How Does a Complex Number Represent a Circle in an Argand Diagram?

AI Thread Summary
The discussion focuses on how a complex number represented in an Argand diagram can describe a circle. The complex number z is defined as z = 1/(3 + it), and participants work through the algebra to show that as t varies, the point T lies on a circle. They derive the equation x^2 + y^2 - (1/3)x = 0, identifying the center of the circle as (1/6, 0) and the radius as 1/6. Participants emphasize the importance of correctly applying formulas and completing the square to find the circle's properties. The conversation highlights common mistakes in algebraic manipulation while solving the problem.
rock.freak667
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[SOLVED] Urgent help needed with complex numbers

Homework Statement


a complex no. z is represented by the point T in an Argand diagram.

z=\frac{1}{3+it}

where t is a variable

show that z+z*=6ZZ*

and that as t varies,T lies on a circle, and state its centre

Homework Equations





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

z=\frac{3-it}{p+t^2}

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 x^2+y^2+2fx+2gy+c=0 ?
 
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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.
 
Dick said:
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

y^2=\frac{t^2}{9+t^2}

and from the eq'n in x

t^2=\frac{3}{x}-9

making

y^2=(\frac{3}{x}-9)(\frac{x^2}{9})

x^2+y^2-\frac{1}{3}x=0

correct?
 
Small error here: Wasn't y = -t/(9+t^2) ? You didnt square denominator.
 
I did,I just left it out when typing

x^2/9 is 1/(9+t^2)^2
 
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?
 
then centre will just be (-1/6,0) and the radius is 1/6
 
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:
x^2+y^2-\frac{1}{3}x=0

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

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

is the eq'n wrong or did I actually not sq. the denominator?
 
  • #10
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.
 
  • #11
ahhh...my brain is idle

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

sorry about my mistake

so the centre is (1/6,0) and radius is 1/6
 
  • #12
No need to apologize. But, that's the problem with trying to memorize too many formulas you don't need.
 
  • #13
yeah,I know but when I learned the equation of a circle, that equation was the one I remembered more than the other one.

But anyhow,thanks!
 
  • #14
rock.freak667 said:
ahhh...my brain is idle

when put in the form x^2+y^2+2fx+2gy+c=0 ,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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