I'm trying to use Cauchy Integral Formula to find the value of the integral
$$\int_\gamma (z-(4+4i))\sin(\frac{1}{z+(3-3i)})dz$$
where $\gamma$ is the circle centered at the origin and with radius 7.
This is my proposed solution: We will start by naming $f(z) = (z-(4+4i))\sin(\frac{1}{z+(3-3i)})$, noticing that $f(z)$ has a simple pole inside $\gamma$ and that both $\sin(z)$ and $f(z)$ are entire.
To solve this problem we will have to use Cauchy's Residue Theorem. First, we start by expressing $\sin(\frac{1}{z+(3-3i)})$ as a series, we known that the Taylor series for $\sin(z)$ is:
$$ \sin(z) = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!} (z)^{2n+1} $$
Therefore we will have that: $$ \sin(\frac{1}{z + (3-3i)}) = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!} \frac{1}{(z + (3-3i))^{2n+1}} $$
We can also see that: $$ (z-(4+4i)) = (z + (3 - 3i) - (3 - 3i) - (4 + 4i)) = [(z + (3 -3i)) - (7+i)] $$
Now we can express $f(z)$ as:
$$ f(z) = [(z + (3 -3i)) - (7-i)]\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!} \frac{1}{(z + (3-3i))^{2n+1}} $$
$$ = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!} \frac{1}{(z + (3-3i))^{2n}} + \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}\frac{-(7+i)}{(z + (3-3i))^{2n+1}} $$
From this last equation its easy to see that:
$$Res(f(z), -(3-3i)) = -(7+i) $$
Now, using Cauchy Residue Theorem we have that
$$\int_\gamma (z-(4+4i))\sin(\frac{1}{z+(3-3i)})dz = -2\pi i(7+i) = (2 -14 i)\pi$$