Show that R

C

f = 0, where C : |z| =

√

2 and

a. f(z) = (z

2 + 2)

(z

2 + 3).

b. f(z) = (z + sin z − 2)

cos z

.

To show that \(\int_{C} f(z) \, dz = 0\) for \(C : |z| = \sqrt{2}\) and given functions \(f(z)\), we use the fact that if \(f(z)\) is analytic (holomorphic) inside and on the contour \(C\), then by Cauchy's theorem, the integral of \(f(z)\) around \(C\) is zero. Cauchy's theorem states that for any closed contour \(C\) in a domain \(D\), if \(f(z)\) is analytic in \(D\), then:

\[
\int_{C} f(z) \, dz = 0
\]

### Part (a): \(f(z) = \frac{z^2 + 2}{z^2 + 3}\)

First, we need to determine if \(f(z)\) is analytic inside and on the contour \(C : |z| = \sqrt{2}\).

1. **Poles of \(f(z)\)**:
   
   The function \(f(z)\) has potential poles where the denominator is zero:
   
   \[
   z^2 + 3 = 0 \implies z^2 = -3 \implies z = \pm i\sqrt{3}
   \]

   The poles \(\pm i\sqrt{3}\) lie outside the contour \(C\) since \(|i\sqrt{3}| = \sqrt{3} > \sqrt{2}\).

2. **Analyticity**:

   Since \(f(z)\) has no poles inside or on \(|z| = \sqrt{2}\), it is analytic in the region enclosed by \(C\). Therefore, by Cauchy's theorem,

   \[
   \int_{C} f(z) \, dz = 0
   \]

### Part (b): \(f(z) = \frac{z + \sin z - 2}{\cos z}\)

Next, we need to determine if \(f(z)\) is analytic inside and on the contour \(C : |z| = \sqrt{2}\).

1. **Poles of \(f(z)\)**:
   
   The function \(f(z)\) has potential poles where the denominator is zero:
   
   \[
   \cos z = 0 \implies z = \left( n + \frac{1}{2} \right)\pi \quad \text{for} \quad n \in \mathbb{Z}
   \]

   We need to check if any of these poles lie inside or on \(|z| = \sqrt{2}\):

   - For \(n = 0\), \(z = \frac{\pi}{2}\) and \(z = -\frac{\pi}{2}\).
   - For \(n = 1\), \(z = \frac{3\pi}{2}\) and \(z = -\frac{3\pi}{2}\).
   - And so on.

   The smallest \(\left|z\right|\) for these poles is \(\frac{\pi}{2}\), which is approximately 1.57. Since \(\sqrt{2} \approx 1.414\), all these poles are outside the contour \(|z| = \sqrt{2}\).

2. **Analyticity**:

   Since \(f(z)\) has no poles inside or on \(|z| = \sqrt{2}\), it is analytic in the region enclosed by \(C\). Therefore, by Cauchy's theorem,

   \[
   \int_{C} f(z) \, dz = 0
   \]

In conclusion, for both parts (a) and (b), \(f(z)\) is analytic inside and on the contour \(C : |z| = \sqrt{2}\), and thus by Cauchy's theorem,

\[
\int_{C} f(z) \, dz = 0
\]

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 difference between a pole and a removable singularity?
2. How do you determine the residue of a function at a given pole?
3. What is the significance of the radius of convergence for a power series in complex analysis?
4. Can you explain the concept of branch cuts in complex functions?
5. What are essential singularities, and how do they differ from poles?
6. How does the maximum modulus principle apply to analytic functions?
7. What is the relationship between conformal mappings and complex analysis?
8. Can you discuss the application of Laurent series in evaluating complex integrals?

**Tip**: When evaluating complex integrals, always identify the singularities of the integrand and check if they lie within the contour of integration. This helps in applying the appropriate theorems, such as Cauchy's theorem or the residue theorem.

Explore how to prove that the contour integral of analytic functions over a circular path is zero using Cauchy's Theorem. Analyze specific examples involving complex analysis and contour integrals. Suitable for undergraduate students.

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