I’m learning complex analysis but I stumbled across this problem which I cannot solve:

Let α1(t)=eit\alpha_1(t)=e^{it} , with t∈[0,2π]t\in[0,2\pi] the circle with radius 1 and centered at the origin.

I need to find an homotopy from this circle to the the curve composed by the circle with radius 1/41/4 and center in 1/21/2, the circle with radius 1/41/4 and center −1/2-1/2, and the line segment between those circles traversed in both directions.

Any help will be appreciated.

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Are you asked for an explicit formula for a homotopy (messy), or is it sufficient to draw a picture of one?

– arkeet

Oct 21 at 2:10

The book asks for an explicit formula, I don’t know how to define it in this case.

– mobzopi

Oct 21 at 2:16

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1 Answer

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I will let you derive the formula for this homotopy, but consider the picture below. For −1/2≤x≤1/2-1/2 \le x \le 1/2, you can do a straight line homotopy in the vertical direction. For x≤−1/2x \le -1/2 you can a straight line homotopy along lines passing through the point −1/2-1/2; similar for x≥1/2x \ge 1/2. This defines the homotopy in 8 pieces (depending on where xx lies relative to the numbers −1,−1/2,−1/4,1/4,1/2,1-1,-1/2,-1/4,1/4,1/2,1, and on the sign of yy).

Ohh thanks a lot, now it makes sense!

– mobzopi

2 days ago

No problem 🙂 Of course this isn’t the only possible homotopy.

– arkeet

2 days ago