] \\ $$$$=\left[\mathrm{2}\sqrt{\mathrm{2}};\:\frac{\pi}{\mathrm{3}}\:+\:\pi\right]=\left[\mathrm{2}\sqrt{\mathrm{2}}\:;\:\frac{\mathrm{4}\pi}{\mathrm{3}}\right]=\mathrm{2}\sqrt{\mathrm{2}}{e}^{\frac{\mathrm{4}\pi}{\mathrm{3}}{i}} \\ $$
Answered by floor(10²Eta[1]) last updated on 09/Sep/20
if −i is a root: (−i)^3 +k(−i)^2 +(8+2(√2))(−i)+8i=0 =i−k−8i+8i−2(√2)i⇒k=(1−2(√2))i f(z)=z^3 +(1−2(√2))iz^2 +(8+2(√2))z+8i =(z+i)(z^2 +az+8)=z^3 +(a+i)z^2 +(8+ai)z+8i { ((a+i=(1−2(√2))i)),((8+ai=8+2(√2))) :} ⇒a=−2(√2)i f(z)=(z+i)(z^2 −2(√2)iz+8)=0 z^2 −2(√2)iz+8=0 z=i(√2)±i(√(10)) the roots are: −i, i(√2)+i(√(10)), i(√2)−i(√(10))
$$\mathrm{if}\:−\mathrm{i}\:\mathrm{is}\:\mathrm{a}\:\mathrm{root}: \\ $$$$\left(−\mathrm{i}\right)^{\mathrm{3}} +\mathrm{k}\left(−\mathrm{i}\right)^{\mathrm{2}} +\left(\mathrm{8}+\mathrm{2}\sqrt{\mathrm{2}}\right)\left(−\mathrm{i}\right)+\mathrm{8i}=\mathrm{0} \\ $$$$=\mathrm{i}−\mathrm{k}−\mathrm{8i}+\mathrm{8i}−\mathrm{2}\sqrt{\mathrm{2}}\mathrm{i}\Rightarrow\mathrm{k}=\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{2}}\right)\mathrm{i} \\ $$$$\mathrm{f}\left(\mathrm{z}\right)=\mathrm{z}^{\mathrm{3}} +\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{2}}\right)\mathrm{iz}^{\mathrm{2}} +\left(\mathrm{8}+\mathrm{2}\sqrt{\mathrm{2}}\right)\mathrm{z}+\mathrm{8i} \\ $$$$=\left(\mathrm{z}+\mathrm{i}\right)\left(\mathrm{z}^{\mathrm{2}} +\mathrm{az}+\mathrm{8}\right)=\mathrm{z}^{\mathrm{3}} +\left(\mathrm{a}+\mathrm{i}\right)\mathrm{z}^{\mathrm{2}} +\left(\mathrm{8}+\mathrm{ai}\right)\mathrm{z}+\mathrm{8i} \\ $$$$\begin{cases}{\mathrm{a}+\mathrm{i}=\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{2}}\right)\mathrm{i}}\\{\mathrm{8}+\mathrm{ai}=\mathrm{8}+\mathrm{2}\sqrt{\mathrm{2}}}\end{cases} \\ $$$$\Rightarrow\mathrm{a}=−\mathrm{2}\sqrt{\mathrm{2}}\mathrm{i} \\ $$$$\mathrm{f}\left(\mathrm{z}\right)=\left(\mathrm{z}+\mathrm{i}\right)\left(\mathrm{z}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{2}}\mathrm{iz}+\mathrm{8}\right)=\mathrm{0} \\ $$$$\mathrm{z}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{2}}\mathrm{iz}+\mathrm{8}=\mathrm{0} \\ $$$$\mathrm{z}=\mathrm{i}\sqrt{\mathrm{2}}\pm\mathrm{i}\sqrt{\mathrm{10}} \\ $$$$\mathrm{the}\:\mathrm{roots}\:\mathrm{are}: \\ $$$$−\mathrm{i},\:\mathrm{i}\sqrt{\mathrm{2}}+\mathrm{i}\sqrt{\mathrm{10}},\:\mathrm{i}\sqrt{\mathrm{2}}−\mathrm{i}\sqrt{\mathrm{10}} \\ $$
Commented by bemath last updated on 09/Sep/20
Commented by bemath last updated on 09/Sep/20
sir how to change your answer to choice answer?
