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Math Help - complex number question

  1. #1
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    complex number question

    if z = a + bi and|z| = 1, let w = (z+1)/(z-1). prove that w is a pure imaginary number of the form bi.

    so from the given i have a^2 + b^2 = 1 since sqrt(a^2 + b^2) = 1 = |z|.

    so i plugged in a + bi for z into w = ((a + bi) + 1) / ((a + bi) - 1) i multiplied the top and bottom by the conjugate of the bottom witch is ((a + bi) + 1) and i got something complicated: (2a^2 + 2abi + 2a + 2bi) / (2abi - 2b^2) when i split this up into separate fractions i don't get a pure imaginary number. how do i do this?
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  2. #2
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    Quote Originally Posted by oblixps View Post
    if z = a + bi and|z| = 1, let w = (z+1)/(z-1). prove that w is a pure imaginary number of the form bi.?
    You should know that \left| z \right| = 1\; \Rightarrow \;z\overline z  = 1 and that \left| {z - 1} \right|^2 is a real number.
    Also that \overline z  - z =  - 2\text{Im} (z).

    \frac{{z + 1}}{{z - 1}} = \frac{{\left( {z + 1} \right)\left( {\overline z  - 1} \right)}}{{\left| {z - 1} \right|^2 }} = \frac{{z\overline z  + \left( {\overline z  - z}\right) - 1}}{{\left| {z - 1} \right|^2 }}

    Now put it all together.
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  3. #3
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    Quote Originally Posted by Plato View Post
    You should know that \left| z \right| = 1\; \Rightarrow \;z\overline z  = 1 and that \left| {z - 1} \right|^2 is a real number.
    Also that \overline z  - z =  - 2\text{Im} (z).

    \frac{{z + 1}}{{z - 1}} = \frac{{\left( {z + 1} \right)\left( {\overline z  - 1} \right)}}{{\left| {z - 1} \right|^2 }} = \frac{{z\overline z  + \left( {\overline z  - z}\right) - 1}}{{\left| {z - 1} \right|^2 }}

    Now put it all together.
    what does the notation \overline z and \ - 2\text{Im} (z) mean? i haven't encountered those before in my class.
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  4. #4
    MHF Contributor red_dog's Avatar
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    We have, as Plato said, z\cdot\overline{z}=1\Rightarrow \overline{z}=\frac{1}{z}

    A complex number w is pure imaginary if and only if \overline{w}=-w.

    \overline{w}=\overline{\left(\frac{z+1}{z-1}\right)}=\frac{\overline{z+1}}{\overline{z-1}}=\frac{\overline{z}+1}{\overline{z}-1}=

    =\frac{\displaystyle\frac{1}{z}+1}{\displaystyle\f  rac{1}{z}-1}=\frac{1+z}{1-z}=-\frac{z+1}{z-1}=-w
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