Solved: Urgent Help Needed with Complex Numbers

[rock.freak667]

[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 $x^2+y^2+2fx+2gy+c=0$ ?

 

[Dick]

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]

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?

 

[Defennder]

Small error here: Wasn't y = -t/(9+t^2) ? You didnt square denominator.

 

[rock.freak667]

I did,I just left it out when typing

x^2/9 is 1/(9+t^2)^2

 

[Dick]

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]

then centre will just be (-1/6,0) and the radius is 1/6

 

[Dick]

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]

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

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

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

is the eq'n wrong or did I actually not sq. the denominator?

 

[Dick]

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]

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

 

[Dick]

No need to apologize. But, that's the problem with trying to memorize too many formulas you don't need.

 

[rock.freak667]

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!

 

[Dick]

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.