a)$\displaystyle\int_{\partial B_2(0)}\dfrac{e^z}{(z+1)(z-3)^2}dz$
Apply the Cauchy integral with $f(z)=\dfrac{e^z}{(z-3)^2}$ at $ z= -1 $. Then:
$$\int_{\partial B_2(0)}\dfrac{e^z}{(z+1)(z-3)^2}dz = 2\pi if(-1) =2 \pi \dfrac{e^{-1}}{(-4)^2}=2\pi i \dfrac{e^{-1}}{16}=\dfrac{1}{8}\pi i e^{-1} $$
b) $\displaystyle\int_{\partial B_2(0)}\dfrac{\sin z}{(z+i)}dz$
Apply the Cauchy integral with $f(z)=\sin z$ at $z=-i$.
$f(z)=\sin z$ is holomorphic inside $|z|=2$. Then
$$\int_{\partial B_2(0)}\dfrac{\sin z}{z+i}dz= 2\pi i f(-i)=2 \pi i \sin(-i)=2\pi i (-i \sinh(1)) = 2 \pi \sinh(1)$$
$\sin(-i)=-i\sinh(1)$
c) $\displaystyle\int_{\partial B_2(-2i)}\dfrac{dz}{(z^2+1)}$
partial fractions:
$\dfrac{1}{z^2+1} = \dfrac{1}{z^2-i^2}= \dfrac{1}{(z+i)(z-i)}=\dfrac{i/2}{z+i}-\dfrac{i/2}{z-i}$
Then
$$\int_{\partial B_2(-2i)}\dfrac{dz}{(z^2+1)} = \dfrac{i}{2} \int_{\partial B_2(-2i)}\dfrac{dz}{(z+i)}-\dfrac{i}{2} \int_{\partial B_2(-2i)}\dfrac{dz}{(z-i)}$$
$C= \partial B_2(-2i)$
Then $f(z)=1$ and $i$ y $-i$ its inside $\mathbb{C}$, then would be:
$$\int_C \dfrac{dz}{z-i} = 2 \pi i f(-i) = 2 \pi i $$
$$\int_C \dfrac{dz}{z-i} = 2 \pi i f(i) = 2 \pi i $$
Then $$\int_{\partial B_2(-2i)}\dfrac{dz}{(z^2+1)}=0$$
d) $\displaystyle\int_{\partial B_1(0)}\dfrac{e^z}{(z-2)^3}dz $
Applying the Cauchy integral with $f(z)=e^z$ at $z=2$. But 2 does not belong to the circle of radius 1, then the integral is 0.
Am I correct? I would like to know if there is any mistake, I am just starting learning about this
The answers to questions a), b), and d) are correct. In the case of c), note that $-i\in B_2(-2i)$, whereas $i\notin B_2(-2i)$. So,$$\frac i2\int_{B_2(-2i)}\frac{\mathrm dz}{z+i}-\frac i2\int_{B_2(-2i)}\frac{\mathrm dz}{z-i}=\frac i2\times2\pi i-\frac i2\times0=-\pi.$$