A Sum of Imaginary Powers | Problem 311

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  • čas přidán 8. 09. 2024
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Komentáře • 12

  • @derwolf7810
    @derwolf7810 Před měsícem +6

    You also could use the fact that the sum of all n-th roots of 1 equals 0, which is also true for the 5-th:
    e^(2πi/5) + e^(4πi/5) + e^(6πi/5) + e^(8πi/5)
    = -e^(0πi/5) + (e^(0πi/5) + e^(2πi/5) + e^(4πi/5) + e^(6πi/5) + e^(8πi/5))
    = -1 + (0)
    = -1

    • @bsmith6276
      @bsmith6276 Před měsícem +4

      I did the same thing. This should have been in video: Third method taking advantage of the circular symmetry!

  • @scottleung9587
    @scottleung9587 Před měsícem

    Nice!

  • @dreael
    @dreael Před měsícem +3

    There is a complete visual geometric way: All 4 terms represent vectors (all of lengt 1 and the angles as discussed) in the complex plane. Adding means that you can draw a figure like the famous turtle graphics. This will result in a regular pentagon without the bottom side, i.e. the last point is at -1+0*i = -1

    • @0over0
      @0over0 Před měsícem

      Beautiful. what a simple, geometric proof!

  • @Nobodyman181
    @Nobodyman181 Před měsícem +3

    Trigonometry+complexivity😅😂❤

  • @mathmachine4266
    @mathmachine4266 Před měsícem

    -1

  • @phill3986
    @phill3986 Před měsícem

    👍😎👏✌️👏😎👍

  • @TheBlueboyRuhan
    @TheBlueboyRuhan Před měsícem +1

    w = 4 not -1
    you have indeterminate form of (z⁵ - 1) / (z - 1) at z = 1
    so its limit approaches 5, not 0

    • @vighnesh153
      @vighnesh153 Před měsícem +5

      z isn't equal 1. z^5 is.

    • @TheBlueboyRuhan
      @TheBlueboyRuhan Před měsícem

      @@vighnesh153 the whole answer is wrong tbh, this video needs a redo

    • @vighnesh153
      @vighnesh153 Před měsícem

      @@TheBlueboyRuhan Could you clarify what part is wrong?