Calculate the integral
$$\int_{\gamma} \frac{dz}{z - 1 - i}$$
For a path given by the line joining $z_1 = 2i$ and $z_2 = 3$.
Initially I parameterized the path as the line $$y = -\frac{2x}{3} + 2$$ and then I used the complexification of this curve:
$$\gamma = t - 2i\left(\frac{t}{3} - 1\right)$$
applied in the integral, considering:
$$z = t - 2i\left(\frac{t}{3} - 1\right) \quad\implies\quad dz = dt -2i\left(\frac{dt}{3} - 0\right) = dt\left(1 - \frac{2i}{3}\right)$$
which gives me:
$$ \int_{0}^{3} \frac{(1 - \frac{2i}{3}) \, dt}{(1 - \frac{2i}{3})t +2i - 1 + i} = (3 - 2i)\int_{0}^{3} \frac{dt}{(3 - 2i)t +3(3i - 1)}. $$
To solve this integral I used the substitution:
$$ w = (3 - 2i)t + 3(3i -1) \quad\implies\quad \frac{dw}{(3 - 2i)} = dt$$
which leads me to:
$$(3 - 2i) \int_{9i - 3}^{3i + 8} \frac{dw}{(3 - 2i)}\, \frac{1}{w} = \frac{(3 - 2i)}{(3 - 2i)} \int_{9i - 3}^{3i + 8} \frac{dw}{w} = \ln \left(\frac{3i + 8}{9i - 3}\right)$$
I would like to know if my approach is correct, and if it's not, what should I do differently!
The function that you are integrating has a primitive, which is $\ln(z-1-i)$, where $\ln$ is the main logarithm. So,\begin{align}\int_\gamma\frac{\mathrm dz}{z-1-i}&=\ln(-1+i)-\ln(2-i)\\&=\ln|-1+i|+\frac{3\pi}4i-\left(\ln|2-i|-\arctan\left(\frac12\right)i\right)\\&=\frac12\ln\left(\frac25\right)+\left(\frac{3\pi}4+\arctan\left(\frac12\right)\right)i.\end{align}
In your answer, you don't say what $\ln$ is.