Menu Close

If-z-1-3-3-3-i-and-z-2-1-3-i-Find-z-1-3-z-2-6-

Tinku Tara 0 Comments
Pin




Question Number 203309 by hardmath last updated on 15/Jan/24
If z_1 = 3 + 3(√3) i and z_2 = -1 − (√3) i Find: (z_1 ^( 3) /z_2 ^( 6) ) = ?
$$\mathrm{If}\:\:\:\mathrm{z}_{\mathrm{1}} =\:\mathrm{3}\:+\:\mathrm{3}\sqrt{\mathrm{3}}\:\boldsymbol{\mathrm{i}}\:\:\:\mathrm{and}\:\:\:\mathrm{z}_{\mathrm{2}} =\:-\mathrm{1}\:−\:\sqrt{\mathrm{3}}\:\boldsymbol{\mathrm{i}} \\ $$$$\mathrm{Find}:\:\:\:\frac{\mathrm{z}_{\mathrm{1}} ^{\:\mathrm{3}} }{\mathrm{z}_{\mathrm{2}} ^{\:\mathrm{6}} }\:=\:? \\ $$
Answered by Rasheed.Sindhi last updated on 16/Jan/24
z_1 =3(1+(√3) i) ∧ z_2 =−(1+(√3) i) ⇒z_1 =−3z_2 (z_1 ^( 3) /z_2 ^( 6) ) =(((−3z_2 )^3 )/z_2 ^6 )=((−27z_2 ^3 )/z_2 ^6 )=((−27)/z_2 ^3 ) z_2 +1=−(√3) i⇒z_2 ^2 +2z_2 +1=−3 z_2 ^2 =−2z_2 −4⇒z_2 ^3 =−2z_2 ^2 −4z_2 =−2(−2z_2 −4)−4z_2 =8 (z_1 ^( 3) /z_2 ^( 6) ) =((−27)/z_2 ^3 )=((−27)/8)=−((27)/8)
$${z}_{\mathrm{1}} =\mathrm{3}\left(\mathrm{1}+\sqrt{\mathrm{3}}\:{i}\right)\:\wedge\:{z}_{\mathrm{2}} =−\left(\mathrm{1}+\sqrt{\mathrm{3}}\:{i}\right) \\ $$$$\Rightarrow{z}_{\mathrm{1}} =−\mathrm{3}{z}_{\mathrm{2}} \\ $$$$\frac{\mathrm{z}_{\mathrm{1}} ^{\:\mathrm{3}} }{\mathrm{z}_{\mathrm{2}} ^{\:\mathrm{6}} }\:=\frac{\left(−\mathrm{3}{z}_{\mathrm{2}} \right)^{\mathrm{3}} }{{z}_{\mathrm{2}} ^{\mathrm{6}} }=\frac{−\mathrm{27}{z}_{\mathrm{2}} ^{\mathrm{3}} }{{z}_{\mathrm{2}} ^{\mathrm{6}} }=\frac{−\mathrm{27}}{{z}_{\mathrm{2}} ^{\mathrm{3}} } \\ $$$$\: \\ $$$${z}_{\mathrm{2}} +\mathrm{1}=−\sqrt{\mathrm{3}}\:{i}\Rightarrow{z}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{2}{z}_{\mathrm{2}} +\mathrm{1}=−\mathrm{3} \\ $$$${z}_{\mathrm{2}} ^{\mathrm{2}} =−\mathrm{2}{z}_{\mathrm{2}} −\mathrm{4}\Rightarrow{z}_{\mathrm{2}} ^{\mathrm{3}} =−\mathrm{2}{z}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{4}{z}_{\mathrm{2}} \\ $$$$\:=−\mathrm{2}\left(−\mathrm{2}{z}_{\mathrm{2}} −\mathrm{4}\right)−\mathrm{4}{z}_{\mathrm{2}} =\mathrm{8} \\ $$$$\frac{\mathrm{z}_{\mathrm{1}} ^{\:\mathrm{3}} }{\mathrm{z}_{\mathrm{2}} ^{\:\mathrm{6}} }\:=\frac{−\mathrm{27}}{{z}_{\mathrm{2}} ^{\mathrm{3}} }=\frac{−\mathrm{27}}{\mathrm{8}}=−\frac{\mathrm{27}}{\mathrm{8}} \\ $$$$ \\ $$
Commented by hardmath last updated on 19/Jan/24
thank you dear ser
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{ser} \\ $$
Answered by behi834171 last updated on 15/Jan/24
z_1 =3(1+(√3)i),z_2 =−(1+(√3)i),z_1 =−3z_2 z_2 =−2((1/2)+((√3)/2)i)=−2(cos(𝛑/3)+i.sin(𝛑/3))= =−2.e^(i.(𝛑/3)) ⇒z_2 ^3 =[−2e^(i.(𝛑/3)) ]^3 =−8e^(i.𝛑) =+8 ⇒(z_1 ^3 /z_2 ^6 )=((z_1 /z_2 ))^3 .(1/z_2 ^3 )=(−3)^3 .(1/8)=−((27)/( 8)) .■ [note! e^(i.𝛑) =cos𝛑+i.sin𝛑=−1]
