Express (√3-¡)^8 in the form a+¡b there a,b € R
Answer please
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
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)
To express (√3-¡)^8 in the form a+¡b, we can first simplify the expression (√3-¡) by rationalizing the denominator.
Step 1: Rationalize the denominator:
Multiply the numerator and denominator by the conjugate of (√3-¡), which is (√3+¡):
[ (√3-¡) * (√3+¡) ] / [ (√3+¡) * (√3+¡) ]
Simplifying the numerator:
(√3)^2 - (¡)^2
3 + 1
4
Simplifying the denominator:
(√3)^2 - (¡)^2
3 - 1
2
So the rationalized form of (√3-¡) is 4/2, which simplifies to 2.
Step 2: Raise 2 to the 8th power:
2^8 = 256
Now, we have (√3-¡)^8 = 256.
To express 256 in the form a+¡b, we can write it as 256 + 0¡.
Therefore, a = 256 and b = 0.
So the expression (√3-¡)^8 can be expressed in the form a+¡b as 256 + 0¡.