# Urgent help needed with complex numbers

by rock.freak667
Tags: complex, numbers, solved, urgent
 HW Helper P: 6,202 1. The problem statement, all variables and given/known data 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 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 $$z=\frac{3-it}{p+t^2}$$ 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 $x^2+y^2+2fx+2gy+c=0$ ?
 Sci Advisor HW Helper Thanks P: 25,228 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.
HW Helper
P: 6,202
 Quote by 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.
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?

 HW Helper P: 2,616 Urgent help needed with complex numbers Small error here: Wasn't y = -t/(9+t^2) ? You didnt square denominator.
 HW Helper P: 6,202 I did,I just left it out when typing x^2/9 is 1/(9+t^2)^2
 Sci Advisor HW Helper Thanks P: 25,228 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?
 HW Helper P: 6,202 then centre will just be (-1/6,0) and the radius is 1/6
 Sci Advisor HW Helper Thanks P: 25,228 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.
 HW Helper P: 6,202 $$x^2+y^2-\frac{1}{3}x=0$$ $$x^2+y^2+2(-\frac{1}{6}x)+2(0)+0=0$$ f=-1/6 g=0 c=0 is the eq'n wrong or did I actually not sq. the denominator?
 Sci Advisor HW Helper Thanks P: 25,228 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.
 HW Helper P: 6,202 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
 Sci Advisor HW Helper Thanks P: 25,228 No need to apologize. But, that's the problem with trying to memorize too many formulas you don't need.
 HW Helper P: 6,202 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!!
 Quote by 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