Let $z_0 \in \mathbb{C}$, $R>0$, $$\gamma_1(t) = z_0 + R e^{it}, \quad t \in [0,2\pi]$$ $$\gamma_2(t) = z_0 + R e^{it}, \quad t \in [\frac{\pi}{2},2\pi + \frac{\pi}{2}].$$ How can I prove that they are not equivalent, that is there is no $\phi: [0,2\pi] \to [\frac{\pi}{2},2\pi + \frac{\pi}{2}] $ such that $\phi$ is $C^{1}$, $\phi$ is bijective, $\phi^{-1}$ is $C^{1}$, $\phi$ is increasing, and $\gamma_2 \circ \phi = \gamma_1?$
Equivalence of $\gamma_1(t) = z_0 + R e^{it}, t \in [0,2\pi]$ and $\gamma_2(t) = z_0 + R e^{it}, t \in [\frac{\pi}{2},2\pi + \frac{\pi}{2}]$
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complex-analysis
2 Answers
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You did not clearly state what you mean by "equivalent". The two curves are clearly not the "same" since $\gamma_1$ begins and ends at $z_0+R$, while $\gamma_2$ begins and ends at $z_0+iR$. If you mean by "equivalent" the existence of an increasing parameter transformation $\phi$ such that $\gamma_2\circ\phi=\gamma_1$ then the two curves are not equivalent, because equivalence leaves the initial and end points of a curve unchanged.
On the other hand, the two curves are equal as $1$-chains, i.e., "equivalent" if it comes to line integrals, because there are two circular arcs $\alpha$, $\beta$ such that $\gamma_1=\alpha+\beta=\gamma_2$ in the sense of addition of chains.
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0In this context by equivalent I mean there exists $\phi: [0,2\pi] \to [\frac{\pi}{2},2\pi + \frac{\pi}{2}] $ such that $\phi$ is $C^{1}$, $\phi$ is bijective, $\phi^{-1}$ is $C^{1}$, $\phi$ is increasing, and $\gamma_2 \circ \phi = \gamma_1.$ – 2017-01-15
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0@Christian Blatter Thanks for the correction. I posted a new answer. I think that the new one is fine. – 2017-01-15
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Suppose the contrary, i.e. there exists such a $φ$. In this case we will reach a contradiction.
If $γ_2φ=γ_1$, then for a certain $t$ we have:
$$γ_2φ(t)=γ_1(t)\Rightarrow z_0+Re^{iφ(t)}=z_0+Re^{it}\Rightarrow\\ \exists k \in \mathbb Z : φ(t)-t=2k\pi$$
We will prove that this $k$ is the same for all $t\in [0,2\pi]$. Indeed, suppose there exist $t_1 $$g(t_1)=2k_1\pi,\ g(t_2)=2k_2\pi,$$ where $g(t)=φ(t)-t, \forall t \in [0,2\pi]$. Since $φ$ is continuous as $C^1$, we get that $g$ is also continuous. $k_1 $$k_1+1\leq k_2\Rightarrow 2k_1+2\leq 2k_2\Rightarrow n\pi \in (2k_1\pi, 2k_2\pi),$$ where $n:=2k_1+1$. We apply the Intermediate Value Theorem and so $\exists t_0\in (0,2\pi):g(t_0)=n\pi \Rightarrow φ(t_0)=t_0+n\pi$. Then $γ_2φ(t_0)=z_0-Re^{it_0}\neq z_0+Re^{it_0}=γ_1(t_0)$. Finally $φ(t)=t+2k\pi, \forall t \in [0,2\pi]$. Hence we do not have $φ([0,2\pi])\subset [\frac{\pi}{2},2\pi + \frac{\pi}{2}]$, impossible. We are done.