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Let $C_R$ denote the upper half of the circle $|z| = R\ \ (R\in \mathbb{R}\ \cap\ (2, \infty))$ taken in counter-clockwise direction.

Show that $|\int_{C_{R}} \frac{2z^2-1}{z^4 + 5z^2+4}dz|\leq \frac{\pi R(2R^2 + 1)}{(R^2 - 1)(R^2-4)}$

Attempt:

Apply the ML-inequality directly. We first get the easy part, L = $\pi R$ (half of circumference)

Next we attempt to bound $|f(z) = \frac{2z^2-1}{z^4 + 5z^2+4}|$

$|f(z)| = |\frac{2z^2-1}{(z^2-1)(z^2-4)}| \leq |\frac{2z^2}{(z^2-1)(z^2-4)}| + |\frac{1}{(z^2-1)(z^2-4)}| $ by $\triangle$-inequality

This is almost what we want, but how do I justify substituiting $z$ with $R$ ?

Any help or insight is deeply appreciated.

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For $z \in C_R$ you have $\lvert z \rvert = R > 2$. The triangle inequality gives an upper bound for the numerator: $$ \lvert 2 z^2 - 1 \rvert \le \lvert 2 z^2 \rvert + \lvert 1 \rvert = 2 R^2 + 1 $$ as well as a lower bound for the denominator: $$ \lvert z^2 - 1 \rvert \ge \lvert z^2 \rvert - \lvert 1 \rvert = R^2 - 1 > 0\\ \lvert z^2 - 4 \rvert \ge \lvert z^2 \rvert - \lvert 4 \rvert = R^2 - 4 > 0 $$ (this is sometimes called "reverse triangle equality", compare e.g. Proving the reverse triangle inequality of the complex numbers).

Putting it all together gives exactly the desired estimate for $\lvert f(z) \rvert$.

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