To evaluate the given integral using the Cauchy Integral Formula, we first recall the formula:

$f^{(n)}(a) = \frac{n!}{2\pi i} \oint_C \frac{f(z)}{(z-a)^{n+1}} dz,$

for a function $f(z)$ analytic inside and on a closed contour $C$. When $n = 0$, it simplifies to:

$f(a) = \frac{1}{2\pi i} \oint_C \frac{f(z)}{z-a} dz.$

The given integral is:

$\oint_C \frac{z}{z-2} dz,$

where $C$ is the circle $|z-2| = \frac{3}{2}$. Let's solve this step-by-step:

Step 1: Identify $f(z)$ and $a$

Here:

• $f(z) = z$,
• $a = 2$.

The function $f(z)$ is analytic everywhere inside and on the circle $C$.

Step 2: Apply Cauchy Integral Formula

Using the formula for $f(a)$:

$\oint_C \frac{z}{z-2} dz = 2\pi i \cdot f(2).$

We now calculate $f(2)$. Substituting $z = 2$ into $f(z) = z$:

$f(2) = 2.$

Step 3: Evaluate the Integral

Now substitute $f(2)$ into the formula:

$\oint_C \frac{z}{z-2} dz = 2\pi i \cdot 2 = 4\pi i.$

Final Answer:

$\oint_C \frac{z}{z-2} dz = 4\pi i.$

Would you like further details on the Cauchy Integral Formula or a breakdown of the process?

Related Questions:

1. What is the Cauchy Integral Theorem, and how does it relate to this formula?
2. How does the integral change if $f(z)$ is a different function, e.g., $f(z) = z^2$?
3. What is the significance of the contour $|z-2| = \frac{3}{2}$ in this problem?
4. Can we evaluate this integral using a residue approach instead of Cauchy’s formula?
5. What happens if $f(z)$ has a singularity within the contour?

Tip:

Always ensure the function $f(z)$ is analytic inside the contour when applying the Cauchy Integral Formula. If not, residues need to be calculated.