A (theory) question at index of a curve

Question

Ιn our attempt to define index of a curve, let $c\colon [a,b]\to C\setminus \{0\}$ be a $C^1$ simple closed complex curve with oriented positive direction and passing through negative real axis one and only one time. Consider that the curve $c$ is parametrized such that $c(a)=c(b)=x<0$ and for some $0<\epsilon_0<\frac{b-a}{2}$ and every $0<\varepsilon\leq \epsilon_0$, $\ \ c(a+\varepsilon),\ c(b-\varepsilon)\in D(x,\frac{|x|}{2})$ and $Im(c(a+\varepsilon))<0\ ,Im(c(b-\varepsilon))>0$. Then we analysed the curve integral $$\int_{c} \frac{dz}{z}=\int_{c|_{[a,a+\varepsilon]}}\frac{dz}{z}+\int_{c|_{[a+\varepsilon,b-\varepsilon]}}\frac{dz}{z}+\int_{c|_{[b-\varepsilon,b]}}\frac{dz}{z}.$$ We claim that $$\int_{c|_{[a,a+\varepsilon]}}\frac{dz}{z}\ \ \mbox{and} \ \ \int_{c|_{[b-\varepsilon,b]}}\frac{dz}{z}\to 0, \ as\ \ \varepsilon \to 0 .$$ My question: Why $$\int_{c|_{[a,a+\varepsilon]}}\frac{dz}{z}\leq \frac{2}{|x|} \|c'\|\varepsilon,$$ where $\|c'\|=\displaystyle\max _{t\in [a,b]} |c'(t)|$. I understand that I have to use the triangle inequality of the curve intergration but something is going wrong. My try $$\left|\int_{c|_{[a,a+\varepsilon]}}\frac{dz}{z}\right|=\left|\int_{a}^{a+\varepsilon} \frac{1}{c(t)} c'(t) dt\right|\leq \max_{t\in[a,b]} \left|\frac{1}{c(t)}\right|\int_{a}^{a+\varepsilon} \|c'\|dt=\max_{t\in[a,b]} \left|\frac{1}{c(t)}\right| \|c'\| \varepsilon.$$ I feel I am to close but at this point I stuck. A little help. Thanks.

Answer 1

You just need to use the construction: $$c(t)\in D(x,|x|/2)$$ for $$a\leq t\leq a+\varepsilon \quad ({\rm Note~ that ~}c(a)=x),$$ which implies that with the same condition on $t$, one has $$|c(t)-x|\leq \frac{|x|}2,$$ hence by triangle inequality $$|c(t)|\geq \frac{|x|}2\Leftrightarrow\frac 1{|c(t)|}\leq \frac 2{|x|}.$$