# roots of z tan z

• Apr 10th 2013, 01:16 AM
sigma1
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
• Apr 10th 2013, 05:56 AM
HallsofIvy
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).