Complex Number Loci: Proof and Circle Variation with z = 1/(3+it)

In summary, a locus of complex numbers is a set of all complex numbers that satisfy a given condition and can be represented geometrically as a curve or shape on the complex plane. To find the locus, one must determine the condition and use algebraic or geometric methods to represent it. The significance of loci of complex numbers lies in their applications in mathematics, physics, and engineering. They can be a straight line if the condition is a linear equation, and not all complex numbers are included in a locus, only those that satisfy the given condition.
  • #1
righteous818
7
0
Given that z = 1/(3+it), it is denoted by T on a argand diagram
1. show that z + z* = 6zz*
Got this part out but the next part i am totally confused

I did abit of loci but i can't figure out this one

2. Show that if t varies T lies on a circle , and state the coordinates of the centre of the circle.

Please help i am clueless
 
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  • #2
righteous818 said:
Given that z = 1/(3+it), it is denoted by T on a argand diagram
1. show that z + z* = 6zz*
Got this part out but the next part i am totally confused

I did abit of loci but i can't figure out this one

2. Show that if t varies T lies on a circle , and state the coordinates of the centre of the circle.

Please help i am clueless

1. $\displaystyle z+ z^{*}= \frac{1}{3+i\ t}+ \frac{1}{3-i\ t}= \frac{6}{9+t^{2}}= z\ z^{*}$
Kind regards

$\chi$ $\sigma$
 
Last edited:
  • #3
i don't understand part 2 can u explain what u did abit more
 
  • #4
righteous818 said:
Given that z = 1/(3+it), it is denoted by T on a argand diagram
1. show that z + z* = 6zz*
Got this part out but the next part i am totally confused

I did abit of loci but i can't figure out this one

2. Show that if t varies T lies on a circle , and state the coordinates of the centre of the circle.

Please help i am clueless

Part 2.

\[z=\frac{1}{3+it}=\frac{3-it}{9+t^2}\]

So putting \(z=x+iy\) we have:

\[x=\frac{3}{9+t^2}\]\[y=-\frac{t}{9+t^2}\]

squaring and adding gives:

\[x^2+y^2=\frac{9}{(9+t^2)^2}+\frac{t^2}{(9+t^2)^2}=\frac{1}{9+t^2}=\frac{x}{3}\]

so:

\[\left(x^2-\frac{x}{3}+\frac{1}{6^2}\right)+y^2=\frac{1}{6^2}\]

or:

\[ \left( x-\frac{1}{6} \right)^2+y^2=\frac{1}{6^2}\]

CB
 
  • #5
chisigma said:
1. $\displaystyle z+ z^{*}= \frac{1}{3+i\ t}+ \frac{1}{3-i\ t}= \frac{6}{9+t^{2}}= {\color{red}6}z\ z^{*}$
Kind regards

$\chi$ $\sigma$

See correction in red.

CB
 
  • #6
righteous818 said:
Given that z = 1/(3+it), it is denoted by T on a argand diagram
1. show that z + z* = 6zz*
Got this part out but the next part i am totally confused

I did abit of loci but i can't figure out this one

2. Show that if t varies T lies on a circle , and state the coordinates of the centre of the circle.

Please help i am clueless

Part 2, method 2.

Given \(z + z^* = 6zz^*\), and letting \(z=x+iy\) we have: \(2x=6(x^2+y^2)\) ..

CB
 
  • #7
how and where did you get 1/36
 
  • #8
righteous818 said:
how and where did you get 1/36

By completing the square:

\[x^2-\frac{x}{3}= \left(x^2-\frac{x}{3}+\frac{1}{6^2}\right)-\frac{1}{6^2}=\left(x-\frac{1}{6}\right)^2-\frac{1}{6^2}\]

And please quote the post that your post is referring to.

CB
 
  • #9
righteous818 said:
Given that z = 1/(3+it), it is denoted by T on a argand diagram
1. show that z + z* = 6zz*
Got this part out but the next part i am totally confused

I did abit of loci but i can't figure out this one

2. Show that if t varies T lies on a circle , and state the coordinates of the centre of the circle.
For part 2, I would use part 1:

If $z + z^* = 6zz^*$ then $zz^* -\frac16z - \frac16z^* = 0.$ Therefore $\bigl(z-\frac16\bigr)\bigl(z^*-\frac16\bigr) = \frac1{36}.$ Thus $\bigl|z-\frac16\bigr|^2 = \frac1{36}.$ Take the square root to get $\bigl|z-\frac16\bigr| = \frac16$, which is the equation of a circle centred at 1/6 with radius 1/6.
 

FAQ: Complex Number Loci: Proof and Circle Variation with z = 1/(3+it)

What is a locus of complex numbers?

A locus of complex numbers refers to the set of all complex numbers that satisfy a given condition. It can be represented geometrically as a curve or shape on the complex plane.

How do you find the locus of a complex number?

To find the locus of a complex number, you must first determine the condition that the complex number must satisfy. Then, you can use algebraic or geometric methods to represent the locus on the complex plane.

What is the significance of loci of complex numbers?

The loci of complex numbers have important applications in mathematics, physics, and engineering. They can be used to solve equations, model physical phenomena, and analyze complex systems.

Can a locus of complex numbers be a straight line?

Yes, a locus of complex numbers can be a straight line. This occurs when the condition for the complex numbers to satisfy is a linear equation in the form of ax + by = c, where a, b, and c are real numbers.

Are all complex numbers included in a locus?

No, not all complex numbers are included in a locus. Only the complex numbers that satisfy the given condition are included in the locus. The remaining complex numbers may be included in a different locus with a different condition.

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