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Thread: roots of z tan z

  1. #1
    Apr 2010

    roots of z tan z

    hello, can someone please show me how the roots of z tan z = k , where k > 0 are real is solved.

    what i have done; z tan z =  z * \frac{sinz}{cosz}

    by Euler equations   (x+iy) * \frac{e^i^z- e^-^i^z}{2i} /  \frac{e^i^z + e^-^i^z}{2}

      (x+iy) * \frac{e^i^z- e^-^i^z}{2i} /  \frac{e^i^z + e^-^i^z}{2}

      (x+iy) * \frac{e^i^z - e^-^i^z} { i(e^i^z + e^-^i^z)}

    would appreciate some help solving this thanks guys
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  2. #2
    MHF Contributor

    Apr 2005

    Re: roots of z tan z

    The first thing I would do is get the denominator real by multiplying numerator and denominator by the conjugate, -i (note that eiz+ e[sup]-iz[/itex] is a real number because it is equal to its conjugate).
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