A Complex Coincidence

#ComplexNumbers#Putnam

Problem Statement

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

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

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

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

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

Prove that $A\cap B=C\cap D$.

Putnam-1987

For a complex number $Z=x+iy$, the quantities $x^2-y^2$ and $2xy$ are $\Re(Z^2)$ and $\Im(Z^2)$, respectively(the real and imaginary parts of the complex number). Thus, the equations $(A)$ and $(B)$, taken together, give

$$ Z^2 = \frac{x-iy}{x^2+y^2}+3i = \frac{1}{Z}+3i, $$

and hence

$$ Z^3-3iZ=1. $$

Comparing the real and imaginary parts of the latter equation gives

$$ x^3-3xy^2+3y=1 $$

and

$$ 3x^2y-3x-y^3=0, $$

which are precisely $(C)$ and $(D)$.

Therefore,

$$ A\cap B=C\cap D. $$

Bibliography & References

[1]
Putnam, 1987