What are the nth roots of -1?

Question

Please show me how to do this!!!

Answer 1

cos((π +2πx)/n ) + isin((π +2πx)/n) ; x = 0,1,2,...,n-1

if -1 = e^(iθ) = cos(θ) + isin(θ), then θ=π+2πx for x = 0,1,2,3,...
then (-1)^(1/n) = e^((π+2πx)i/n) = cos((π +2πx)/n ) + isin((π +2πx)/n) for x = 0,1,2,...,n-1

so, for example, if n = 2,
(-1)^(1/2) = cos(π /2)+ isin(π/2) = i
and cos(3π/2) + isin(3π/2) = -i

or if n = 4
(-1)^(1/4) = cos(π/4) + isin(π/4) ; [when x=0]
and cos(3π/4) + isin(3π/4) ; [when x=1]
and cos(5π/4) + isin(5π/4) ; [when x=2]
and cos(7π/4) + isin(7π/4) ; [when x=3]

or the four roots are: ±√2/2 ± i√2/2

Answer 2

Draw the real and imaginary axises on a piece of paper. Draw a circle with radius = 1 around the origin.

Put a dot at (-1, 0). That the first root of -1.

Put dots at (0, 1) and (0, -1). Those are the 2nd roots.

Put dots at (-1, 0), and 120 degree away on the circle, and -120 degrees away. Those are the 3rd roots, there are three of them.

The n-th roots of -1 will be spaced around the circle, symmetrically around the point (-1, 0).

Answer 3

One of my favorite things!

It's easiest to see this using polar represetnation:

-1 = exp(i*pi + k*2*pi) = cos(pi) + i*sin(pi), because of the periodicity of cos and sin, you get the same value for every k an integer.

Now take the root:
(-1)^(1/n) = exp(i*pi/n = k*2*pi/n) , now for k = 0 to n-1 you get n complex numbers. These are evenly spaced on the unit circle.

Ex:

for n = 2 you get exp(i*pi/2) and exp(i*3*pi/2) or +i and -i

The roots look like spokes.

Answer 4

You should write -1 = 1 If need the nth roots there will be n values = 1< pi/n +(2k/n)pi with
k=0,1.......(n-1)
you can go back to binomial form using
x=r cos@
y=r sin@

Answer 5

Cubicroot of -1 = -1
same for 5th 7th 9th and every other odd numbered root.
since -1 * -1 * -1 = -1 and -1 * -1 * -1 * -1 * -1 = -1

even roots of -1 dont exist.

Answer 6

You would need imaginary numbers to find roots of a negative number. Is that what you want?