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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) ) = ?

Ifz1=3+33iandz2=−1−3iFind:z13z26=?

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)

z1=3(1+3i)∧z2=−(1+3i)⇒z1=−3z2z13z26=(−3z2)3z26=−27z23z26=−27z23z2+1=−3i⇒z22+2z2+1=−3z22=−2z2−4⇒z23=−2z22−4z2=−2(−2z2−4)−4z2=8z13z26=−27z23=−278=−278

Commented by hardmath last updated on 19/Jan/24

thank you dear ser

thankyoudearser

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]

z1=3(1+3i),z2=−(1+3i),z1=−3z2z2=−2(12+32i)=−2(cosπ3+i.sinπ3)==−2.ei.π3⇒z23=[−2ei.π3]3=−8ei.π=+8⇒z13z26=(z1z2)3.1z23=(−3)3.18=−278.◼[note!ei.π=cosπ+i.sinπ=−1]

Commented by hardmath last updated on 19/Jan/24

thank you dear ser

thankyoudearser

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)

z1−3=33iz12−6z1+9=−27⇒z12=6z1−36⇒z13=6z12−36z1=6(6z1−36)−36z1=−216z2+1=−3iz22+2z2+1=−3⇒z22=−2z2−4⇒z23=−2z22−4z2=−2(−2z2−4)−4z2=8⇒z26=64▸z13z26=−21664=−278

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