I have this problem in my homework
Problem: Let $γ$ be the unit circle. Show that $$\left | \int _{\gamma}\frac{\sin z}{z^{5}} dz\right |\leq 2\pi e.$$
At the class suggestion, the teacher told us to find the arc length which is equal to $2 \pi$ and to delimit $| \sin z |$ , and I have this doubt that I hope is not very silly
In Marsden's book on complex variables, I found this Proposition 2.1.3 Let $f$ be continuous in an open set $A$ and let $γ$ be a curve $C^1$ by sections in $A$. If there exists a constant $M ≥ 0$ such that
$| f (z) | ≤ M $ for every point $z$ in $γ$ , then
$$ \left | \int _{\gamma}f\right |\leq M l(\gamma).$$
Can I use that proposition? and in what open set A should my function be continuous? Is that I am somewhat confused
This is what I know to do.
- Find $$\int _0^{2\pi }\left |ie^{it} \right |dt=2\pi.$$
$$ \begin{align} \left | \frac{\sin z}{z^{5}} \right |&=\frac{\left | \sin z \right |}{\left | z^{5} \right |}\\&=\left | \sin z \right | \\&=\left | \frac{e^{iz}-e^{-iz}}{2i} \right |\\&\leq 1/2 \left ( \left | e^{iz} \right |+\left | e^{-iz} \right | \right )\\&=1/2(e^{-y}+e^{y}) \\&\leq 1/2(e+e)\\&=e. \end{align} $$ But as I repeat, I don't know if it is used well or how to specify that my conditions are met for the proposition could you explain me please.
Is it enough to be able to use the proposition if I find $|f| \leq M $? or should I verify that it is continuous in open $A$, and what would that open be? Thank you so much.