$$\mathrm{sir}\:\mathrm{how}\:\mathrm{to}\:\mathrm{change}\:\mathrm{your}\:\mathrm{answer}\:\mathrm{to}\:\mathrm{choice} \\ $$$$\mathrm{answer}? \\ $$
Commented by floor(10²Eta[1]) last updated on 09/Sep/20
all answers are wrong the correct one should be: ((√2)+(√(10)))e^((iπ)/2) and ((√2)−(√(10)))e^((iπ)/2)
$$\mathrm{all}\:\mathrm{answers}\:\mathrm{are}\:\mathrm{wrong}\:\mathrm{the}\:\mathrm{correct}\:\mathrm{one} \\ $$$$\mathrm{should}\:\mathrm{be}: \\ $$$$\left(\sqrt{\mathrm{2}}+\sqrt{\mathrm{10}}\right)\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{2}}} \:\mathrm{and}\:\left(\sqrt{\mathrm{2}}−\sqrt{\mathrm{10}}\right)\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{2}}} \\ $$
Answered by malwan last updated on 09/Sep/20
z=−i⇒ k = 2(√2)+ i using Hornors rule we get z^2 +2(√2)z+8=0 z=((−2(√2)±(√(8−32)))/2) =((−2(√2)±2(√6)i)/2) =2(√2)(− (1/2) ± ((√3)/2) i) z_1 = −i z_2 = [2(√2) ; 0][1 ; ((2π)/3)]=[2(√2) ; ((2π)/3)]=2(√2)e^(i((2π)/3)) z_3 =[2(√2) ; 0][1;((4π)/3)]=[2(√2);((4π)/3)]=2(√2)e^(i((4π)/3))
$${z}=−{i}\Rightarrow\:{k}\:=\:\mathrm{2}\sqrt{\mathrm{2}}+\:{i} \\ $$$${using}\:{Hornors}\:{rule}\:{we}\:{get} \\ $$$${z}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{2}}{z}+\mathrm{8}=\mathrm{0} \\ $$$${z}=\frac{−\mathrm{2}\sqrt{\mathrm{2}}\pm\sqrt{\mathrm{8}−\mathrm{32}}}{\mathrm{2}}\:=\frac{−\mathrm{2}\sqrt{\mathrm{2}}\pm\mathrm{2}\sqrt{\mathrm{6}}{i}}{\mathrm{2}} \\ $$$$=\mathrm{2}\sqrt{\mathrm{2}}\left(−\:\frac{\mathrm{1}}{\mathrm{2}}\:\pm\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{i}\right) \\ $$$${z}_{\mathrm{1}} \:=\:−{i} \\ $$$${z}_{\mathrm{2}} \:=\:\left[\mathrm{2}\sqrt{\mathrm{2}}\:;\:\mathrm{0}\right]\left[\mathrm{1}\:;\:\frac{\mathrm{2}\pi}{\mathrm{3}}\right]=\left[\mathrm{2}\sqrt{\mathrm{2}}\:;\:\frac{\mathrm{2}\pi}{\mathrm{3}}\right]=\mathrm{2}\sqrt{\mathrm{2}}{e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \\ $$$${z}_{\mathrm{3}} \:=\left[\mathrm{2}\sqrt{\mathrm{2}}\:;\:\mathrm{0}\right]\left[\mathrm{1};\frac{\mathrm{4}\pi}{\mathrm{3}}\right]=\left[\mathrm{2}\sqrt{\mathrm{2}};\frac{\mathrm{4}\pi}{\mathrm{3}}\right]=\mathrm{2}\sqrt{\mathrm{2}}{e}^{{i}\frac{\mathrm{4}\pi}{\mathrm{3}}} \\ $$
Commented by bemath last updated on 09/Sep/20
thank you sir. but z_3 =z_2 ^− z_3 =2(√2) e^(−((2π)/3)i )
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{but}\:\mathrm{z}_{\mathrm{3}} =\overset{−} {\mathrm{z}}_{\mathrm{2}} \\ $$$$\mathrm{z}_{\mathrm{3}} =\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{e}^{−\frac{\mathrm{2}\pi}{\mathrm{3}}\mathrm{i}\:} \\ $$
Commented by malwan last updated on 09/Sep/20
If z=[r;θ]⇒z^(−) =[r;2π−θ]=[r;−θ] but Ifz=[r;π−θ]⇒z^(−) =[r;2π−(π−θ)] =[r;π+θ] so z=[2(√2);((2π)/3)]⇒z^(−) =[2(√2);((4π)/3)]
$${If}\:{z}=\left[{r};\theta\right]\Rightarrow\overline {{z}}=\left[{r};\mathrm{2}\pi−\theta\right]=\left[{r};−\theta\right] \\ $$$${but}\:{Ifz}=\left[{r};\pi−\theta\right]\Rightarrow\overline {{z}}=\left[{r};\mathrm{2}\pi−\left(\pi−\theta\right)\right] \\ $$$$=\left[{r};\pi+\theta\right] \\ $$$${so}\:{z}=\left[\mathrm{2}\sqrt{\mathrm{2}};\frac{\mathrm{2}\pi}{\mathrm{3}}\right]\Rightarrow\overline {{z}}=\left[\mathrm{2}\sqrt{\mathrm{2}};\frac{\mathrm{4}\pi}{\mathrm{3}}\right] \\ $$

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