let z = √3 - i
for polar form r = √((√3)^2 + 1) = 2
tanØ = -1/√3
Ø = -π/3 or - 30° or 330°
so z = 2(cos -240, sin -240)
check: z = 2(√3/2) , -1/2) = (√3, -1) , looks good
z^8 = 2^8(cos 2640, sin 2640) or 2^8(cos -240, sin -240)
= 256(-.5, √3/2)
= (128 , 128√3)
(√3 - i)^8 = (128 + 128√3)
check my arithmetic
Express (√3-¡)^8 in the form a+¡b there a,b € R
Answer please
2 answers
check last 3 lines
= 256(-.5, √3/2)
= (128 , 128√3)
(√3 - i)^8 = (128 + 128√3)
should be:
= 256(-.5, √3/2)
= (-128 , 128√3)
(√3 - i)^8 = (-128 + 128√3)
= 256(-.5, √3/2)
= (128 , 128√3)
(√3 - i)^8 = (128 + 128√3)
should be:
= 256(-.5, √3/2)
= (-128 , 128√3)
(√3 - i)^8 = (-128 + 128√3)