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Urgent help needed with complex numbers

by rock.freak667
Tags: complex, numbers, solved, urgent
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rock.freak667
#1
Jun5-08, 10:31 PM
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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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Dick
#2
Jun5-08, 10:53 PM
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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.
rock.freak667
#3
Jun5-08, 11:06 PM
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Quote Quote by Dick View Post
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.
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?

Defennder
#4
Jun5-08, 11:11 PM
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Urgent help needed with complex numbers

Small error here: Wasn't y = -t/(9+t^2) ? You didnt square denominator.
rock.freak667
#5
Jun5-08, 11:14 PM
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I did,I just left it out when typing

x^2/9 is 1/(9+t^2)^2
Dick
#6
Jun5-08, 11:16 PM
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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?
rock.freak667
#7
Jun5-08, 11:21 PM
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then centre will just be (-1/6,0) and the radius is 1/6
Dick
#8
Jun5-08, 11:26 PM
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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.
rock.freak667
#9
Jun5-08, 11:31 PM
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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?
Dick
#10
Jun5-08, 11:36 PM
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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.
rock.freak667
#11
Jun5-08, 11:41 PM
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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
Dick
#12
Jun5-08, 11:44 PM
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No need to apologize. But, that's the problem with trying to memorize too many formulas you don't need.
rock.freak667
#13
Jun5-08, 11:46 PM
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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!!
Dick
#14
Jun5-08, 11:48 PM
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Quote Quote by rock.freak667 View Post
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
Now that I can agree with.


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