Let Ω = C ∖ [ − 1 , 1 ], i.e. deleting ``the line'' only, is there a function f : Ω → C such that f satisfies f ( z ) 2 = 1 − z 2 and is continuous on this region?My guess is that such a function would exist, but requires a piecewise definition. A candidate solution I have been working on is f ( z ) = e 1 2 log ⁡ ( 1 − z 2 ) . The problem with this solution is having the domain, as I realize it is possible to have 1 − z 2 &gt; 1, where my solution is well-defined. Is there a way to work around this or should I try something else?

Question

Let   Ω  =      C    ∖  [  −  1  ,  1  ], i.e. deleting ``the line&#039;&#039; only, is there a function   f  :  Ω  →      C   such that   f satisfies   f  (  z      )    2    =  1  −      z    2   and is continuous on this region?My guess is that such a function would exist, but requires a piecewise definition. A candidate solution I have been working on is   f  (  z  )  =      e                  1        2            log      ⁡      (      1      −              z        2            )      . The problem with this solution is having the domain, as I realize it is possible to have   1  −      z    2    &amp;gt;  1, where my solution is well-defined. Is there a way to work around this or should I try something else?

Kaydence Washington · Accepted Answer

Using the principal branch of   log we can define   η  (  w  )  =      1    −    w   on       C    ∖  [  1  ,  ∞  ), and we can then define   ϕ  (  z  )  =  η  (            1      z        ) on       C    ∖  [  0  ,  1  ]. Note that   ϕ is analytic.Define   f  :      C    ∖  [  −  1  ,  1  ]  →      C   by   f  (  z  )  =  i      z    2    ϕ  (  z  )  ϕ  (  −  z  ) and note that   f  (  z      )    2    =  1  −      z    2  .

Let <mi mathvariant="normal">Ω = <mrow class="MJX-TeXAtom-ORD"> <mi math

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