Let f ( z ) = z ( z − a ) ( z − b ) ( z − c ) ( z − d ) be a complex rational function and let S = { z ∈ C : | z | = 1 } be the unit circle.My question:For which (non-zero complex numbers) a , b , c and d there exist an arc C of S such that f ( C ) ⊆ S ?

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Let  f  (  z  )  =                    z        (        z        −        a        )                    (        z        −        b        )        (        z        −        c        )        (        z        −        d        )            be a complex rational function and let   S  =  {  z  ∈      C    :      |    z      |    =  1  } be the unit circle.My question:For which (non-zero complex numbers)   a  ,  b  ,  c and   d there exist an arc   C of   S such that   f  (  C  )  ⊆  S ?

Alexzander Guerrero · Accepted Answer

Let             f      ¯        (  z  )  =  z  (  z  −            a      ¯        )      /    (  z  −            b      ¯        )  (  z  −            c      ¯        )  (  z  −            d      ¯        ).If       |    z      |    =  1, then             z      ¯        =      z          −      1      , and so             f      (      z      )        ¯    =            f      ¯        (      z          −      1        ), and finally       |    f  (  z  )            |        2    =  f  (  z  )            f      ¯        (      z          −      1        ).Your hypothesis says that   f  (  z  )            f      ¯        (      z          −      1        )  =  1 is true on a tiny arc on the unit circle, so it is true on all of       C   (as a corollary, the whole unit circle is sent inside the unit circle, not just a tiny arc).Rewriting the identity, you get   (  z  −  a  )  (      z          −      1        −            a      ¯        )  =  (  z  −  b  )  (      z          −      1        −            b      ¯        )  (  z  −  c  )  (      z          −      1        −            c      ¯        )  (  z  −  d  )  (      z          −      1        −            d      ¯        )After multiplying by       z    3   you get the equality of two polynomials       z    2    (  z  −  a  )  (  1  −            a      ¯        z  )  =  (  z  −  b  )  (  1  −            b      ¯        z  )  (  z  −  c  )  (  1  −            c      ¯        z  )  (  z  −  d  )  (  1  −            d      ¯        z  ).By looking at the coefficients of       z    0  ,      z    1   and       z    2   this implies that two of   b  ,  c  ,  d have to be 0, and the third has to be equal to   a, so   f  (  z  )  =      z          −      1       is the only solution.

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