express the roots of unity in standard form a+bi.

1.) cube roots of unity

2.) fourth roots of unity

3.) sixth roots of unity

4.) square roots of unity

See:

http://www.jiskha.com/display.cgi?id=1303349576

To express the roots of unity in the standard form of a+bi, we need to use Euler's formula, which states e^(iθ) = cos(θ) + i·sin(θ). From this, we can obtain the general formula for the roots of unity:

n-th root of unity = cos(2πk/n) + i·sin(2πk/n),

where n is the number of roots we want to find, and k ranges from 0 to n-1. Now, let's apply this formula to each case:

1) Cube roots of unity:
n = 3, so we have:
- For k = 0: cos(2π·0/3) + i·sin(2π·0/3) = cos(0) + i·sin(0) = 1 + 0i = 1.
- For k = 1: cos(2π·1/3) + i·sin(2π·1/3) = cos(2π/3) + i·sin(2π/3) = -1/2 + (√3/2)i.
- For k = 2: cos(2π·2/3) + i·sin(2π·2/3) = cos(4π/3) + i·sin(4π/3) = -1/2 - (√3/2)i.

Hence, the cube roots of unity are: 1, -1/2 + (√3/2)i, and -1/2 - (√3/2)i.

2) Fourth roots of unity:
n = 4, so we have:
- For k = 0: cos(2π·0/4) + i·sin(2π·0/4) = cos(0) + i·sin(0) = 1 + 0i = 1.
- For k = 1: cos(2π·1/4) + i·sin(2π·1/4) = cos(π/2) + i·sin(π/2) = 0 + 1i = i.
- For k = 2: cos(2π·2/4) + i·sin(2π·2/4) = cos(π) + i·sin(π) = -1 + 0i = -1.
- For k = 3: cos(2π·3/4) + i·sin(2π·3/4) = cos(3π/2) + i·sin(3π/2) = 0 - 1i = -i.

Hence, the fourth roots of unity are: 1, i, -1, and -i.

3) Sixth roots of unity:
n = 6, so we have:
- For k = 0: cos(2π·0/6) + i·sin(2π·0/6) = cos(0) + i·sin(0) = 1 + 0i = 1.
- For k = 1: cos(2π·1/6) + i·sin(2π·1/6) = cos(π/3) + i·sin(π/3) = 1/2 + (√3/2)i.
- For k = 2: cos(2π·2/6) + i·sin(2π·2/6) = cos(2π/3) + i·sin(2π/3) = -1/2 + (√3/2)i.
- For k = 3: cos(2π·3/6) + i·sin(2π·3/6) = cos(π) + i·sin(π) = -1 + 0i = -1.
- For k = 4: cos(2π·4/6) + i·sin(2π·4/6) = cos(4π/3) + i·sin(4π/3) = -1/2 - (√3/2)i.
- For k = 5: cos(2π·5/6) + i·sin(2π·5/6) = cos(5π/3) + i·sin(5π/3) = 1/2 - (√3/2)i.

Hence, the sixth roots of unity are: 1, 1/2 + (√3/2)i, -1/2 + (√3/2)i, -1, -1/2 - (√3/2)i, and 1/2 - (√3/2)i.

4) Square roots of unity:
n = 2, so we have:
- For k = 0: cos(2π·0/2) + i·sin(2π·0/2) = cos(0) + i·sin(0) = 1 + 0i = 1.
- For k = 1: cos(2π·1/2) + i·sin(2π·1/2) = cos(π) + i·sin(π) = -1 + 0i = -1.

Hence, the square roots of unity are: 1 and -1.