Please help with algebra II?

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  • 2 months ago

    i^14 = (i^2)^7 = (-1)^7 = -1

    i^19 = (i^18)(i) = (i^2)^9 (i) = -1(i) = - i

    i^25 = (i^24)(i) = (i^2)^12 (i) = 1(i) = i

    i^37 = (i^36)(i) = (i^2)^13 (i) = -1(i) = -i

    i^42 = (i^2)^21 = (-1)^21 = -1

    i^57 = (i^56)(i) = (i^2)^(28) (i) = (1)(i) = i

    i^64 = (i^2)^32 = 1

    i^71 = i^(70) (i) = (i^2)^35 (i) = -1(i) = -i

  • 2 months ago

    powers of i are in a cycle of 4:

    i⁰ = 1

    i¹ = i

    i² = -1

    i³ = -i

    Then i⁴ is 1 again.

    So a "power of i" can be simplified to one of those 4 values by finding the remainder when you divide the exponent by 4.

    i¹⁴

    14 / 4 = 3 R 2

    Since the remainder is 2, it simplifies to -1.

    Do the same for the other 7 values and find out which ones have a remainder of 3 which will make it simplify to -i.

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