Posers and Puzzles

Posers and Puzzles

  1. Argentina
    Joined
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    18 Jan '06 07:371 edit
    You have at most an hour to reply:
    Which is the square root of -i ?

    P.S.: I dont know the answer. but could intenti to find it! And this isn't a college duty! 😀

    -J

    EDIT: well, you deserve more time... it's too late in the night! 🙂
  2. Standard memberTheMaster37
    Kupikupopo!
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    18 Jan '06 08:55
    Originally posted by CrazyLilTing
    You have at most an hour to reply:
    Which is the square root of -i ?

    P.S.: I dont know the answer. but could intenti to find it! And this isn't a college duty! 😀

    -J

    EDIT: well, you deserve more time... it's too late in the night! 🙂
    Aside from sqrt(-i)?

    i*sqrt(i) would work too, is the same.
  3. Standard memberXanthosNZ
    Cancerous Bus Crash
    p^2.sin(phi)
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    18 Jan '06 09:10
    Google knows all.

    http://www.google.com/search?&q=sqrt(-i)

    Also 0.707106781 = sqrt(2)/2
  4. Joined
    31 Oct '05
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    18 Jan '06 14:21
    Originally posted by XanthosNZ
    Google knows all.

    http://www.google.com/search?&q=sqrt(-i)

    Also 0.707106781 = sqrt(2)/2
    You forgot the (-sqrt(2)/2i)

    sqrt(-i) = sqrt(2)/2 - sqrt(2)/2i
  5. Standard memberXanthosNZ
    Cancerous Bus Crash
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    18 Jan '06 15:23
    Originally posted by Balla88
    You forgot the (-sqrt(2)/2i)

    sqrt(-i) = sqrt(2)/2 - sqrt(2)/2i
    No I didn't. As the google page says the answer is 0.707... - 0.707...i

    I also pointed out that 0.707... = sqrt(2)/2 as the squareroot form is much more useful (and precise). I did not state that the answer was 0.707...

    Also be careful of your brackets. sqrt(2)/2i is easily read incorrectly. Better would be sqrt(2)/2 * i.
  6. Joined
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    18 Jan '06 15:35
    It is read incorrectly only by people that do not know math. They should be more careful then.
  7. Standard memberroyalchicken
    CHAOS GHOST!!!
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    18 Jan '06 19:03
    -i = exp{i*[(-pi)/2]}, whose square root is exp{i*[(-pi)/4]} = [sqrt(2)/2]*(1-i), as they said.

    I only contributed to this thread so I could take the opportunity to be paranthetically unambiguous in as many places as possible, thus doing all I can to keep plonkers like XanthosNZ at bay.

    This is also my first use of the word 'plonker' in any context, ever.
  8. Standard memberleisurelysloth
    Man of Steel
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    20 Jan '06 06:521 edit
    Originally posted by XanthosNZ
    Google knows all.

    http://www.google.com/search?&q=sqrt(-i)

    Also 0.707106781 = sqrt(2)/2
    Google is wise. Wikipedia is wiser!

    http://en.wikipedia.org/wiki/Complex_numbers#Geometric_interpretation_of_the_operations_on_complex_numbers

    It's best to think of complex numbers geometrically as suggested in the Wikipedia page. Then it becomes clear that there are two answers to the question. Both:

    1/sqrt(2) - i/sqrt(2)

    and

    - 1/sqrt(2) + i/sqrt(2)

    are solutions.

    EDIT: Hey Russ! How 'bout you stop fiddling around with that analysis board, and get us a decent equation editor in the forums. Jeez, you'd think this was a chess site or somethin'. 😛
  9. Standard memberleisurelysloth
    Man of Steel
    rushing to and fro
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    20 Jan '06 07:15
    Originally posted by royalchicken
    -i = exp{i*[(-pi)/2]}, whose square root is exp{i*[(-pi)/4]} = [sqrt(2)/2]*(1-i), as they said.

    I only contributed to this thread so I could take the opportunity to be paranthetically unambiguous in as many places as possible, thus doing all I can to keep plonkers like XanthosNZ at bay.

    This is also my first use of the word 'plonker' in any context, ever.
    Don't let it happen again! 😠
  10. Standard memberroyalchicken
    CHAOS GHOST!!!
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    20 Jan '06 12:17
    Originally posted by leisurelysloth
    Google is wise. Wikipedia is wiser!

    http://en.wikipedia.org/wiki/Complex_numbers#Geometric_interpretation_of_the_operations_on_complex_numbers

    It's best to think of complex numbers geometrically as suggested in the Wikipedia page. Then it becomes clear that there are two answers to the question. Both:

    1/sqrt(2) - i/sqrt(2)

    and

    - 1 ...[text shortened]... ent equation editor in the forums. Jeez, you'd think this was a chess site or somethin'. 😛
    True, but we'll assume the original poster asked for the principal value 😛.
  11. Standard memberark13
    Enola Straight
    mouse mouse mouse
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    20 Jan '06 17:57
    Originally posted by royalchicken
    -i = exp{i*[(-pi)/2]}, whose square root is exp{i*[(-pi)/4]} = [sqrt(2)/2]*(1-i), as they said.

    I only contributed to this thread so I could take the opportunity to be paranthetically unambiguous in as many places as possible, thus doing all I can to keep plonkers like XanthosNZ at bay.

    This is also my first use of the word 'plonker' in any context, ever.
    It's a right of passage.
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