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

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  • Опубликовано: 30 окт 2024

Комментарии • 5

  • @camgere
    @camgere День назад +6

    Or you could just start from -i = e^i(-pi/2)

  • @andreaslarsson4076
    @andreaslarsson4076 22 часа назад

    Well, just visualize it. Angles are added. 270/2 = 135 = 90+45. So that’s -1+i right there. Lengths are multiplied so scale by sqrt(2). (-1+i)/sqrt(2). The other solution is just a mirror. I.e. -45 degrees.

  • @ianboard544
    @ianboard544 17 часов назад

    If you think of -i in polar form it's a hell of a lot easier.

  • @walterwen2975
    @walterwen2975 15 часов назад

    Oxford entrance exam question: √(- i) =?
    - i = (- 2i)/2 = (1 - 2i - 1)/2 = (1 - 2i + i²)/2 = [(1 - i)²]/2 = [(1 - i)/√2]²
    √(- i) = √{[(1 - i)/√2]²} = ± (1 - i)/√2 = ± (√2 - i√2)/2
    Final answer:
    √(- i) = (√2 - i√2)/2 or √(- i) = (- √2 + i√2)/2

  • @RealQinnMalloryu4
    @RealQinnMalloryu4 День назад

    (x ➖ 1ix+1i).