if |f(z)| ≤ M for all z on a (piecewise smooth) curve C with
length L, then

R
C
f

≤ M · L. Now let C be the upper half circle |z| = R > 2,Imz > 0, show that
lim
R→∞ Z
C
2z
2 − 5
(z
2 + 1)(z
2 + 4) dz = 0.

To solve the problem, we need to show that:

\[
\lim_{R \to \infty} \int_{C} \frac{2z}{(z^2 - 5)(z^2 + 1)(z^2 + 4)} \, dz = 0
\]

where \( C \) is the upper half-circle \( |z| = R \) for \( R > 2 \) and \( \text{Im}(z) > 0 \).

1. **Function Behavior on the Contour**:
   First, note that \( C \) is the upper half-circle of radius \( R \). Parametrize \( C \) by \( z = Re^{i\theta} \) where \( \theta \) ranges from 0 to \( \pi \). Thus, \( dz = iRe^{i\theta} \, d\theta \).

2. **Bound the Function on \( C \)**:
   Consider the integrand \( f(z) = \frac{2z}{(z^2 - 5)(z^2 + 1)(z^2 + 4)} \). We need to find an upper bound for \( |f(z)| \) on \( C \).

   \[
   |f(z)| = \left| \frac{2z}{(z^2 - 5)(z^2 + 1)(z^2 + 4)} \right|
   \]

   Since \( |z| = R \) on \( C \),

   \[
   |f(z)| = \left| \frac{2z}{(z^2 - 5)(z^2 + 1)(z^2 + 4)} \right| \le \frac{2R}{(|z^2 - 5|)(|z^2 + 1|)(|z^2 + 4|)}
   \]

   For large \( R \),

   \[
   |z^2| = R^2 \quad \text{and} \quad |z^2 - 5| \approx |R^2|, \quad |z^2 + 1| \approx |R^2|, \quad |z^2 + 4| \approx |R^2|
   \]

   Therefore,

   \[
   |f(z)| \le \frac{2R}{R^2 \cdot R^2 \cdot R^2} = \frac{2R}{R^6} = \frac{2}{R^5}
   \]

3. **Estimate the Integral**:
   We can now use this bound to estimate the integral over \( C \):

   \[
   \left| \int_{C} \frac{2z}{(z^2 - 5)(z^2 + 1)(z^2 + 4)} \, dz \right| \le \int_{C} \left| \frac{2z}{(z^2 - 5)(z^2 + 1)(z^2 + 4)} \right| \, |dz|
   \]

   Since \( |dz| = R \, d\theta \) and \( \theta \) ranges from 0 to \( \pi \) (length of \( C \) is \( \pi R \)),

   \[
   \int_{C} \left| \frac{2z}{(z^2 - 5)(z^2 + 1)(z^2 + 4)} \right| \, |dz| \le \int_{0}^{\pi} \frac{2}{R^5} R \, d\theta = \frac{2}{R^4} \int_{0}^{\pi} d\theta = \frac{2\pi}{R^4}
   \]

4. **Take the Limit as \( R \to \infty \)**:

   \[
   \left| \int_{C} \frac{2z}{(z^2 - 5)(z^2 + 1)(z^2 + 4)} \, dz \right| \le \frac{2\pi}{R^4}
   \]

   As \( R \to \infty \),

   \[
   \frac{2\pi}{R^4} \to 0
   \]

Thus,

\[
\lim_{R \to \infty} \int_{C} \frac{2z}{(z^2 - 5)(z^2 + 1)(z^2 + 4)} \, dz = 0
\]

This completes the proof. If you have any questions or need further details, please let me know!

Here are 8 related questions for further exploration:
1. What is the residue theorem and how is it applied in complex analysis?
2. How do you parametrize different types of contours in the complex plane?
3. What are some common techniques for estimating integrals in complex analysis?
4. How does the concept of analyticity play a role in contour integration?
5. What are singularities, and how are they classified in complex functions?
6. Can you explain the concept of winding number in relation to contour integration?
7. How do you apply Jordan's lemma in contour integration problems?
8. What are some applications of complex contour integration in physics and engineering?

**Tip**: When working with complex integrals, always consider the nature of the singularities and the behavior of the function at infinity to determine the most efficient way to evaluate the integral.

Explore how to evaluate the limit of a complex contour integral over the upper half circle |z| = R > 2, Im(z) > 0. Understand the application of contour integration, estimation techniques, and complex function behavior in mathematical analysis.

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