• Jul 29th 2012, 06:16 PM
righteous818
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.

• Jul 29th 2012, 07:15 PM
Prove It
Quote:

Originally Posted by righteous818
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.

Are you using z* to represent the conjugate of z?
• Jul 29th 2012, 07:16 PM
righteous818
yes i am
• Jul 29th 2012, 07:52 PM
Prove It
Quote:

Originally Posted by righteous818
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.

Q.1

\displaystyle \displaystyle \begin{align*} z &= \frac{1}{3 + i\,t} \\ &= \frac{1}{3 + i\,t} \cdot \frac{3 - i\,t}{3 - i\,t} \\ &= \frac{3 - i\,t}{9 + t^2} \\ &= \frac{3}{9 + t^2} - i\left( \frac{t}{9 + t^2} \right) \\ \\ \overline{z} &= \frac{3}{9 + t^2} + i\left(\frac{t}{9 + t^2}\right) \\ \\ z + \overline{z} &= \frac{3}{9 + t^2} - i\left(\frac{t}{9 + t^2}\right) + \frac{3}{9 + t^2} + i\left(\frac{t}{9 + t^2}\right) \\ &= \frac{6}{9 + t^2} \\ \\ \\ 6z\overline{z} &= 6\left[ \frac{3}{9 + t^2} - i\left(\frac{t}{9 + t^2}\right) \right]\left[ \frac{3}{9 + t^2} + i\left(\frac{t}{9 + t^2}\right) \right] \\ &= 6\left[ \frac{9}{\left(9 + t^2\right)^2} + \frac{t^2}{\left(9 + t^2\right)^2} \right] \\ &= 6\left[ \frac{9 + t^2}{\left(9 + t^2\right)^2} \right] \\ &= \frac{6}{9 + t^2} \\ &= z + \overline{z} \end{align*}

Q.2

Since we know \displaystyle \displaystyle \begin{align*} z \end{align*} lies on the circle, so does \displaystyle \displaystyle \begin{align*} \overline{z} \end{align*}. Also, so will the points that are these reflected in the Imaginary axis, \displaystyle \displaystyle \begin{align*} -\frac{3}{9 + t^2} + i\left(\frac{t}{9 + t^2}\right) \end{align*} and \displaystyle \displaystyle \begin{align*} -\frac{3}{9 + t^2} - i\left(\frac{t}{9 + t^2}\right) \end{align*}.

If we join the diagonally opposite points, we get two lines that cross at the centre of the circle.

Line 1: Points \displaystyle \displaystyle \begin{align*} \left( \frac{3}{9 + t^2}, -\frac{t}{9 + t^2}\right) \end{align*} and \displaystyle \displaystyle \begin{align*} \left(-\frac{3}{9 + t^2}, \frac{t}{9 + t^2}\right) \end{align*} lie on the line, so

\displaystyle \displaystyle \begin{align*} m &= \frac{\frac{t}{9 + t^2} - \left(-\frac{t}{9 + t^2}\right)}{-\frac{3}{9 + t^2} - \frac{3}{9 + t^2}} \\ &= \frac{\frac{2t}{9 + t^2}}{-\frac{6}{9 + t^2}} \\ &= -\frac{t}{3} \\ \\ y &= m\,x + c \\ -\frac{t}{9 + t^2} &= -\frac{t}{3}\left( \frac{3}{9 + t^2} \right) + c \\ -\frac{t}{9 + t^2} &= -\frac{t}{9 + t^2} + c \\ c &= 0 \\ \\ y &= -\frac{t}{3}\,x \end{align*}

Line 2: Points \displaystyle \displaystyle \begin{align*} \left( \frac{3}{9 + t^2}, \frac{t}{9 + t^2} \right) \end{align*} and \displaystyle \displaystyle \begin{align*} \left(-\frac{3}{9 + t^2}, -\frac{t}{9 + t^2}\right) \end{align*} lie on the line, so

\displaystyle \displaystyle \begin{align*} m &= \frac{-\frac{t}{9 + t^2} - \frac{t}{9 + t^2}}{-\frac{3}{9 + t^2} - \frac{3}{9 + t^2}} \\ &= \frac{-\frac{2t}{9 + t^2}}{-\frac{6}{9 + t^2}} \\ &= \frac{t}{3} \\ \\ y &= m\,x + c \\ \frac{t}{9 + t^2} &= \frac{t}{3}\left(\frac{3}{9 + t^2}\right) + c \\ \frac{t}{9 + t^2} &= \frac{t}{9 + t^2} + c \\ c &= 0 \\ \\ y &= \frac{t}{3}\,x \end{align*}

Both points have a y intercept of (0, 0), so this is clearly the point where the two lines cross. Therefore the centre of the circle is (0, 0).
• Jul 29th 2012, 08:00 PM
righteous818
shouldn it be reflected on the real axis
• Jul 29th 2012, 08:04 PM
Prove It
Quote:

Originally Posted by righteous818
shouldn it be reflected on the real axis

No, conjugates are already reflected in the real axis.
• Jul 29th 2012, 08:05 PM
righteous818