Exercise 8.24

Question

Exercise 24: 
    Let $\gamma$ be as in Exercise 23, and assume in addition that the range of $\gamma$ does not intersect the negative real axis. Prove that $\Ind(\gamma)=0$.

analambanomenos · Accepted Answer

Following the hint, for $0\le c<\infty$ let $\gamma_c(t)=\gamma(t)+c$. Then $\gamma_c$ is also a continuous differentiable closed curve in $\mathbf C$ such that $\gamma_c(t)
e 0$ for all
    $t\in[a,b]$, so we can define $\Ind(\gamma_c)$. Let $\{c_n\}$ be a sequence of nonnegative real numbers converging to $c$. Then $$\frac{\gamma_{c_n}'(t)}{\gamma_{c_n}}=\frac{\gamma'(t)}
    {\gamma(t)+c_n}ightarrow\frac{\gamma'(t)}{\gamma(t)}\hbox{ as }nightarrow\infty,$$ and the convergence is uniform for $t\in[a,b]$. Hence by Theorem 7.16 $$\Ind(\gamma_{c_n})=
    \frac{1}{2\pi i}\int_a^b\frac{\gamma'(t)}{\gamma(t)+c_n}\,dtightarrow\frac{1}{2\pi i}\int_a^b\frac{\gamma'(t)}{\gamma(t)}\,dt=\Ind(\gamma),$$ so that $\Ind(\gamma_{c})$ is a continuous
    function of $c$. Since $\Ind(\gamma_c)$ has only integer values, this forces $\Ind(\gamma_c)=\Ind(\gamma)$ for all $0\le c<\infty$.

Since $[a,b]$ is compact, $\min\gamma(t)>-\infty$ for $t\in[a,b]$. Hence, for large enough $c$ we have $\min\big|\gamma(t)+c\big|$ as large as we like. In that case, we have
    $$\big|\Ind(\gamma_c)\big|\le\frac{1}{2\pi}\int_a^b\bigg|\frac{\gamma'(t)}{\gamma(t)+c}\bigg|\,dt\le\frac{1}{2\pi}\Bigg(\frac{\max\big|\gamma'(t)\big|}{\min\big|\gamma(t)+c\big|}\Bigg)(b-a)
    ightarrow 0$$ as $cightarrow\infty$. Hence $\Ind(\gamma)=\Ind(\gamma_c)=0$ for large enough $c$.

Exercise 8.24

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Exercise 8.24

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