$${z}_{\mathrm{1}} =\mathrm{3}\left(\mathrm{1}+\sqrt{\mathrm{3}}{i}\right),{z}_{\mathrm{2}} =−\left(\mathrm{1}+\sqrt{\mathrm{3}}{i}\right),{z}_{\mathrm{1}} =−\mathrm{3}{z}_{\mathrm{2}} \\ $$$$\boldsymbol{{z}}_{\mathrm{2}} =−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\boldsymbol{{i}}\right)=−\mathrm{2}\left(\boldsymbol{{cos}}\frac{\boldsymbol{\pi}}{\mathrm{3}}+\boldsymbol{{i}}.\boldsymbol{{sin}}\frac{\boldsymbol{\pi}}{\mathrm{3}}\right)= \\ $$$$=−\mathrm{2}.\boldsymbol{{e}}^{\boldsymbol{{i}}.\frac{\boldsymbol{\pi}}{\mathrm{3}}} \Rightarrow\boldsymbol{{z}}_{\mathrm{2}} ^{\mathrm{3}} =\left[−\mathrm{2}\boldsymbol{{e}}^{\boldsymbol{{i}}.\frac{\boldsymbol{\pi}}{\mathrm{3}}} \right]^{\mathrm{3}} =−\mathrm{8}\boldsymbol{{e}}^{\boldsymbol{{i}}.\boldsymbol{\pi}} =+\mathrm{8} \\ $$$$\Rightarrow\frac{\boldsymbol{{z}}_{\mathrm{1}} ^{\mathrm{3}} }{\boldsymbol{{z}}_{\mathrm{2}} ^{\mathrm{6}} }=\left(\frac{\boldsymbol{{z}}_{\mathrm{1}} }{\boldsymbol{{z}}_{\mathrm{2}} }\right)^{\mathrm{3}} .\frac{\mathrm{1}}{\boldsymbol{{z}}_{\mathrm{2}} ^{\mathrm{3}} }=\left(−\mathrm{3}\right)^{\mathrm{3}} .\frac{\mathrm{1}}{\mathrm{8}}=−\frac{\mathrm{27}}{\:\mathrm{8}}\:\:.\blacksquare \\ $$$$\left[\boldsymbol{{note}}!\:\:\:\:\:\boldsymbol{{e}}^{\boldsymbol{{i}}.\boldsymbol{\pi}} =\boldsymbol{{cos}\pi}+\boldsymbol{{i}}.\boldsymbol{{sin}\pi}=−\mathrm{1}\right] \\ $$
Commented by hardmath last updated on 19/Jan/24
thank you dear ser
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{ser} \\ $$
Answered by Rasheed.Sindhi last updated on 16/Jan/24
z_1 −3=3(√3) i z_1 ^2 −6z_1 +9=−27⇒z_1 ^2 =6z_1 −36 ⇒z_1 ^3 =6z_1 ^2 −36z_1 =6(6z_1 −36)−36z_1 =−216 z_2 +1=−(√3) i z_2 ^2 +2z_2 +1=−3⇒z_2 ^2 =−2z_2 −4 ⇒z_2 ^3 =−2z_2 ^2 −4z_2 =−2(−2z_2 −4)−4z_2 =8 ⇒z_2 ^6 =64 ▶(z_1 ^3 /z_2 ^6 )=((−216)/(64))=−((27)/8)
$${z}_{\mathrm{1}} −\mathrm{3}=\mathrm{3}\sqrt{\mathrm{3}}\:{i} \\ $$$${z}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{6}{z}_{\mathrm{1}} +\mathrm{9}=−\mathrm{27}\Rightarrow{z}_{\mathrm{1}} ^{\mathrm{2}} =\mathrm{6}{z}_{\mathrm{1}} −\mathrm{36} \\ $$$$\Rightarrow{z}_{\mathrm{1}} ^{\mathrm{3}} =\mathrm{6}{z}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{36}{z}_{\mathrm{1}} =\mathrm{6}\left(\mathrm{6}{z}_{\mathrm{1}} −\mathrm{36}\right)−\mathrm{36}{z}_{\mathrm{1}} =−\mathrm{216} \\ $$$$\: \\ $$$${z}_{\mathrm{2}} +\mathrm{1}=−\sqrt{\mathrm{3}}\:{i} \\ $$$${z}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{2}{z}_{\mathrm{2}} +\mathrm{1}=−\mathrm{3}\Rightarrow{z}_{\mathrm{2}} ^{\mathrm{2}} =−\mathrm{2}{z}_{\mathrm{2}} −\mathrm{4} \\ $$$$\Rightarrow{z}_{\mathrm{2}} ^{\mathrm{3}} =−\mathrm{2}{z}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{4}{z}_{\mathrm{2}} =−\mathrm{2}\left(−\mathrm{2}{z}_{\mathrm{2}} −\mathrm{4}\right)−\mathrm{4}{z}_{\mathrm{2}} =\mathrm{8} \\ $$$$\:\Rightarrow{z}_{\mathrm{2}} ^{\mathrm{6}} =\mathrm{64} \\ $$$$\: \\ $$$$\blacktriangleright\frac{{z}_{\mathrm{1}} ^{\mathrm{3}} }{{z}_{\mathrm{2}} ^{\mathrm{6}} }=\frac{−\mathrm{216}}{\mathrm{64}}=−\frac{\mathrm{27}}{\mathrm{8}} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *