It is not specified in the question whether the open or closed unit disk is to be dissected.In fact, the answers are different for these two cases.
In the following, by f^n(x) I men the function f applied to x n times, not the nth power of f(x). Similarly, by f^-1 I mean the inverse function of f, and by f^-n(x) I mean (f^-1)^n(x).
case 1: the open unit disk.
This may be dissected
Construction:
Denote the disk by U the point (0, 1) by p, and the function dilating the plane by a factor of two about P by f. For any x in U, |x-p| > 0, |f(x)-p| = 2 * |x-p|. By induction |f^n(x)-p| = 2^n * |x-p| for any positive integer n. So for each x in U there exists a positive integer m such that f^m(x) is not in U. For any x in U, let g(x) to be the least positive integer n such that f^n(x) is not in U. Let the set S_0 be the set of all x in U with g(x) ~ 0 mod 2, and S_1 the set of all x in U with g(x) ~ 1 mod to. Then {S_0, S_1} is a partition of the unit disk into two similar sets: It is easy to show that S_0 = f(S_1), using the fact that f(U) entirely contains U.
case 2: the closed unit disk.
This may not be dissected into two similar sets (note that this includes pairs of sets similar about the origin).
Outline proof: by contradiction
Let U be the closed unit disk, and let S_O and S_1 be sets, and f a similarity of the plane, centred at p, such that {S_0, S_1} is a partition of U and S_0 = f(S_1).
Claim 1: p is not contained in U
Proof: contradiction
Assume p to be contained in U. Since p = f(p), p is in S_0 iff p is in S_1. But p is required to lie in exactly one of these two sets, which is the required contradiction.
For similar reasons, f may not be the identity function.
Let U_1 = f(U) and U_-1 = f^-1(U). Let R_-1 = U - U_-1 and R_1 = U - U_1. Let R be the intersection of R_-1 and R_1.
Claim 2: R is non-empty
Proof: In three cases
case 1: U does not intersect U_-1
then U_1 does not intersect U, and so R_-1 and R_1 are both equal to U, as is R. Hence R is non-empty, as required.
case 2: U intersects U_-1 in exactly 1 point.
then U_1 intersects U in exactly 1 point. R is U minus those two points, and so is again non-empty, as required.
case 3: U intersects U_-1 in at least two points.
Let C_0 be the boundary of U, C_-1 the boundary of U_-1 and C_1 the boundary of U_1. Using claim 1, it may be seen that C_-1 and C_0 intersect at exactly two points, which I will call x and y. Indeed, the boundary of the intersection of U_-1 and U consists of x, y and the two open arcs A_-1 of C_-1 and A_0 of C_0. Some elementary geometry shows that the angles subtended by these arcs at the centres of their respective circles sum to less than 2 pi. Two arcs of C_0 are not in R: The first, A_0, is not contained in R_-1. The second, f(A_-1), is not contained in R_1. But C_0 subtends an angle of 2 pi at 0, whereas the sum of the angles subtended by these two arcs is less than 2 pi. So at least one point of C_0 is in neither arc, and so is contained in R. Hence R is non-empty, as required.
Claim 3: Any point in R_-1 must be contained in S_1
Proof: contradiction
Let x be contained in R_-1 but not S_1. Then x is contained in S_0. This implies f(x) is contained in S_1, which is contained in U. But f(R_-1) = f(U-U_-1) = f(U) - f(U_-1) = U_1 - U, which has an empty intersection with U. However, x is contained in this empty intersection! This is a contradiction, as required.
For similar reasons, R_1 is contained in S_0
So R is contained in both S_0 and S_1 and is nonempty
Hence {S_0, S_1} is not a partition of U, which is the original desired contradiction.
Note that the problem becomes simpler (and has a different answer) if we redefine a dissection as follows: a pair of open sets S_0 and S_1 such that:
1: They have empty intersection
2: The union of their closures is the closed unit disk
This only allows for dissections of closed figures. Now let S be any closed convex subset of the plane, and let f be any similarity, centre p, of the plane mapping S onto a subset of itself. Let n(x) (for x in S - {p}) be the least positive integer such that f^-n(x) is on the boundary of or outside S.
Let g(x) = -1 if f^(-n(x))(x) is on the boundary or if x = p, and be congruent to n(x) mod 2 otherwise. If we then define S_-1, S_0 and S_1 in the natural way, we have that (S_0, S_1) is a dissection of S
Here S could be U, the closed unit disk.
Thankyou
sfwc
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