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Urgent help needed with complex numbers

  1. Jun 5, 2008 #1

    rock.freak667

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    [SOLVED] Urgent help needed with complex numbers

    1. The problem statement, all variables and given/known data
    a complex no. z is represented by the point T in an Argand diagram.

    [tex]z=\frac{1}{3+it}[/tex]

    where t is a variable

    show that z+z*=6ZZ*

    and that as t varies,T lies on a circle, and state its centre
    2. Relevant equations



    3. The attempt at a solution

    Did the first part easily.

    Need help with the 2nd part with the circle

    so far I multiplied z by z*/z* to get

    [tex]z=\frac{3-it}{p+t^2}[/tex]

    Do I now say that let z=x+iy and then find |z| and the modulus of the otherside (with t) and put that in the form [itex]x^2+y^2+2fx+2gy+c=0[/itex] ?
     
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  3. Jun 5, 2008 #2

    Dick

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    p is 9, right? Sure, you now have x=3/(9+t^2) and y=(-t)/(9+t^2). Eliminate the t in favor of x and y and write the quadratic form.
     
  4. Jun 5, 2008 #3

    rock.freak667

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    whoops sorry,p=9.


    so then

    [tex]y^2=\frac{t^2}{9+t^2}[/tex]

    and from the eq'n in x

    [tex]t^2=\frac{3}{x}-9[/tex]

    making

    [tex]y^2=(\frac{3}{x}-9)(\frac{x^2}{9})[/tex]

    [tex]x^2+y^2-\frac{1}{3}x=0[/tex]

    correct?
     
  5. Jun 5, 2008 #4

    Defennder

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    Small error here: Wasn't y = -t/(9+t^2) ? You didnt square denominator.
     
  6. Jun 5, 2008 #5

    rock.freak667

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    I did,I just left it out when typing

    x^2/9 is 1/(9+t^2)^2
     
  7. Jun 5, 2008 #6

    Dick

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    y^2=t^2/(9+t^2)^2. But everything else is correct, so I'll take that as a typo. Ok, so what's the center and radius?
     
  8. Jun 5, 2008 #7

    rock.freak667

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    then centre will just be (-1/6,0) and the radius is 1/6
     
  9. Jun 5, 2008 #8

    Dick

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    Yeah, Defennder is on your tail so you are rushing it right? Don't. You have a sign error in the center. Fix it quick! If y=0 then x=0 and x=1/3 are both on the curve.
     
    Last edited: Jun 5, 2008
  10. Jun 5, 2008 #9

    rock.freak667

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    [tex] x^2+y^2-\frac{1}{3}x=0 [/tex]

    [tex] x^2+y^2+2(-\frac{1}{6}x)+2(0)+0=0[/tex]

    f=-1/6
    g=0
    c=0

    is the eq'n wrong or did I actually not sq. the denominator?
     
  11. Jun 5, 2008 #10

    Dick

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    What are you doing? Just complete the square. x^2-2(x/6)=(x-1/6)^2-(1/6)^2. x-1/6 not x+1/6.
     
  12. Jun 5, 2008 #11

    rock.freak667

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    ahhh....my brain is idle

    when put in the form [itex]x^2+y^2+2fx+2gy+c=0[/itex] ,the centre is (-f,-g)

    sorry about my mistake

    so the centre is (1/6,0) and radius is 1/6
     
  13. Jun 5, 2008 #12

    Dick

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    No need to apologize. But, that's the problem with trying to memorize too many formulas you don't need.
     
  14. Jun 5, 2008 #13

    rock.freak667

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    yeah,I know but when I learnt the equation of a circle, that equation was the one I remembered more than the other one.

    But anyhow,thanks!!
     
  15. Jun 5, 2008 #14

    Dick

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    Now that I can agree with.
     
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