Solve 2 cosh ⁡ z + sinh ⁡ z = i for z

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Solve   2  cosh  ⁡  z  +  sinh  ⁡  z  =  i for z

gozaderaradiox5 · Accepted Answer

2  cosh  ⁡  (  z  )  +  sinh  ⁡  (  z  )  =  i      e    z    +      e          −      z        +                    e        z            −              e                  −          z                      2    =  i  2      e    z    +  2      e          −      z        +      e    z    −      e          −      z        =  2  i  3      e    z    +      e          −      z        =  2  i  3      e    z    +  (      e          z            )          −      1        =  2  iLet       e    z    =  x  3  x  +      x          −      1        =  2  i  3      x    2    +  1  =  2  i  x  3      x    2    −  2  i  x  +  1  =  0We solve equations of the form   a      x    2    +  b  x  +  c  =  0 with      x          1      ,      2        =            −      b      ±                        b          2                −        4        a        c                    2      a      By plugging in a=3, b=−2i and c=1:      x          1      ,      2        =            2      i      ±              −        4        −        4        ⋅        3        ⋅        1              6    =            2      i      ±              −        4        −        12              6    =            2      i      ±              −        16              6    =            2      i      ±      4      i        6    =      i    3    ±            2      i        3  The two roots are therefore      x    1    =  i  ,        x    2    =  −      i    3  Since we required       e    z    =  x,   z  =  Ln  (  x  ) (Ln being the complex-valued logarithm). We are lazy and throw both values in Wolfram Alpha and get      z    1    =  ln  ⁡  (  i  )  =            i      π        2    ,        z    2    =  ln  ⁡      (    −          i      3        )    =  −  ln  ⁡  (  3  )  −            i      π        2  And now we have       z    1   and       z    2  , which are the solutions to the initial equation. Notice that the solution to that complex logarithm has arbitrary multiples of   2  π  i in it

Solve 2 cosh ⁡

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