(i) TS(x,y) = T(x+1,y) = (-y,x+1) but ST(x,y) = S(-y,x) = (-y+1,x).

(ii) T(x,y) \neq (x,y), but T^2(x,y) = T(-y,x) = T(-x,y), from which we see that T^4(x,y)=(x,y), i.e. T^4=\mathrm{id}. So T^{4n}(x,y) = T^{4}T^{4(n-1)}(x,y) = T^{4(n-1)}(x,y) = \ldots = (x,y).

(iii) We need to apply S when the first coordinate is -x (so that we get \pm(x-1)), so let’s first try ST^2(x,y) = S(-x,y) = (-x+1,y). But we’ve already seen that T^2 simply changes the sign of the first coordinate, so T^2ST^2(x,y) = (x-1,y).

To see that we need at least five applications, note that the only way to change the sign of the first coordinate is by applying T twice, and we need to apply S once to get the \pm1 term, but then we need to change the sign of the first coordinate one last time, which needs another two applications of T, so five is minimal.

(iv) S and U let us make the first coordinate anything we want, since S adds 1 and U subtracts 1. So we can simply apply S^b or U^{-b} (depending on if b is positive or negative), and apply T, so that we have (0,b). Then we just apply either S^a or S^{-a} (depending of if a is positive or negative).

(v) C is given by y=x^2+2x+2. Applying S maps x to x+1, so we get y = (x+1)^2+2(x+1)+2 = x^2+2x+1+2x+2+2 = x^2+4x+5 = (x+4)(x+1). Applying T instead replaces x with -y and y with x, and so we get x = (-y)^2+2(-y)+2 = y^2-2y+2.

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