Solved: Urgent Help Needed with Complex Numbers

In summary, the conversation discusses a problem involving complex numbers on an Argand diagram and how to show that the points lie on a circle. The first part of the problem was easily solved, but the second part required finding the center and radius of the circle. The conversation ends with the correct solution being (1/6,0) for the center and a radius of 1/6.
  • #1
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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  • #2
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.
 
  • #3
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?
 
  • #4
Small error here: Wasn't y = -t/(9+t^2) ? You didnt square denominator.
 
  • #5
I did,I just left it out when typing

x^2/9 is 1/(9+t^2)^2
 
  • #6
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?
 
  • #7
then centre will just be (-1/6,0) and the radius is 1/6
 
  • #8
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:
  • #9
[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.
 

What are complex numbers?

Complex numbers are numbers that contain both a real part and an imaginary part. They are represented in the form of a + bi, where a is the real part and bi is the imaginary part with i being the imaginary unit equal to √(-1).

How do you add or subtract complex numbers?

To add or subtract complex numbers, you simply combine the real parts and the imaginary parts separately. For example, (3 + 2i) + (5 + 4i) = (3 + 5) + (2i + 4i) = 8 + 6i. Similarly, (3 + 2i) - (5 + 4i) = (3 - 5) + (2i - 4i) = -2 - 2i.

How do you multiply complex numbers?

To multiply complex numbers, you use the FOIL method just like you would with binomials. For example, (3 + 2i)(5 + 4i) = 15 + 12i + 10i + 8i^2 = 15 + 22i - 8 = 7 + 22i.

How do you divide complex numbers?

To divide complex numbers, you use the conjugate of the denominator to rationalize the expression. This means that you multiply the numerator and denominator by the conjugate of the denominator. For example, (3 + 2i) / (5 + 4i) = (3 + 2i)(5 - 4i) / (5 + 4i)(5 - 4i) = (15 - 12i + 10i - 8i^2) / (25 - 16i^2) = (7 - 2i) / 41.

What are the applications of complex numbers?

Complex numbers have many applications in various fields such as engineering, physics, and economics. They are used to represent wave functions, electrical circuits, and financial models. They are also used in solving differential equations and finding roots of polynomials.

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