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Loci of Complex Number

righteous818

New member
Jul 30, 2012
4
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.

Please help i am clueless
 

chisigma

Well-known member
Feb 13, 2012
1,704
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.

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:

righteous818

New member
Jul 30, 2012
4
i dont understand part 2 can u explain what u did abit more
 

CaptainBlack

Well-known member
Jan 26, 2012
890
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.

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
 

CaptainBlack

Well-known member
Jan 26, 2012
890
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
Jan 26, 2012
890
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.

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
 

righteous818

New member
Jul 30, 2012
4
how and where did you get 1/36
 

CaptainBlack

Well-known member
Jan 26, 2012
890
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
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
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.