2007-04-20 06:16:01 · 5 answers · asked by Ritz 1 in Science & Mathematics ➔ Mathematics
1 = 1<2kpi in polar notation
so 1^1/5 = 1< 2kpi/5 from k= 0 to k = 4and in binomial form
1^1/5 =cos(2kpi/5)+i sin(2kpi/5)
for K=0 you get 1
The others are complex conjugates
2007-04-20 07:48:30 · answer #1 · answered by santmann2002 7 · 1⤊ 0⤋
1
2007-04-20 06:18:19 · answer #2 · answered by DAN G 1 · 0⤊ 2⤋
1, -1, and 3 other complex numbers.
2007-04-20 06:33:07 · answer #3 · answered by Jared Z 3 · 0⤊ 1⤋
hi
Xi=cos( 2k pi i/5) + i sin ( 2k pi i/5)
k = 0 to 4
k=0
X0=1
k=1;4
cos(72*) +/- i sin(72*)
= (V5-1)/4 +/- i [root(10 + 2V5)] / 4
k=2;3
cos(144*) +/- i sin(144*)
= - cos( 36*) +/- i sin( 36*)
= - (V5+1)/4 +/- i [root(10 - 2V5)] / 4
bye
2007-04-20 09:23:21 · answer #4 · answered by railrule 7 · 0⤊ 0⤋
Not possible. Four of them are complex, and transcendental to boot.
2007-04-20 06:20:23 · answer #5 · answered by Anonymous · 0⤊ 2⤋