Evaluate the following integrals (1) $$ \int_{c} \frac{z d z}{\left(9-z^{2}\right)(z+i)} d z, c:|z|=2 $$ (8mks (11) $$ \int_{c} \frac{2 z^{2}+1}{(z+2)(z-3)^{2}} d z, c:|z|=5 $$ (smks

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**(1)** Evaluate 
$$
\int_{c} \frac{z}{\left(9-z^{2}\right)(z+i)}\,dz,\quad c: |z|=2.
$$

1. **Factor the denominator:**  
   We write 
   $$
   9-z^2 = (3-z)(3+z).
   $$
   Thus, the poles of the integrand occur at:
   $$
   z=3,\quad z=-3,\quad \text{and}\quad z=-i.
   $$
   
2. **Locate poles with respect to the contour $$ |z|=2 $$:**  
   - $$z=3$$ and $$z=-3$$ have modulus 3, so they lie outside the circle.  
   - $$z=-i$$ has modulus 1 and lies inside the circle.

3. **Compute the residue at $$ z=-i $$:**  
   Since the pole at $$ z=-i $$ is a simple pole, the residue is given by
   $$
   \operatorname{Res}(f, -i) = \lim_{z\to -i} (z+i)\frac{z}{(9-z^2)(z+i)} = \frac{-i}{9-(-i)^2}.
   $$
   First, compute $$(-i)^2$$:
   $$
   (-i)^2 = -1.
   $$
   Therefore,
   $$
   9-(-1)=10,
   $$
   and the residue becomes
   $$
   \operatorname{Res}(f, -i)= \frac{-i}{10}.
   $$

4. **Apply the Residue Theorem:**  
   The integral is
   $$
   \int_{c} \frac{z}{(9-z^2)(z+i)}\,dz = 2\pi i \left(\operatorname{Res}(f, -i)\right) = 2\pi i \left(-\frac{i}{10}\right).
   $$
   Since 
   $$
   i\cdot (-i) = -i^2 = 1,
   $$
   it follows that
   $$
   2\pi i \left(-\frac{i}{10}\right) = \frac{2\pi}{10} = \frac{\pi}{5}.
   $$

---

**(2)** Evaluate 
$$
\int_{c} \frac{2 z^{2}+1}{(z+2)(z-3)^{2}}\,dz,\quad c: |z|=5.
$$

1. **Identify the poles of the integrand:**  
   The denominator $$(z+2)(z-3)^2$$ provides:
   - A simple pole at $$z=-2$$,
   - A double pole at $$z=3$$.

2. **Check which poles lie within $$ |z|=5 $$:**  
   Both $$z=-2$$ and $$z=3$$ have moduli less than 5, so they are inside the contour.

3. **Residue at $$ z=-2 $$ (simple pole):**  
   For a simple pole, 
   $$
   \operatorname{Res}(f, -2)= \lim_{z\to -2} (z+2)\frac{2z^2+1}{(z+2)(z-3)^2} = \frac{2(-2)^2+1}{(-2-3)^2}.
   $$
   Calculate the numerator:
   $$
   2(4)+1=8+1=9.
   $$
   And the denominator:
   $$
   (-5)^2 =25.
   $$
   Thus,
   $$
   \operatorname{Res}(f, -2)= \frac{9}{25}.
   $$

4. **Residue at $$ z=3 $$ (double pole):**  
   For a double pole at $$ z=3 $$, the residue is given by
   $$
   \operatorname{Res}(f, 3)= \lim_{z\to3}\frac{d}{dz}\left[(z-3)^2 \frac{2z^2+1}{(z+2)(z-3)^2}\right] = \lim_{z\to3}\frac{d}{dz}\left(\frac{2z^2+1}{z+2}\right).
   $$
   
   Let 
   $$
   g(z)=\frac{2z^2+1}{z+2}.
   $$
   Its derivative using the quotient rule is
   $$
   g'(z)=\frac{(4z)(z+2) - (2z^2+1)(1)}{(z+2)^2}.
   $$
   Evaluate at $$z=3$$:
   $$
   g'(3)=\frac{4(3)(3+2) - (2\cdot3^2+1)}{(3+2)^2} = \frac{12\cdot5 - (18+1)}{25} = \frac{60 - 19}{25} = \frac{41}{25}.
   $$
   So,
   $$
   \operatorname{Res}(f, 3)=\frac{41}{25}.
   $$

5. **Sum of the residues and applying the Residue Theorem:**  
   Sum the residues:
   $$
   \operatorname{Res}(f, -2) + \operatorname{Res}(f, 3) = \frac{9}{25} + \frac{41}{25} = \frac{50}{25}=2.
   $$
   Therefore, the integral is
   $$
   \int_{c} \frac{2 z^{2}+1}{(z+2)(z-3)^{2}}\,dz = 2\pi i\cdot 2 = 4\pi i.
   $$

---

**Final Answers:**  
(1) $$\displaystyle \frac{\pi}{5}$$  
(2) $$\displaystyle 4\pi i$$
The integrals evaluate to: 1. $$ \frac{\pi}{5} $$ 2. $$ 4\pi i $$

Evaluate the following integrals
(1) (8mks
(11) (smks

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Evaluate the following integrals (1) (8mks (11) (smks