Upper bounds for moduli of contour integral.

Question

Without evaluating the integral, show that $$\left | \int_{C}\frac{dz}{\left ( z^2-1 \right )} \right |\leq \frac{\pi}{3}$$ where $C$is the arc of a circle $\left | z \right |=2$, from $z=2$ to $z=2i$ that lies in the first quadrant.I know $$\left | \int_{C} f\left ( z \right )dz\right |\leq ML$$ $$L=\pi$$ $$\left | z^2-1 \right |\geq \left | \left | z \right |^2-\left ( 1 \right ) \right |=3$$ hence $$\left | \int_{C}\frac{dz}{\left ( z^2-1 \right )} \right |\leq \frac{\pi}{3}$$ I have an uneasy feeling that my inequality is incorrect.

Answer 1

The development in the OP was fine. As an alternative, we can parameterize the curve $C$ by letting $z=2e^{i\phi}$, $0\le \phi\le \pi/2$. Then, we have the estimates

$$\begin{align} \left|\int_C\frac{1}{z^2-1}\,dz\right|&=\left|\int_0^{\pi/2}\frac{i2e^{i\phi}}{4e^{i2\phi}-1}\,d\phi\right|\\\\ &\le 2\int_0^{\pi/2}\frac{1}{|4e^{i2\phi}-1|}\,d\phi\\\\ &\le 2\int_0^{\pi/2}\frac{1}{||4e^{i2\phi}|-1|}\,d\phi\\\\ &=2\int_0^{\pi/2}\frac{1}{3}\,d\phi\\\\ &=\frac\pi3 \end{align}$$

as was to be shown!