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How do I find each of these complex numbers and Write the complex number in its usual form? All you have to do is show me one. Thank you.
Complex Number 1: i
Complex Number 2: i2 (so this just means i•i )
Complex Number 3: i3 (so this just means i•i•i )
Complex Number 4: i4 (in other words, i•i•i•i )

2007-12-14 18:45:32 · 7 answers · asked by Anonymous in Science & Mathematics Mathematics

7 answers

Complex Number 1
i = 1 /_90 ° = cos 90 ° + i sin 90 °

Complex Number 2
i ² = - 1 = 1 /_180 °

Complex Number 3
i ³ = 1 /_270 ° = cos 270° + i sin 270°

Complex Number 4
i^4 = (i²) (i²) = (-1) (-1) = 1 = cos 360° + i sin 360°

2007-12-15 02:41:03 · answer #1 · answered by Como 7 · 2 1

For i^2, remember that i = sqrt(-1). So, i^2 = ( sqrt(-1) )^2 = -1.

You can find i^3 by writing it as i^2 x i.

2007-12-15 02:48:19 · answer #2 · answered by Doug 2 · 0 0

1: i

2: i^2 = - 1

3: i^3 = i^2i = (-1)i = - i

4: i^4 = (i^2)(i^2) = (-1)(-1) = 1

and this pattern repeats for higher exponents

2007-12-15 02:50:44 · answer #3 · answered by kindricko 7 · 1 0

1. i
2. i^2
3. i^3
4. i^4

2007-12-15 02:55:29 · answer #4 · answered by Jun Agruda 7 · 3 0

i= sqrt(-1)

so...
1) i, or sqrt(-1)

2) -1

3) -sqrt(-1)

4) +1

2007-12-15 02:49:54 · answer #5 · answered by WhatWasThatNameAgain? 5 · 0 0

i = sqrt(-1)
i^2 = sqrt(-1) * sqrt(-1) = -1
i^3 = sqrt(-1) * sqrt(-1) * sqrt(-1)
= (-1) * sqrt(-1) = -i
i^4 = [sqrt(-1) * sqrt(-1)] * [sqrt(-1) * sqrt(-1)]
= -1 * -1
= 1

2007-12-15 03:00:42 · answer #6 · answered by Jeremy H 2 · 0 0

i = i

i² = i*i = -1

i³ = i² * i = (-1) * i = -i

i^4 = i² * i² = (-1) * (-1) = 1

2007-12-15 02:48:30 · answer #7 · answered by gudspeling 7 · 2 0

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