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

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Homework Help Overview

The discussion revolves around the representation of a complex number in an Argand diagram, specifically focusing on the relationship between the variable \( t \) and the geometric representation of the complex number as it varies. The original poster seeks to demonstrate that as \( t \) changes, the point \( T \) lies on a circle, and they are tasked with identifying the center of that circle.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss manipulating the expression for \( z \) and its conjugate to derive the equation of a circle. There are attempts to express \( z \) in terms of \( x \) and \( y \) and to eliminate \( t \) from the equations. Questions arise regarding the correctness of the transformations and the implications for the center and radius of the circle.

Discussion Status

The conversation is active, with participants providing feedback on each other's calculations and interpretations. Some participants have offered corrections and clarifications regarding the center and radius of the circle, while others have pointed out potential errors in the algebraic manipulations. There is a recognition of the need to complete the square to accurately determine the center of the circle.

Contextual Notes

There is a mention of specific values, such as \( p = 9 \), which may influence the calculations. The participants are also navigating through potential typographical errors and misunderstandings related to the algebraic forms used to describe the circle.

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.

[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

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

[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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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

[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?
 
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:
[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?
 
  • #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 [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
 
  • #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 [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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