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Continuity of an arc in the complex plane

  1. Nov 29, 2014 #1
    Hello everyone,

    I have a rather simple question. I have the curve

    ##
    C(t) =
    \begin{cases}
    1 + it & \text{if}~ 0 \le t \le 2 \\
    (t-1) + 2i & \text{if }~ 2 \le t \le 3
    \end{cases}
    ##

    which is obviously formed from the two curves. This curve is regarded as an arc if the functions ##x(t)## and ##y(t)## are continuous real mappings. Clearly each individual curve has continuous real mappings. But here is my concern: what if the two curves did not coincide somewhere in the complex plane, that is, they were not joined somewhere? Would the curve ##C(t)## no longer be an arc?
     
  2. jcsd
  3. Nov 29, 2014 #2
    Another question I have relates to an arc being a simple arc. The definition of this concept is the following:

    An arc is simple arc if, on ##a < t < b##, ##t_1 \ne t_2## implies ##C(t_1) \ne C(t_2)##, e.g. the function is one-to-one.

    So, according to this definition, if a curve is composed of smaller curves, such as the one given in the first post, then no individual curve can cross itself, but it is possible for one of the other curves to cross another curve. Is this right?
     
  4. Nov 29, 2014 #3

    mfb

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    Staff: Mentor

    Right, an arc cannot have "gaps".
    Why not? C(t)=sin(2t) + i sin(3t) for 0<t<2pi
    In fact, those cases ("define it in two parts") are just used for notation - they have no special mathematical meaning.
     
  5. Nov 29, 2014 #4
    Doesn't the condition ##t_1 \ne t_2 \implies C(t_1) \ne C(t_2)## means that the curve never crosses itself?
     
  6. Nov 29, 2014 #5

    mfb

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    2016 Award

    Staff: Mentor

    A "simple arc" never does that by definition in post 2, but a general curve can (violating this condition if it does).
     
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