# Loci of Complex Number

#### righteous818

##### New member
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 cant 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.

#### chisigma

##### Well-known member
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 cant 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.

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:

#### righteous818

##### New member
i dont understand part 2 can u explain what u did abit more

#### CaptainBlack

##### Well-known member
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 cant 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.

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}$

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

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

#### CaptainBlack

##### Well-known member
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

#### CaptainBlack

##### Well-known member
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 cant 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.

Part 2, method 2.

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

CB

#### righteous818

##### New member
how and where did you get 1/36

#### CaptainBlack

##### Well-known member
how and where did you get 1/36
By completing the square:

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

CB

#### Opalg

##### MHB Oldtimer
Staff member
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 cant 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.