2007-07-16 08:16:33 · 9 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
The imaginary number i follows an easy pattern :
i^1 = i
i^2 = -1
i^3 = -i
i^4 = 1
so i^5 just begins to repeat again with i.
2007-07-16 08:25:30 · answer #1 · answered by Don E Knows 6 · 0⤊ 0⤋
1
2007-07-16 15:23:33 · answer #2 · answered by Anonymous · 0⤊ 0⤋
i^5 = i² x i² x i = -1 x -1 x i = i
2007-07-17 02:29:53 · answer #3 · answered by Como 7 · 0⤊ 0⤋
i^5 = i^4 * i = 1*i = i
By the way, i^5 is not a complex number; rather, it is a pure imaginary number. A complex number takes the form a +bi. If a = 0, you have a pure imaginary number. If b = 0 you have a real number.
2007-07-16 15:20:55 · answer #4 · answered by ironduke8159 7 · 2⤊ 0⤋
Powers of i go through a cycle:
i^1 = i
i^2 = -1
i^3 = -i
i^4 = 1
i^5 = i
i^6 = -1
i^7 = -i
i^8 = 1
and so on.
2007-07-16 15:20:41 · answer #5 · answered by lithiumdeuteride 7 · 2⤊ 0⤋
i = sqrt(-1)
so, i^2 = -1
which means i^4 = -1*-1 = 1
Therefore, i^5 = i^4 * i
= 1i
=i
2007-07-16 15:21:44 · answer #6 · answered by Wendy E 1 · 0⤊ 0⤋
i^5 = i
Bramble
2007-07-16 15:22:37 · answer #7 · answered by Bramble 7 · 0⤊ 0⤋
"i" = sqrt(-1)
[sqrt(-1)]^5 =
sqrt(-1) * sqrt (-1) * sqrt(-1) * sqrt(-1) * sqrt(-1) =
sqrt(-1) = i
2007-07-16 15:21:36 · answer #8 · answered by miggitymaggz 5 · 1⤊ 0⤋
rad(-1) * rad(-1) * rad(-1) * rad(-1) * rad(-1)
= -1 * -1 * rad(-1)
= rad(-1) i.e: i
2007-07-16 15:22:20 · answer #9 · answered by gebobs 6 · 0⤊ 0⤋