Show that this integral tends to zero

Question

Could somebody justify the second inequality please? What is the thought process? What is the significance of $|dz|$? This is in the context of contour integration. Sorry for the image, on mobile so would be hard to type.

Answer 1

You want to find upper bound for $$\int_{arc} \frac {|e^{5iz}|}{|z^2 +1|} dz$$ so first found the upper bound for integrand $$\frac {|e^{5iz}|}{|z^2 +1|}$$ clearly if $z=x + iy$ then $|e^{5iz}| = |e^{5ix - 5y}| = |e^{-5y} e^{i5x}| =e^{-5y}= e^{-5 imz} $ so $$\frac {|e^{5iz}|}{|z^2 +1|} = \frac {e^{-5 imz}}{|z^2+1|}$$ now use the fact that $|z^2+1| \ge | |z^2| - 1| \ge R^2 - 1$ as $R\ge 1$ so $$\frac{1}{z^2 +1} \le \frac {1} {| |z^2| - 1|} \le \frac{1}{R^2-1}$$ so $$\int_{arc} \frac {|e^{5iz}|}{|z^2 +1|} dz\le \int_{arc} \frac {1}{R^2 - 1} dz \le \frac {1}{R^2 - 1} \int_{arc} dz \le\frac {1}{R^2 - 1} \int_{arc} |dz| $$ now $\int_{arc} |dz|$ represents arc length of the contour curve which is circemfrence of half circle of radius $R$ so $\int_{arc} |dz|$ = $ \pi R $ $\space$ or You can observe $z = Re^{i\theta}$ with $0\leq \theta\leq \pi$: $$dz = iRe^{i\theta}d\theta\Rightarrow |dz| = R d\theta$$ now $$\int_{arc} |dz| = \int _{0}^\pi R \ d\theta = \pi R$$ So $$\frac {1}{R^2 - 1} \int_{arc} |dz| =\frac {\pi R}{R^2 - 1}$$

Answer 2

This inequality comes as follows:

• $z = Re^{it}$ with $0\leq t\leq \pi$: $$dz = iRe^{it}dt\Rightarrow |dz| = R dt$$
• reverse triangle inequality $$|z^2 +1| \geq ||z|^2 - 1| = R^2-1 \text{ for } R \geq 1$$
• $z=Re z + i Im z$ $$\Rightarrow |e^{5iz}| = |e^{5i(Re z + i Im z)}| = e^{-5Im z}$$

All together

$$\int_{arc}\frac{|e^{5iz}|}{|z^2+1|}dz \leq \underbrace{\int_{arc}}_{\int_0^{\pi}}\frac{e^{-5Im z}}{R^2-1}\underbrace{Rdt}_{|dz|}$$