Roots of unity

The purpose of this task is to explore working with complex numbers. Given n , find the n-th roots of unity.

ALGOL 68

FOR root FROM 2 TO 10 DO
  printf(($g(4)$,root));
  FOR n TO root -1 DO
    printf(($xg(5,3)g(5,3)"i"$,complex exp( 0 I 2*pi*n/root)))
  OD;
  printf($l$)
OD

Output:

 +2 -1.00+.000i
 +3 -.500+.866i -.500-.866i
 +4 +.000+1.00i -1.00+.000i -.000-1.00i
 +5 +.309+.951i -.809+.588i -.809-.588i +.309-.951i
 +6 +.500+.866i -.500+.866i -1.00+.000i -.500-.866i +.500-.866i
 +7 +.623+.782i -.223+.975i -.901+.434i -.901-.434i -.223-.975i +.623-.782i
 +8 +.707+.707i +.000+1.00i -.707+.707i -1.00+.000i -.707-.707i -.000-1.00i +.707-.707i
 +9 +.766+.643i +.174+.985i -.500+.866i -.940+.342i -.940-.342i -.500-.866i +.174-.985i +.766-.643i
+10 +.809+.588i +.309+.951i -.309+.951i -.809+.588i -1.00+.000i -.809-.588i -.309-.951i +.309-.951i +.809-.588i

J

   rou=: [: ^ i. * (o. 0j2) % ]

   rou 4
1 0j1 _1 0j_1

   rou 5
1 0.309017j0.951057 _0.809017j0.587785 _0.809017j_0.587785 0.309017j_0.951057

The computation can also be written as a loop, shown here for comparison only.

rou1=: 3 : 0
 z=. 0 $ r=. ^ o. 0j2 % y [ e=. 1
 for. i.y do.
  z=. z,e
  e=. e*r
 end.
 z
)

Python

Interpreter: Python 2.5 Function nthroots() returns all n-th roots of unity.

from cmath import exp, pi
def nthroots(n):
    return [exp(k * 2j * pi / n) for k in range(1, n + 1)]

Example:

>>> f = nthroots
>>> for n in range(1, 4):
...    print map(lambda c: ("%.3f " + ("+" if c.imag > 0 else "-") + " %.3gj") % (c.real, abs(c.imag)), f(n))
['1.000 - 2.45e-016j']
['-1.000 + 1.22e-016j', '1.000 - 2.45e-016j']
['-0.500 + 0.866j', '-0.500 - 0.866j', '1.000 - 2.45e-016j']