i'm having a bit of a misunderstanding with the Complex version of the estimation lemma, i'm looking to prove that
$$\sup_{\gamma} \left|\frac{sin(z)}{z^2} \right| \leq e$$
where $\gamma = e^{it} ~ (0 \leq t \leq 2\pi)$ ie, the circle centred at 0 or radius 1
tell me if i my logic is correct or not please.
$\gamma$ attains a maximum at 0 or $2 \pi$ with $e^{i0}=1$, substituting that into $$\left| \frac{sin(1)}{1^2} \right|\leq e$$
a similiar question was answered in Using the Estimation Lemma
and for some reason i just can't see where they got $|f(z)| \leq 1$, this may just be me having an odd 5 min
any help would be greatly appreciated. Thanks!
Best Answer
Well, using Euler's formula,
$$ |\sin(z)|=|\frac{e^{iz}-e^{-iz}}{2i}|\leq \frac{|e^{iz}|+|e^{-iz}|}{2}=\frac{e^{Re(iz)}+e^{Re(-iz)}}{2}=\frac{e^{-Im(z)}+e^{-Im(z)}}{2} $$
Now, for $z=e^{it},$ with $t\in\mathbb{R}$, we have that $Im(z)\in [-1,1]$ and applying this to the above, we get, by monotonicity of the exponential on the real line, that
$$ |\sin(z)|\leq \frac{e^1+e^1}{2}=e, $$
establishing the desired, since $\frac{1}{|z|^2}=1$ for $z=e^{it}$.