Re: square root of 3i

From: Dan (30pack_at_sbcglobal.net)
Date: 07/08/04 

Date: Thu, 8 Jul 2004 03:39:21 +0000 (UTC)

On 07 Jul 2004, Daniel W. Johnson wrote:
>Dan <30pack@sbcglobal.net> wrote:
>
>> Thanks for the correction, but it still does not answer my question--
>> In the third quadrant in the Cartesian coordinate system where the
>> values of x,y are [-x, -y] and (my) sqrt(-3) =
>>
>> (((sqrt(3*4+2))-2)/2) * -1 = -.870828693...
>
>First, how do you get sqrt(-3) to be anything other than sqrt(3)i or
>-sqrt(3)i?
>
>> Where--
>> (-x)= -.870828693...
>> (-y)= -2.129171307...
>> Completing the square -- -x + -y = -3.
>>
>> Checking the square --
>> -3 =((((.870828693...* 2 +2)^2)-2)/4)*-1
>>
>> Checking Cartesian 3rd quadrant sqrt(-3) against (descartesian)
>> sqrt(-3)--
>
>What is "descartesian", and how is it different from "Cartesian"?

I would say "descartesian" is where y is imaginary and x is real.
"Cartesian" is where x and y are reals.
  
>> ((((sqrt(3*4+2))-2)/2) * -1) / (sqrt(3/2) + sqrt(3/2 )i) =
>> -.355514325... + .355514325...i = SAME DIDGETS IN THE
>> DECIMAL EXPANSION --->oo on the real side and on the imaginary side.
>
>Let a be the real number ((((sqrt(3*4+2))-2)/2) * -1)
>Let b be the real number sqrt(3/2)
>Note that 1/(1+i) = (1-i)/2
>
>Why would you be surprised that a/(b*(1+i)) = (a/b)*((1-i)/2) =
>(a/(b*2)) * (1-i) = (a/(b*2)) - (a/(b*2))i has real and imaginary parts
>with the same absolute value?

You are right, my blunder!

Thanks,

Dan
  
>Daniel W. Johnson
>panoptes@iquest.net
>http://members.iquest.net/~panoptes/
>039 53 36 N / 086 11 55 W

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