I am struggling with the following question using the Estimation Lemma:

Let $ \gamma$ describe the semi-circle $Re^{it}$, where $ 0 \le t \le \pi$, and $ R \gt 3$. Show that

$$\int_\gamma {e^{3iz}\over (z^2+4)(z^2+9)}dz \le {\pi R \over (R^2-4)(R^2-9)}$$

I understand that the RHS of this equation is the same as $ML(\gamma)$ where $M$ is the upper bound and $L(\gamma)$ is the contour length. I have also calculated that in this case $$L(\gamma) =\pi R$$ using $ L(\gamma) = \int_a^b |\gamma'(t)|)dt $ but don't really know where to begin with the M part. In my course we do very few examples in the lectures so I'm finding it hard to figure out how to approach it.



I assume that you actually meant $$ \left |\int_\gamma {e^{3iz}\over (z^2+4)(z^2+9)}dz \right |\le {\pi R \over (R^2-4)(R^2-9)} $$

Using $ |e^w| = e^{\operatorname{Re} w}$ you can conclude that $$|e^{3iz}| \le 1 \, ,$$ note that $3iz$ lies in the left half-plane, so its real part is negative.

And the denominator can be estimated from below using the (reverse) triangle inequality: $$ |(z^2+4)(z^2+9)| \ge (|z|^2-4)(|z|^2-9) = (R^2-4)(R^2-9) \, . $$

These two inequalities together give the wanted upper bound $M$ for the integrand.

  • $\begingroup$ Yes you are quite right I missed out the absolute value lines, I'm still getting the hang of the formatting for math equations on here! Is it just generally accepted that R can be subbed in for |z| in this kind of case then? Or is that only because R is the radius of the semi-circle in this example? $\endgroup$ – Hannah Apr 19 '15 at 12:58
  • $\begingroup$ @Hannah: The latter. For all $z$ on the arc $\gamma$, $|z| = |Re^{it}| = R$. $\endgroup$ – Martin R Apr 19 '15 at 13:02
  • $\begingroup$ That makes sense, thanks. For a similar question I'm trying to find just the upper bound M for $$\left | {e^{2z} \over z} \right | $$ where $ z = e^{i \theta}, \theta \in R $ . Using a similar idea I'm thinking that $ | e^{2z} | \le 1$ and the denominator $ |z| = |e^{i\theta} | \le 1$. It seems a bit too simple, but is the upper bound just 1 then? I feel like there should be some Rs involved somewhere.. $\endgroup$ – Hannah Apr 19 '15 at 13:13
  • $\begingroup$ @Hannah: Generally you have $|e^z| = e^{\operatorname{Re} z}$, which is $\le 1$ if and only ${\operatorname{Re}} \, z \le 0$, i.e. if $z$ lies in the left half-plane. $\endgroup$ – Martin R Apr 19 '15 at 13:17

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