maths

Let z^n = -1
z^n = cos(2kπ + π) + i sin(2kπ + π) where k = 0, 1, 2,... , n - 1
z = cos(2kπ + π)/n + i sin(2kπ + π)/n
For example, for n = 2
z^2 = -1
z = (-1)^(1/2) = cos π/2 + i sin π/2 = i
or cos (3π/2) + i sin(3π/2) = -i. There are 2 roots, i and -i
For n = 3, z = (-1)^(1/3) = cos(2kπ + π)/3 + i sin(2kπ + π)/3 for k = 0, 1 or 2
There are 3 roots : cos(π/3) + i sin(π/3) = 1/2 + (√3/2)i
cos(π) + i sin(π) = -1
cos(5π/3) + i sin(5π/3) = 1/2 - (√3/2)i
For n = 4, z = (-1)^(1/4) = cos(2kπ + π)/4 + i sin(2kπ + π)/4 for k = 0, 1, 2 or 3
There are 4 roots : cos(π/4) + i sin(π/4) = √2/2 + (√2/2) i
cos(3π/4) + i sin(3π/4) = -√2/2 + (√2/2) i
cos(5π/4) + i sin(5π/4) = -√2/2 - (√2/2) i
cos(7π/4) + i sin(7π/4) = √2/2 - (√2/2) i
You can continue the same for the rest of the items

1.(-1)^(1/2)=-1/2
2.(-1)^(1/3)=-1/3
3.(-1)^(1/4)=-1/4
4.(-1)^(1/5)=-1/5
5.(-1)^(1/6)=-1/6
6.(-1)^(1/7)=-1/7
7.(-1)^(1/8)=-1/8
8.(-1)^(1/9)=-1/9

This is the answer of the maths questions!!!

全部都math error

1. -1
2. -1
3. -1
4. -1
5. -1
6. -1
7. -1
8. -1

全部都係-1...
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