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.

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?

yes i am

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

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

\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).

shouldn it be reflected on the real axis

Originally Posted by righteous818
shouldn it be reflected on the real axis
No, conjugates are already reflected in the real axis.