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Complex Proof

  1. Dec 1, 2017 #1
    1. The problem statement, all variables and given/known data
    ##z## is a complex number such that ##z = \frac{a+bi}{a-bi}##, where ##a## and ##b## are real numbers. Prove that ##\frac{z^2+1}{2z} = \frac{a^2-b^2}{a^2+b^2}##.

    2. Relevant equations


    3. The attempt at a solution
    I calculated
    \begin{equation*}

    \begin{split}

    z = \frac{a+bi}{a-bi} &= \frac{a+bi}{a-bi}\times \frac{a+bi}{a+bi} \\

    &= \frac{a^2+2abi-b^2}{a^2+b^2} \\

    &= \frac{a^2-b^2}{a^2+b^2}+\frac{2ab}{a^2+b^2}i.

    \end{split}

    \end{equation*}
    But sticking that ugly thing into ##\frac{z^2+1}{2z}## gives me something nasty. I'm sure there is a much simpler way!!!
     
  2. jcsd
  3. Dec 1, 2017 #2

    PeroK

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    What could you do with ##\frac{z^2+1}{2z}##?
     
  4. Dec 1, 2017 #3

    kuruman

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    It might be prettier if you defined ##u=a+bi## in which case ##z=u/u^*##. Put that ratio in ##\frac{z^2+1}{2z}## and see what you get.
     
  5. Dec 1, 2017 #4
    Great thank you very much!!!
     
  6. Dec 2, 2017 #5

    haruspex

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    Another way might be by noting that ##\frac {z^2+1}{2z}=\frac 1{\frac 1{z+i}+\frac 1{z-i}}##
     
  7. Dec 2, 2017 #6

    PeroK

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    I think, as aluded to in post #2, the key is:

    ##\frac{z^2+1}{z}=z+\frac{1}{z}##
     
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