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Math Help - A complex equation

  1. #1
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    A complex equation

    Can anyone help me show this equation?


    For a complex z != 0

    Thanks
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by changing_seasons View Post
    Can anyone help me show this equation?


    For a complex z != 0

    Thanks
    \left | \bar{z} - \frac{1}{z} \right | = \left | |z| - \frac{1}{|z|} \right |

    Let z = a + bi. Then \bar{z} = a - bi and |z| = | \bar{z} | = \sqrt{a^2 + b^2}. So the LHS becomes:
    \left | \bar{z} - \frac{1}{z} \right | = \left | a - bi - \frac{1}{a + bi} \right | = \left |\frac{(a - bi)(a^2 + b^2) - (a - bi)}{a^2 + b^2} \right |

    = \left | \frac{(a^2 + b^2 - 1)}{a^2 + b^2} \cdot (a - bi) \right | = \frac{a^2 + b^2 - 1}{\sqrt{a^2 + b^2}}

    Now for the RHS:
    \left | |z| - \frac{1}{|z|} \right | = \left | \sqrt{a^2 + b^2} - \frac{1}{\sqrt{a^2 + b^2}} \right |

    Since this is real we don't need the outside | | bars.

    = \frac{(a^2 + b^2) - 1}{\sqrt{a^2 + b^2}}

    which is the same as the LHS.

    -Dan
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  3. #3
    Senior Member TriKri's Avatar
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    \left|\overline{z}-\frac{1}{z}\right|\ =\ \left||z|-\frac{1}{|z|}\right|\ \Leftrightarrow\ /z \neq 0/\ \Leftrightarrow

    \left|\overline{z}-\frac{1}{z}\right|\cdot|z|\ =\ \left||z|-\frac{1}{|z|}\right|\cdot|z|\ \Leftrightarrow

    \left|\overline{z}\cdot z-1\right|\ =\ \left||z|^2-1\right|\ \Leftrightarrow

    \left||z|^2-1\right|\ =\ \left||z|^2-1\right|\ \Leftrightarrow

    1\ =\ 1, which is true for all z

    z \not = 0, else z can be anything.

    -Kristofer.
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  4. #4
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    Quote Originally Posted by topsquark View Post

    = \left | \frac{(a^2 + b^2 - 1)}{a^2 + b^2} \cdot (a - bi) \right | = \frac{a^2 + b^2 - 1}{\sqrt{a^2 + b^2}}
    Thanks for the help, but could you please elaborate on this final step?

    NEVER MIND I got it!
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by changing_seasons View Post
    Thanks for the help, but could you please elaborate on this final step?

    NEVER MIND I got it!
    I'll do it anyway, in case someone wants it. Because I'm just that nice a guy!

    \left | \frac{(a^2 + b^2 - 1)}{a^2 + b^2} \cdot (a - bi) \right |

    Note that the first factor (the fraction) is real. So we can bring it outside the | |:

    \left | \frac{(a^2 + b^2 - 1)}{a^2 + b^2} \cdot (a - bi) \right | = \frac{a^2 + b^2 - 1}{a^2 + b^2} \cdot |a - bi|

    = \frac{a^2 + b^2 - 1}{a^2 + b^2} \cdot \sqrt{a^2 + b^2}

    = \frac{a^2 + b^2 - 1}{\sqrt{a^2 + b^2}}

    -Dan
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  6. #6
    Senior Member TriKri's Avatar
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    Quote Originally Posted by topsquark View Post
    Note that the first factor (the fraction) is real. So we can bring it outside the | |:

    \left | \frac{(a^2 + b^2 - 1)}{a^2 + b^2} \cdot (a - bi) \right | = \frac{a^2 + b^2 - 1}{a^2 + b^2} \cdot |a - bi|
    What about if a^2 + b^2 - 1 < 0?

    -Kristofer
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  7. #7
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    Here is another way.
    z \not= 0\quad  \Rightarrow \quad \frac{1}{z} = \frac{{\bar z}}{{\left| z \right|^2 }}\quad \& \quad \left| z \right| = \left| {\bar z} \right|

    \begin{array}{rcl}<br />
 \left| {\bar z - \frac{1}{z}} \right| & = & \left| {\bar z - \frac{{\bar z}}{{\left| z \right|^2 }}} \right| \\ <br />
  & = & \frac{{\left| {\bar z} \right|}}{{\left| z \right|}}\left| {\left| z \right| - \frac{1}{{\left| z \right|}}} \right| \\ <br />
  & = & \left| {\left| z \right| - \frac{1}{{\left| z \right|}}} \right| \\ <br />
 \end{array}<br />
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by TriKri View Post
    What about if a^2 + b^2 - 1 < 0?

    -Kristofer
    Unless I'm missing something (always possible!) I don't see how that condition would affect anything.

    -Dan

    Ah! Gotcha. Then you wouldn't be able to remove the | | bars. (I hadn't thought of that.) But since they are there on both sides of the equality we are trying to show both sides would still be equal; you would simply add a negative sign to both the LHS and the RHS to make it correct.

    -Dan
    Last edited by topsquark; December 8th 2006 at 05:40 PM. Reason: Eureka moment ;)
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