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Complex numbers - polar form - does this work (indices) ?

  1. Sep 26, 2007 #1

    JPC

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    Complex numbers - polar form - does this work (indices) ???

    hey

    i havent studied in class complex numbers yet, but i know some of the basis , and i was wondering if something i saw in complex numbers was true :

    polar form :
    let 'a' be the angle
    and x the length (dont know how to call it in english)
    : (x, a)

    ok so now is this true :

    -> ( (x, a) )^n = (x^n, a * n)

    -> sqroot( (x, a) ) = (sqroot(x), a/2) or (sqroot(x), pi + a/2)
     
  2. jcsd
  3. Sep 26, 2007 #2

    dextercioby

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    The first formula is true. While for the second, the one following "or" is false.
     
  4. Sep 26, 2007 #3

    mathman

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    JPC is correct. For angle a, to get nth roots, use (a+2kpi)/n for k=0,1,..,n-1.

    Therefore for sqrt, need k=0,1.
     
  5. Sep 28, 2007 #4

    JPC

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    ok , and now , is this true for cube root , i only checked it on xroot(3, i) yet :

    polar :
    xroot(3, (x, a) ) =

    ( xroot(3, x) , a / 3 )
    or
    ( xroot(3, x) , pi - a/3 )
    or
    ( xroot(3, x) , a + pi )

    im especially not sure for the "( xroot(3, x) , a + pi )", because i cant really find a rule for it.
    finding cube root of i :

    (a+ bi)^3 = i :

    a^3 - 3ab² = 0
    3a²b - b^3 = 1

    a ( a² - 3b²) = 0
    a = 0 or a² = 3b²
    a = sqroot(3) * b , or a = - sqroot(3)* b or a = 0

    i replace in the second equation a² by 3b²
    and find b = 1 / 2
    and find a= sqroot(3) / 2

    now, if a=0 , then we end up in the second equation with :
    -b^3 = 1
    b= -1

    giving me 3 solutions :

    + sqroot(3) / 2 + 0.5i
    or
    - sqroot(3) / 2 + 0.5i
    or
    -i
     
  6. Sep 28, 2007 #5

    mathman

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    Your notation is very confusing:

    In any case if (x,a) represents a number in polar coordinates, the three cube roots are
    (y,a/3), (y,(a+2pi)/3),(y,(a+4pi)/3), where y is the cube root of x.
     
  7. Sep 29, 2007 #6

    JPC

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    Oh yeah, works completely with cube root of i and others, thanks

    c : an Imaginary Number)

    Now maybe ill try xroot(4, c), with c = i

    would it end up with 4 solutions , (do sqroot of c once, then do sqroot to each of the first sqroot answers) ?
    or would there only be 2 answers ? (out of the 4 answers of that way, 2 by 2 they end up the same)
     
  8. Sep 29, 2007 #7

    mathman

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    Your notation is confusing - I have never seen anything like it. Could you explain what it means? If c is supposed to be angle, it has to be a real number.
     
  9. Sep 30, 2007 #8

    JPC

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    well like square root of a number c is :
    xroot(2, c)

    if u have xroot(a, b) means :

    .....___
    a \/ b

    this notation is from my HP calculator
     
  10. Oct 1, 2007 #9

    mathman

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    It looks like xroot(a,b) means (in standard notation) b1/a. Here b is any complex number, but the form of b has not be specified. There are two usual ways:

    1) b=x+iy, where x and y are real.

    2) b=r(eiz), where r is non-negative and z is an angle between 0 and 2pi.

    The question I was trying to answer previously iinvolves using the second representation.
     
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