Show that $\gamma (t) = \big( (1 + a \cos t) \cos t, (1 + a \cos t) \sin t \big), t \in [0, 2\pi]$, where $a$ is a constant, is a simple closed curve if $|a| < 1$ , but that if $|a| > 1$ its complement is the disjoint union of three connected subsets of $\Bbb R^2$, two of which are bounded and one is unbounded. What happens if $a = \pm 1$?
Hopf's theorem states that the total signed curvature of a simple closed curve in $\Bbb R^2$ is $\pm 2 \pi$. But if a curve has total signed curvature of $\pm 2 \pi$ that does not mean it is a simple closed one. Right? So any simple way to solve this one? Have to prove that it has a period in order to be closed. And to be simple it must not have self-intersections..I try to solve $γ(t_1)=γ(t_2)$ but i get to $$((1+acost_1)cost_1)=((1+acost_2)cost_2$$
cant solve for $t_1$ or $t_2$
Partial answer.$\gamma (t)$ has period $2\pi$ and is continuous...(I)... If $|a|<1$ then $1+a\cos t$ is never $0$, so $\gamma (t)=\gamma (u)\implies (\cos t=\cos u\land \sin t=\sin u)\implies (t-u)/2\pi\in Z,$ so for $v\in [0,1]$ the function $g(v)=\gamma (2\pi v)$ is a homeomorphic embedding of the unit circle of $C$ into $R^2$. (II). If $|a|>1$ there are $2$ values of $t\in (0,2\pi)$ for which $\cos t=-1/a$ and $\gamma (t)=(0,0).$