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Posers and Puzzles

Posers and Puzzles

  1. 18 Jan '06 07:37 / 1 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 member TheMaster37
    Kupikupopo!
    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 member XanthosNZ
    Cancerous Bus Crash
    18 Jan '06 09:10
    Google knows all.

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

    Also 0.707106781 = sqrt(2)/2
  4. 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 member XanthosNZ
    Cancerous Bus Crash
    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. 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 member royalchicken
    CHAOS GHOST!!!
    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 member leisurelysloth
    Man of Steel
    20 Jan '06 06:52 / 1 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 member leisurelysloth
    Man of Steel
    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 member royalchicken
    CHAOS GHOST!!!
    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 member ark13
    Enola Straight
    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.