Can you solve this? | iota maths problem | Oxford entrance exam question

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  • Опубліковано 30 жов 2024

КОМЕНТАРІ • 10

  • @TheMathManProfundities
    @TheMathManProfundities 6 днів тому +4

    There should be no ± in the solution as the question uses radical symbols and as such refers only to the principal roots. √i + √(-i) = {(√2 + i√2) + (√2 - i√2)}/2 = √2.

    • @knutthompson7879
      @knutthompson7879 5 днів тому

      Yes. If you are going to include non-principal roots, there would be 4 total answers. +/-sqrt(2) and +/-sqrt(2)i. But unless specified, it is just the principle, so just sqrt(2).

  • @daniel_feiglin
    @daniel_feiglin 4 дні тому +1

    Draw a picture!!!! This is simple geometry. sqrt(i) = (1+i)/sqrt(2). Similarly, sqrt(-i)=(1-i)/sqrt(2). Add them up and
    you get sqrt(2). Big deal!

  • @mwinfield1969
    @mwinfield1969 5 днів тому

    I just wrote i and -i in exponential form. Then raised both to power 1/2. then converted both to sin/cos form.

  • @ELALFAMentalidad
    @ELALFAMentalidad 3 дні тому

    Could you please lower the volume of the music?

  • @RealQinnMalloryu4
    @RealQinnMalloryu4 5 днів тому

    (x ➖ 1ix+1i )

  • @RadekBuczkowski-h2y
    @RadekBuczkowski-h2y 5 днів тому

    The result in the video is not quite correct.
    Answer (all roots):
    𝓍 ∈ { √𝟸,﹣√𝟸, √𝟸 𝒊,﹣√𝟸 𝒊 }
    Answer (only the principal root):
    𝓍 ﹦ √𝟸
    I personally think that square roots of complex number constants do not only refer to the principal root, since so many people calculate all roots. But that is debatable. So the video was half-right in either case.