Does a r g ( z − 1 z + 1 ) = π 3 represent a circle or arc of a circle?By substituting x+iy=z I do get equation of a circle. But I feel it should be one particular arc of the circle (the locus) depending on the direction of rotation of (z−1) to (z+1) by Coni's method.Can someone clarify this issue? Thanks.

Question

Does   a  r  g      (                            z          −          1                          z          +          1                      )    =            π      3       represent a circle or arc of a circle?By substituting x+iy=z I do get equation of a circle. But I feel it should be one particular arc of the circle (the locus) depending on the direction of rotation of (z−1) to (z+1) by Coni&#039;s method.Can someone clarify this issue? Thanks.

Pranav Greer · Accepted Answer

z  =  x  +  i  y    ⟹                      z        −        1                    z        +        1              =                    (        x        −        1        +        i        y        )        (        x        +        1        −        i        y        )                    (        x        +        1                  )          2                +                  y          2                      =                              x          2                +                  y          2                −        1        +        i        (        2        y        )                    (        x        +        1                  )          2                +                  y          2                    arg                    2        y                              x          2                +                  y          2                −        1              =            π      3       if       x    2    +      y    2    −  1  &amp;gt;  0In that case,       x    2    +      y    2    −  1  =  2      3    y    ⟺        x    2    +  (  y  −      3        )    2    =      2    2  WLOG any point on       x    2    +  (  y  −      3        )    2    =      2    2   can be represented as   2  cos  ⁡  t  ,      3    +  2  sin  ⁡  tNow       x    2    +      y    2    −  1  =  ⋯  =  6  +  4      3    sin  ⁡  t which will be   &amp;gt;  0    ⟺    sin  ⁡  t  &amp;gt;  −                    3            2      So, its not the full circle, but an arc is represented by arg                    z        −        1                    z        +        1              =            π      3

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