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Mathematics 21 Online
OpenStudy (anonymous):

(i)^58

OpenStudy (ash2326):

@fabo What's your question?

OpenStudy (anonymous):

simplify

OpenStudy (ash2326):

Could you post the question? it just says \[(i)\ ^{58}\]

OpenStudy (ash2326):

Sorry I get it, i the root of -1. Do you know what's \(i^4\)

OpenStudy (anonymous):

no I dont how to to do it

OpenStudy (ash2326):

ok you know what's i?

OpenStudy (anonymous):

yes a cpmplex number

OpenStudy (ash2326):

yes, right \[i=\sqrt{-1}\] so \[i^2=???\]

OpenStudy (anonymous):

we know i^58=(i^2)^29 ..so since i=root(-1) i^2 should be -1..so i^58=(-1)^29=-1

OpenStudy (imtiaz7):

(i)^58 = (i^2)^5 * (i^2)^4 * (i^2)^3 * i^2 = (-1)^5 * (-1)^4 * (-1)^3 *-1 = -1 * 1 *-1 *-1 = -1 * 1 = -1

OpenStudy (loser66):

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