1. **State Cauchy’s integral theorem:** Cauchy’s integral theorem states that if a function $f(z)$ is analytic (holomorphic) inside and on a simple closed contour $C$, then the integral of $f(z)$ around $C$ is zero: $$\int_C f(z) \, dz = 0$$ 2. **Evaluate $\int_C \frac{dz}{z+4}$ where $C$ is the circle $|z|=2$:** Since the circle $|z|=2$ is centered at the origin with radius 2, the point $z=-4$ lies outside this circle. The function $f(z) = \frac{1}{z+4}$ is analytic inside and on $C$ because $z=-4$ is not inside $|z|=2$. By Cauchy’s integral theorem, the integral is zero: $$\int_C \frac{dz}{z+4} = 0$$ 3. **Define Taylor series:** The Taylor series of a function $f(z)$ about a point $z=a$ is an infinite sum of terms calculated from the derivatives of $f$ at $a$: $$f(z) = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!} (z - a)^n$$ It represents $f(z)$ as a power series valid in a neighborhood around $a$. 4. **Find the residue of $f(z) = \frac{z e^z}{(z - a)^3}$ at $z = a$:** The residue at $z=a$ is the coefficient of $\frac{1}{z-a}$ in the Laurent expansion. Since the pole is of order 3, the residue is given by: $$\text{Res}_{z=a} f(z) = \frac{1}{2!} \lim_{z \to a} \frac{d^2}{dz^2} \left[(z - a)^3 \frac{z e^z}{(z - a)^3}\right] = \frac{1}{2} \frac{d^2}{dz^2} (z e^z) \bigg|_{z=a}$$ Calculate derivatives: $$\frac{d}{dz} (z e^z) = e^z + z e^z = e^z (1 + z)$$ $$\frac{d^2}{dz^2} (z e^z) = \frac{d}{dz} [e^z (1 + z)] = e^z (1 + z) + e^z = e^z (2 + z)$$ Evaluate at $z=a$: $$e^a (2 + a)$$ Therefore, $$\text{Res}_{z=a} f(z) = \frac{1}{2} e^a (2 + a)$$ 5. **Using Cauchy’s integral formula, evaluate $\int_C \frac{\sin \pi z^2 + \cos \pi z^2}{(z-2)(z-3)} dz$ where $C$ is $|z|=4$:** The function has simple poles at $z=2$ and $z=3$, both inside $|z|=4$. Rewrite the integral as: $$\int_C \frac{f(z)}{(z-2)(z-3)} dz$$ where $f(z) = \sin \pi z^2 + \cos \pi z^2$ is entire. By Cauchy’s integral formula for multiple poles: $$\int_C \frac{f(z)}{(z-a)(z-b)} dz = 2\pi i [\frac{f(a)}{a-b} + \frac{f(b)}{b-a}]$$ Substitute $a=2$, $b=3$: $$= 2\pi i \left[ \frac{f(2)}{2-3} + \frac{f(3)}{3-2} \right] = 2\pi i \left[-f(2) + f(3)\right]$$ Calculate $f(2)$ and $f(3)$: $$f(2) = \sin(4\pi) + \cos(4\pi) = 0 + 1 = 1$$ $$f(3) = \sin(9\pi) + \cos(9\pi) = 0 + (-1) = -1$$ Therefore, $$\int_C = 2\pi i (-1 - 1) = -4\pi i$$ 6. **Expand $f(z) = \frac{7z - 2}{z(z-2)(z+1)}$ as Laurent series in $1 < |z+1| < 3$:** Rewrite $f(z)$ using partial fractions: $$\frac{7z - 2}{z(z-2)(z+1)} = \frac{A}{z} + \frac{B}{z-2} + \frac{C}{z+1}$$ Solving for $A,B,C$ gives: $$A = 1, B = 2, C = 4$$ So, $$f(z) = \frac{1}{z} + \frac{2}{z-2} + \frac{4}{z+1}$$ Expand around $z = -1$ with $w = z+1$: $$\frac{1}{z} = \frac{1}{w - 1} = -\sum_{n=0}^\infty w^n$$ $$\frac{2}{z-2} = \frac{2}{w - 3} = -\frac{2}{3} \sum_{n=0}^\infty \left(\frac{w}{3}\right)^n$$ $$\frac{4}{z+1} = \frac{4}{w}$$ Thus, $$f(z) = 4 w^{-1} - \sum_{n=0}^\infty w^n - \frac{2}{3} \sum_{n=0}^\infty \left(\frac{w}{3}\right)^n$$ This is the Laurent series valid for $1 < |w| < 3$. 7. **Evaluate $\int_C \frac{z-1}{(z-2)(z+1)^2} dz$ where $C$ is $|z - i|=2$ using Cauchy’s residue theorem:** Poles inside $C$ are at $z=2$ and $z=-1$. Check if these points lie inside $|z - i|=2$: Distance from $i$ to 2 is $|2 - i| = \sqrt{(2)^2 + (-1)^2} = \sqrt{5} < 2.24$, inside. Distance from $i$ to $-1$ is $|-1 - i| = \sqrt{(-1)^2 + (-1)^2} = \sqrt{2} < 2$, inside. Calculate residues: - At $z=2$ (simple pole): $$\text{Res}_{z=2} = \lim_{z \to 2} (z-2) \frac{z-1}{(z-2)(z+1)^2} = \frac{2-1}{(2+1)^2} = \frac{1}{9}$$ - At $z=-1$ (pole order 2): $$\text{Res}_{z=-1} = \lim_{z \to -1} \frac{d}{dz} \left[(z+1)^2 \frac{z-1}{(z-2)(z+1)^2}\right] = \lim_{z \to -1} \frac{d}{dz} \left( \frac{z-1}{z-2} \right)$$ Derivative: $$\frac{d}{dz} \left( \frac{z-1}{z-2} \right) = \frac{(z-2) - (z-1)}{(z-2)^2} = \frac{-1}{(z-2)^2}$$ Evaluate at $z=-1$: $$\text{Res}_{z=-1} = \frac{-1}{(-1-2)^2} = \frac{-1}{9}$$ Sum of residues: $$\frac{1}{9} - \frac{1}{9} = 0$$ By residue theorem, $$\int_C = 2\pi i \times 0 = 0$$ 8. **Evaluate $\int_{-\infty}^\infty \frac{x^2}{(x^2 + a^2)(x^2 + b^2)} dx$:** Assuming $a,b > 0$, use contour integration or known integral formula: $$\int_{-\infty}^\infty \frac{x^2}{(x^2 + a^2)(x^2 + b^2)} dx = \frac{\pi}{a + b}$$ **Final answers:** 2. $0$ 4. $\frac{1}{2} e^a (2 + a)$ 5. $-4 \pi i$ 7. $0$ 8. $\frac{\pi}{a + b}$