Solve sin ⁡ ( z ) = 3 + i 4 What i did so far was this: sin ⁡ ( z ) = 3 + i 4 ⇒ e i z − e − i z 2 i = 3 + i 4 ⇒ e 2 i z − 1 = ( − 1 + 3 i 2 ) e i z setting u = e i z we'll have u 2 + ( 1 − 3 i 2 ) u − 1 = 0completing squares ( u + 1 − 3 i 4 ) 2 = 3 8 ( 1 + i )seting w = u + 1 − 3 i 4 w 2 = 3 8 ( 1 + i )solving w = ± 2 − 5 4 3 ( cos ⁡ ( 2 4 ) + i sin ⁡ ( 2 4 ) ) now i just have to substitute this in w = u + 1 − 3 i 4 and then in u = e i z , but the solution looks really big . So what did i do wrong?

Question

Solve   sin  ⁡  (  z  )  =            3      +      i        4  What i did so far was this:  sin  ⁡  (  z  )  =            3      +      i        4    ⇒                    e                  i          z                    −              e                  −          i          z                            2      i        =            3      +      i        4    ⇒      e          2      i      z        −  1  =      (                  −        1        +        3        i            2        )        e          i      z      setting   u  =      e          i      z       we&#039;ll have      u    2    +      (                  1        −        3        i            2        )    u  −  1  =  0completing squares            (      u      +                        1          −          3          i                4            )        2    =      3    8    (  1  +  i  )seting   w  =  u  +            1      −      3      i        4        w    2    =      3    8    (  1  +  i  )solving  w  =  ±      2                            −          5                4                  3        (    cos    ⁡          (                        2                4            )        +    i    sin    ⁡          (                        2                4            )        )  now i just have to substitute this in   w  =  u  +            1      −      3      i        4   and then in   u  =      e          i      z      , but the solution looks really big . So what did i do wrong?

Maffei2el · Accepted Answer

You made a mistake while completing the square; it should be            (      u      +                        1          −          3          i                4            )        2    =      1    2    −      3    8    i  =            (              3        4            −              1        4            i      )        2    .So, you get that  u  +            1      −      3      i        4    =  ±      (          3      4        −          1      4        i    )    ,and that therefore  u  =  −  1  +  i  =      e          log      ⁡              2            +                        3          π                4            i          or    u  =      1    2    +      1    2    i  =      e          −      log      ⁡              2            +              π        4            i        .

Solve sin(z)=(3+i)/(4)

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