Write the three cube roots of 8 cis (-30°).
2 e^i(-30/3) = 2 cis -10
2 e^i(-30/3 +360/3)= 2 cis 110
2 e^i(-30/3 -360/3) = 2 cis -130
This is probably silly, but where is the 2 in the beginning of all of those from?
2*2*2 = 8
Ohmygoodness. Hhaha. Thanks!
To find the cube roots of a complex number, we need to convert it to its trigonometric form, which includes magnitude and argument.
Given complex number: 8 cis(-30°)
Step 1: Convert the complex number to its rectangular form.
Using Euler's formula: e^(iθ) = cos(θ) + i sin(θ)
8 cis(-30°) = 8 (cos(-30°) + i sin(-30°))
= 8 cos(-30°) + 8i sin(-30°)
= 8 (√3/2 - i/2)
Step 2: Find the magnitude.
The magnitude (r) of a complex number in rectangular form is given by:
r = √(a^2 + b^2)
In this case, a = 8 * √3/2 = 4√3 and b = -8 * 1/2 = -4.
r = √((4√3)^2 + (-4)^2)
= √(48 + 16)
= √64
= 8
Step 3: Find the principal argument (θ).
The principal argument (θ) of a complex number in rectangular form is given by:
θ = arctan(b/a)
In this case, a = 4√3 and b = -4.
θ = arctan((-4)/(4√3))
= arctan(-1/√3)
= -30°
Step 4: Find the cube roots.
The cube roots can be found by using De Moivre's formula:
z^(1/n) = r^(1/n) * [cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)]
In this case, n = 3.
1st cube root:
k = 0
z^(1/3) = 8^(1/3) * [cos((-30° + 2(0)π)/3) + i sin((-30° + 2(0)π)/3)]
= 2 [cos(-10°) + i sin(-10°)]
= 2 cis(-10°)
2nd cube root:
k = 1
z^(1/3) = 8^(1/3) * [cos((-30° + 2(1)π)/3) + i sin((-30° + 2(1)π)/3)]
= 2 [cos(50°) + i sin(50°)]
= 2 cis(50°)
3rd cube root:
k = 2
z^(1/3) = 8^(1/3) * [cos((-30° + 2(2)π)/3) + i sin((-30° + 2(2)π)/3)]
= 2 [cos(170°) + i sin(170°)]
= 2 cis(170°)
Therefore, the three cube roots of 8 cis (-30°) are:
1) 2 cis(-10°)
2) 2 cis(50°)
3) 2 cis(170°)