Question Number 203309 by hardmath last updated on 15/Jan/24

Answered by Rasheed.Sindhi last updated on 16/Jan/24

Commented by hardmath last updated on 19/Jan/24

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]](https://www.tinkutara.com/question/Q203311.png)
Commented by hardmath last updated on 19/Jan/24

Answered by Rasheed.Sindhi last updated on 16/Jan/24
