0
$\begingroup$

Let $a,b,c\in\mathbb C$, and define $f(x,y)=ae^{i(x+y)}-b(e^{ix}+e^{iy})+c$ for $x,y\in[-\pi,\pi]$.

For a "generic" triple $a,b,c$ the set $\{f(x,y)=0\}$ consists of two points, but occasionally the zero set is a curve.

Is there any simple explanation of this fact? Can one define precise conditions for the degenerate situation to occur?

Example of the degenerate situation: $a=c=1,b=-2$. In this case the zero set is $ x=t,y=-i\log \biggr(-{1+2e^{it}\over {2+e^{it}}}\biggl)$

  • 0
    Added the example.2012-11-25

1 Answers 1

2

With $z = e^{ix}$ and $w = e^{iy}$ we have $f(x,y) = azw - bz - bw + c$. So $f(z,w) = 0$ for $z = \dfrac{bw-c}{aw-b}$. The fractional linear transformation $g: w \to \dfrac{bw-c}{aw-b}$ preserves the unit circle iff $g(\zeta) \overline{g(1/\overline{\zeta})} =1 $ for all complex $\zeta$ for which this is defined. That simplifies to $\frac{b\zeta-c}{a\zeta-b} = \frac{\overline{a}-\overline{b} \zeta}{\overline{b}-\overline{c}\zeta}$ Assuming $b \ne 0$, the condition is $a = \dfrac{b \overline{c}}{\overline{b}}$

EDIT: There is also the case $b=0$, $|a|=|c|$.