To evaluate the integral $\oint_{C} \frac{\sin(\pi z^2) + 2z}{(z - 1)(z - 2)} \, dz,$ where $C$ is the circle $|z| = 4$, we need to analyze the integrand, particularly the singularities within the contour.

Step 1: Identify the Singularities

The integrand is $f(z) = \frac{\sin(\pi z^2) + 2z}{(z - 1)(z - 2)}.$ The function has singularities where the denominator is zero, which occurs at $z = 1$ and $z = 2$.

Since $|z| = 4$ is a large enough contour to enclose both singularities at $z = 1$ and $z = 2$, both of these points are inside $C$.

Step 2: Apply the Residue Theorem

The residue theorem states that $\oint_{C} f(z) \, dz = 2\pi i \sum \text{Residues of } f(z) \text{ inside } C.$ Therefore, we need to find the residues of $f(z)$ at $z = 1$ and $z = 2$.

Step 3: Calculate the Residue at $z = 1$

To find the residue of $f(z)$ at $z = 1$, we isolate the term that becomes singular at $z = 1$: $\text{Residue at } z = 1 = \lim_{z \to 1} (z - 1) \frac{\sin(\pi z^2) + 2z}{(z - 1)(z - 2)}.$ Simplifying, we get $= \lim_{z \to 1} \frac{\sin(\pi z^2) + 2z}{z - 2}.$ Substituting $z = 1$ directly, $= \frac{\sin(\pi \cdot 1^2) + 2 \cdot 1}{1 - 2} = \frac{\sin(\pi) + 2}{-1} = \frac{0 + 2}{-1} = -2.$ Thus, the residue at $z = 1$ is $-2$.

Step 4: Calculate the Residue at $z = 2$

Similarly, the residue at $z = 2$ is $\text{Residue at } z = 2 = \lim_{z \to 2} (z - 2) \frac{\sin(\pi z^2) + 2z}{(z - 1)(z - 2)}.$ This simplifies to $= \lim_{z \to 2} \frac{\sin(\pi z^2) + 2z}{z - 1}.$ Substituting $z = 2$ directly, $= \frac{\sin(\pi \cdot 2^2) + 2 \cdot 2}{2 - 1} = \frac{\sin(4\pi) + 4}{1} = 0 + 4 = 4.$ Thus, the residue at $z = 2$ is $4$.

Step 5: Sum of the Residues

The sum of the residues inside $C$ is $-2 + 4 = 2.$

Step 6: Final Answer

By the residue theorem, $\oint_{C} \frac{\sin(\pi z^2) + 2z}{(z - 1)(z - 2)} \, dz = 2\pi i \cdot 2 = 4\pi i.$

Answer

$\oint_{C} \frac{\sin(\pi z^2) + 2z}{(z - 1)(z - 2)} \, dz = 4\pi i.$

Would you like more details on any step or have any questions?

1. How do you calculate residues for more complex functions?
2. Why does the residue theorem apply to closed contours like circles?
3. How would the result change if $C$ enclosed different points?
4. What happens to the integral if $\sin(\pi z^2)$ were replaced by a polynomial?
5. Can the residue theorem be used for contours that are not circular?

Tip: In complex analysis, always check the location of singularities in relation to the contour before applying the residue theorem.