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

1. Oct 18, 2010

### Raziel2701

My book just gives me what each individual piece is but doesn't explain anything.

2. Oct 19, 2010

### lanedance

there would be a heap of ways, but firstly why do you want & how do you want to...

3. Oct 19, 2010

### lanedance

if its for a line integral the standrad way would be to consider each side of the square as a line separately

4. Oct 19, 2010

### HallsofIvy

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.

5. Oct 19, 2010

### Raziel2701

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...

6. Oct 19, 2010

### Raziel2701

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.