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Math Help - Please help complex numbers

  1. #1
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    Please help complex numbers

    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
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  2. #2
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    Re: Please help complex numbers

    Quote Originally Posted by righteous818 View Post
    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
    Are you using z* to represent the conjugate of z?
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  3. #3
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    Re: Please help complex numbers

    yes i am
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    Re: Please help complex numbers

    Quote Originally Posted by righteous818 View Post
    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
    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).
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  5. #5
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    Re: Please help complex numbers

    shouldn it be reflected on the real axis
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    Re: Please help complex numbers

    Quote Originally Posted by righteous818 View Post
    shouldn it be reflected on the real axis
    No, conjugates are already reflected in the real axis.
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  7. #7
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    Re: Please help complex numbers

    strange this is from an examination board an they gave the anser in their markscheme as the centre (1/6 , 0)
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  8. #8
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    Re: Please help complex numbers

    but how can you assume that the points reflected on the imaginary axis is on the circle , arent if u are doin that then you already assume that the centre is (0,0)
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