Show that $$\left| \oint \frac{\sin z}{z^{5}}dz \right| \leq 2e\pi $$
My attempt:
I solved the integral without the absolute value, but I don't know how to get to the inequality. I also don't know if my calculations are correct.
$$f^{(n)}(z_0)=\frac{n!}{2\pi i}\oint_C \frac{f(z)}{(z-z_0)^{n+1}}\,dz$$
Here, we have $f(z)=\sin z$, $z_0=0$, $n=2$. Therefore, we have $$ \oint_{|z|=1} \frac{\sin z}{z^{5}}\,dz=\pi i\left.\frac{d^4\sin z}{4!}\right|_{z=0}=0 $$
Thank you
Assuming that the contour in question is the unit circle, we have $|\sin(z)| \le e$ for $|z| = 1$, and $|z|^5 = 1$. Therefore,
$$\left|\oint_C \frac{\sin z}{z^5}\right| \le \text{length}(C) \sup_{|z|=1}|\sin z| = 2\pi e$$
Of course, since $\sin(z)/z^5 = z^{-4} + \tfrac{1}{6}z^{-2} + h(z)$, where $h$ is analytic on the unit disk, the integral is in fact zero. So this is not a very useful bound!
Edit: $\text{length}(C) = 2\pi$ is the length of the unit circle. For the bound, notice that $\sin(z) = \tfrac{1}{2}(e^{iz} - e^{-iz})$, so $|\sin(z)| \le \tfrac12(|e^{iz}| + |e^{-iz}|)$. But also, $|e^{x+iy}| = e^x$, and if $z$ is on the unit circle, then the real and imaginary parts of $z$ are between -1 and 1. Therefore $|e^{iz}| \le e$, and $|\sin(z)| \le e$.