A William Lowell Putnam Problem and its Solution

Problem:

Curves A, B, C and D are defined in the plane as follows:

$A = \{ (x, y): x^{2}-y^{2}=\frac{x}{x^{2}+y^{2}}\}$ –

$B = \{ (x, y): 2xy + \frac{y}{x^{2}+y^{2}}=3\}$

$C = \{ (x, y): x^{3}-3xy^{2}+3y=1\}$

$D = \{ (x, y): 3x^{2}y -3x - y^{3}=0\}$ .

Prove that $A \bigcap B = C \bigcap D$

(1987 William Lowell Putnam Mathematics Competition)

Solution:

Let $z= x+iy$ . The equations defining A and B are the real and imaginary parts of the equation $z^{2}=z^{-1}+3i$ , and similarly the equations defining C and D are the real and imaginary parts of $z^{3}-3iz=1$ . Hence, for all real x and y, we have $(x,y) \in A \bigcap B$ if and only if $z^{2}=z^{-1}+3i$ . This is equivalent to $z^{3}-3iz=1$ , that is, $(x, y) \in C \bigcap D$ .