Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

How do I do this... 2i^19 - 5i^40 + 4i^97 + 9i^122?

The directions say to simplify.

8 Answers

Relevance
  • 4 years ago

    i^1 = i

    i^2 = -1

    i^3 = -i

    i^4 = 1

    i^5 = i

    i^6 = -1

    i^7 = -i

    i^8 = 1

    See the pattern?

    2i^19 - 5i^40 + 4i^97 + 9i^122 =

    2i^3 - 5i^4 + 4i^3 + 9i^2 =

    -2i - 5 - 4i - 9 =

    -14 - 6i

  • ?
    Lv 7
    4 years ago

    i⁰ = 1+

    i¹ = i

    i² = -1

    i³ = -i

    i⁴ = 1

    so take these exponents mod 4:

    2i^19 - 5i^40 + 4i^97 + 9i^122

    = 2i³-5+4i+9i²

    = -2i-5+4i-9

    = -14+2i

  • alex
    Lv 7
    4 years ago

    Hint:

    i^2 = −1

    i^4 = 1

  • ?
    Lv 7
    4 years ago

     

    i^2 = −1

    i^4 = 1

    2i^19 − 5i^40 + 4i^97 + 9i^122

    = 2(i^4)^4(i^2)(i) − 5(i^4)^10 + 4(i^4)^24(i) + 9(i^4)^30(i^2)

    = 2(1)(−1)(i) − 5(1) + 4(1)(i) + 9(1)(−1)

    = −2i − 5 + 4i − 9

    = −14 + 2i

    - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

    Alternate method:

    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 the pattern continues...

    i^(4k) = 1

    i^(4k+1) = i

    i^(4k+2) = −1

    i^(4k+3) = −i

    2i^19 − 5i^40 + 4i^97 + 9i^122

    = 2i^(4*4+3) − 5i^(4*10) + 4i^(4*24+1) + 9i^(4*30+2)

    = 2(−i) − 5(1) + 4(i) + 9(−1)

    = −2i − 5 + 4i − 9

    = −14 + 2i

  • How do you think about the answers? You can sign in to vote the answer.
  • 4 years ago

    i = √-1

    i^2 = -1

    i^3 = -√-1 = -i

    i^4 = 1

    And the patter repeats. So i^19 is the same as i^3 (since 19/4 has a remainder of 3). The same reasoning applies to the next terms also.

    2i^19 - 5i^40 + 4i^97 + 9i^122

    2i^3 - 5i^4 + 4i^1 + 9i^2

    2(-i) - 5(1) + 4(i) + 9(-1)

    -2i - 5 + 4i -9

    2i-14

  • J
    Lv 7
    4 years ago

    Reduce exponents mod 4, because i,i^2,i^3,i^4 cycle since i^4=1, so

    2i^19 - 5i^40 + 4i^97 + 9i^122 =

    2i^3 - 5i^0 + 4i^1 + 9i^2 =

    -2i - 5 + 4i - 9 =

    2i - 14

  • ?
    Lv 7
    4 years ago

    The powers of i cycle every 4 powers; i.e., i = sqrt(-1), i^2 = -1, i^3 = -i, and i^4 = 1. Divide that exponent of i by 4; whatever the remainder is will correspond to one of the aforementioned powers of i up to i^4 = 1, which results if the exponent is divisible by 4. So:

    2i^19 - 5i^40 + 4i^97 + 9i^122

    = 2(i^3) - 5(i^4) + 4(i^1) + 9(i^2)

    = 2(-i) - 5(1) + 4(i) + 9(-1)

    = -2i - 5 + 4i - 9

    = -14 + 2i. Final.

  • ?
    Lv 7
    4 years ago

    i^19 = i^18*i = (i^2)^9 * 1 = (-1)^9 * i = - i so the first term reduces to 2*(-1) = -2. i^40 = (i^2)^20 = (-1)^20 = + 1 so the second term is 5*1 = + 5; i^122 = (i^2)^61 = (-1)^61 = -1 so the third term is 9*(-1) = -9 so altogether you have -2 + 5 - 9 = -11 + 5 = -6

    Source(s): Retired Maths Teacher
Still have questions? Get your answers by asking now.