simplifying complex numbers: simplify i^27

alsacecat

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I am a 46 year old returnee to college and I need assurance that I am on the right track. The problem is:

Simplify i^27.

I worked out -1. Am I close?
 

stapel

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How did you arrive at your answer? You noted the repeating pattern:

. . . . .i<sup>0</sup> = 1
. . . . .i<sup>1</sup> = i
. . . . .i<sup>2</sup> = -1
. . . . .i<sup>3</sup> = -i
. . . . .i<sup>4</sup> = 1
. . . . .i<sup>5</sup> = i
. . . . .i<sup>6</sup> = -1
. . . . .i<sup>7</sup> = -i
. . . . .i<sup>8</sup> = 1

...so that every fourth power (i<sup>0</sup>, i<sup>4</sup>, i<sup>8</sup>, etc) returns you to "1". How did you relate this to "i<sup>27</sup> = i<sup>24+3</sup> = i<sup>24</sup> i<sup>3</sup>"?

Thank you.

Eliz.
 

alsacecat

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looks like i have to do it again. thanks for shedding some light on this.
 

alsacecat

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ok so now i still have -1, by doing this

i^24+3=i^24i^3=-1. Better?
 

stapel

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alsacecat said:
ok so now i still have -1, by doing this

i^24+3=i^24i^3=-1. Better?
It was explained and demonstrated previously that i<sup>24</sup> = 1. So you have been given:

. . .i<sup>27</sup> = i<sup>24+3</sup>
. . . .. .= i<sup>24</sup> i<sup>3</sup>
. . . .. .= (1)(i<sup>3</sup>)
. . . .. .= i<sup>3</sup>

Is i<sup>3</sup> equal to -1?

Eliz.
 
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