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How do you parametrize the unit square in the complex plane?

  1. Oct 18, 2010 #1
    My book just gives me what each individual piece is but doesn't explain anything.
     
  2. jcsd
  3. Oct 19, 2010 #2

    lanedance

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    there would be a heap of ways, but firstly why do you want & how do you want to...
     
  4. Oct 19, 2010 #3

    lanedance

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    if its for a line integral the standrad way would be to consider each side of the square as a line separately
     
  5. Oct 19, 2010 #4

    HallsofIvy

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    The "individual pieces", I suspect, are the four sides of the square.

    The "unit square" in the complex plane is probably the square with vertices at 0, 1, 1+ i, and i. The bottom side, from 0 to 1, could be parameterized as "t" with 0< t< 1. Obviously, when t= 0, that gives the point "0" and when t= 1 that gives the point "1".

    The right side, from 1= 1+ 0i to 1+ i could be parameterized by "1+ ti" with 0< t< 1. When t= 0 that gives the point 1+0i= 1 and when t= 1, it gives 1+ 1i= 1+ i.

    The top side, from 1+ i to i= 0+ i could be parameterized by "(1- t)+ i" with 0< t< 1. When t= 0 that gives the point (1- 0)+ i= 1+ i and when t= 1 it gives (1- 1)+ i= i.

    Finally, the left side, from i to 0 could be parameterized by "(1- t)i" with 0< t< 1. When t= 0 that gives the point (1- 0)i= i and when t= 1 it gives (1- 1)i= 0.
     
  6. Oct 19, 2010 #5
    I found a formula in my calc book. It parametrizes the lines from 0 to 1, and although I don't know how my complex analysis book parametrized the unit square, this formula works all right.

    I needed the parametrization for some contour integrals and the book is unfortunately not self-contained so they just throw results instead of procedures. I really hate this book...
     
  7. Oct 19, 2010 #6
    Yeah I got the same parametrization as you did HallsofIvy, the book does it differently though, but I don't care, my method works just as well.
     
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