Prove that $\displaystyle\left|\int_L \frac{z^3}{z^2+1} dz\right| \le \frac{9\pi}{8}$ where $L=\big(z: |z|=3, Re(z) \ge 0\big)$

Question

I tried solving this by using the fact that the length of L is half the circumference of a circle with radius 3, which would mean that if $$\left|\frac{z^3}{z^2+1}\right| \le \frac{3}{8}$$ this problem is solved. however, It appears to me as though this cannot be true: $$\left|\frac{z^3}{z^2+1}\right| = \frac{\left|z\right|^3}{|z^2+1|} \ge \frac{27}{10}$$ using the triangle inequallity. Is there something wrong with the way I'm going about this or is the statement really false?

Answer 1

I believe the inequality you stated is not true. Firstly, if $|z|=3$, by the reverse triangle inequality we get that $|z^2+1|=|z^2-(-1)| \geq | |z^2|-1| = |3^2-1| = 8$. Taking reciprocals gives us that $\dfrac{1}{|z^2+1|} \leq \dfrac{1}{8}$.

If we multiply both sides of the inequality by $|z^3|$, we obtain for $|z|=3$ that $$\left|\frac{z^3}{z^2+1}\right|\leq \dfrac{27}{8}$$

Now if we apply standard integral estimates, we get $\displaystyle\left|\int_L \dfrac{z^3}{z^2+1}dz\right| \leq \dfrac{27}{8}\cdot\operatorname{length}(L)$. As you said, the length of $L$ is half the circumference of a circle with radius $3$. That is, $\operatorname{length}(L) = \dfrac{6\pi}{2} = 3\pi$.

Thus, $$\displaystyle\left|\int_L \dfrac{z^3}{z^2+1}dz\right| \leq \dfrac{27}{8}\cdot3\pi= \dfrac{81\pi}{8} $$

Answer 2

You can just be more explicit and do the parametrization $z = 3e^{it}, -\pi/2\leq t \leq\pi/2:$

$$\left|\int_L\dfrac{z^3}{z^2+1}dz\right| = \left|\int_{-\pi/2}^{\pi/2}\dfrac{81e^{3it}i}{9e^{2it}+1}dt\right|\leq\dfrac{81}{8}\int_{-\pi/2}^{\pi/2}dt=\dfrac{81}{8}\pi$$ so the naive triangle-inequality bound gives something different than what you asserted.