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Math Help - limit of complex variable

  1. #1
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    limit of complex variable

    lim z->1
    [z(conjugate z) + 2z - (conjugate z) -2 ]
    ---------------------------------------
    z^2 - 1

    answer given is z / 2

    but i stuck at

    2+(conjugate z)
    ----------------
    z + 1

    can anyone help pls...
    thank you...
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  2. #2
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    Quote Originally Posted by kennytsh View Post
    lim z->1
    [z(conjugate z) + 2z - (conjugate z) -2 ]
    ---------------------------------------
    z^2 - 1

    answer given is z / 2

    but i stuck at

    2+(conjugate z)
    ----------------
    z + 1

    can anyone help pls...
    thank you...
    It can be written:

     \frac{z\bar{z} - \bar{z}}{(z+1)(z-1)} + \frac{2z - 2}{(z+1)(z-1)}

     \frac{\bar{z}(z-1)}{(z+1)(z-1)} + \frac{2(z-1)}{(z+1)(z-1)}

     \frac{\bar{z}}{z+1} + \frac{2}{z+1} = \frac{\bar{z} + 2}{z+1}

    Now multiply top and bottom by z:

     \frac{\bar{z}z + 2z}{z^2+z}

    And remember that  \bar{z}z = |z|^2

    PS: The answer can't be  \frac{z}{2} ...
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  3. #3
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    Hello, kennytsh!

    \lim_{z\to1} \frac{z\overline{z} + 2z - \overline{z} - 2}{z^2-1}

    Answer given: \frac{z}{2} . . . . I don't agree

    but stuck at: . \frac{\overline{z}+2}{z+1}\quad {\color{blue}\leftarrow\:\text{Right!}}

    We have: . \frac{z\overline{z} + 2z - \overline{z} - 2}{z^2-1} \;=\;\frac{z(\overline{z}+2) - (\overline{z}+2)}{z^2-1}\;=\;\frac{(z-1)(\overline{z}+2)}{(z-1)(z+1)} \;=\;\frac{\overline{z}+2}{z+1}


    Now we have: . \lim_{z\to1}\frac{\overline{z}+2}{z+1} \;=\;\lim_{z\to1} \frac{a-bi + 2}{a + bi + 1}


    But if z\to1, then: . \begin{array}{c}a \to 1 \\ b\to 0 \end{array}

    . . So we have: . \lim_{\begin{array}{c}^{a\to1}\\ [-2mm]^{b\to0}\end{array}} \frac{a - bi + 2}{a + bi + 1} \;=\;\frac{1 - 0 + 2}{1 + 0+ 1} \;=\;\frac{3}{2}

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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, kennytsh!


    We have: . \frac{z\overline{z} + 2z - \overline{z} - 2}{z^2-1} \;=\;\frac{z(\overline{z}+2) - (\overline{z}+2)}{z^2-1}\;=\;\frac{(z-1)(\overline{z}+2)}{(z-1)(z+1)} \;=\;\frac{\overline{z}+2}{z+1}


    Now we have: . \lim_{z\to1}\frac{\overline{z}+2}{z+1} \;=\;\lim_{z\to1} \frac{a-bi + 2}{a + bi + 1}


    But if z\to1, then: . \begin{array}{c}a \to 1 \\ b\to 0 \end{array}

    . . So we have: . \lim_{\begin{array}{c}^{a\to1}\\ [-2mm]^{b\to0}\end{array}} \frac{a - bi + 2}{a + bi + 1} \;=\;\frac{1 - 0 + 2}{1 + 0+ 1} \;=\;\frac{3}{2}

    Surely it's simply to state that  \bar{z}z = |z|^2 , evaluate the limit of just z.
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