Evaluate $\oint_C \frac{3}{z + 1 + i}dz$ along the circle $|z| = 2$ 
Evaluate $$\oint_C \frac{3}{z + 1 + i}dz$$ along the circle $|z| = 2$

The solution that I saw read as follows:

The integrand is not holomorphic at $z = -1-i$ and this point $(-1, -1)$ lies within $C$. Put $z-z_0 = re^{it}$ where $z_0 = -1-i$ so $dz = ire^{it}$. So$$\oint \frac{3}{z+1+i}dz = \oint \frac{3}{re^{it}}ire^{it}dt = 3i\int_0^{2\pi}dt = 6\pi i$$

This was my lecturer's solution to the above problem. I can see that the lecturer "set" $z+1+i = re^{it}$ but how is this even possible? Because firstly $r$ is not even defined and secondly $2 \in C$ but $2+ 1 + i$ is certainly not in $C$.
For example a parametrization of the circle $C = \{z \in \mathbb{C} \ | \ |z| = a\}$ is given by the holomorphic function $f : [0, 2\pi] \to C$ defined by $f(t) = ae^{it}$, then for any point $z \in C$ we have $z= f(t) = ae^{it}$ for some $t \in [0, 2\pi]$.
What my lecturer did above does not seem correct. What would a correct solution look like for this? Taking the parameterizatin $z = 2e^{it}$ does not help in this case I think because we'd then end up with $$\frac{6ie^{it}}{2e^{it}+1+i}$$ as the integrand (which we'd be integrating from $t=0$ to $t=2\pi$
 A: Note that by the Cauchy Theorem,
$$\oint_{|z|=2} \frac{3}{z+1+i}dz=\oint_{\gamma} \frac{3}{z+1+i}dz$$
for any simple positively oriented curve $\gamma$ which contain the point $-1-i$. Now when the lecturer "set" $z+1+i=re^{it}$, he actually changed the path of integration by replacing the closed curve $|z|=2$ with a the circle centered at $-1-i$ with arbitrary radius $r>0$.
A: You can do it the hard way and stick with contour integration 
$$\int\limits_{\gamma}f(z)dz=\int\limits_{a}^{b}f(\gamma(t))\gamma'(t)dt$$
and given $\gamma(t)=2e^{it}$ we should have
$$\int\limits_{|z|=2} \frac{3}{z+1+i}dz=\int\limits_{0}^{2\pi}\frac{3}{2e^{it}+1+i}\cdot 2ie^{it} dt=...$$
but, here is the trick $\left(2e^{it}+1+i\right)'=2ie^{it}$ and another one, thus
$$...=\int\limits_{0}^{2\pi}\frac{3}{2e^{it}+1+i} d\left(2e^{it}+1+i\right)=
3\operatorname{Log}\left(2e^{it}+1+i\right)\biggr\rvert_{0}^{2\pi}=\\
3\left(\ln\left|2e^{i2\pi}+1+i\right|+iArg\left(2e^{i2\pi}+1+i\right) -\ln\left|2e^{i\cdot 0}+1+i\right|+iArg\left(2e^{i\cdot 0}+1+i\right)\right)=\\
3\left(\ln\left|3+i\right|+iArg\left(2e^{i2\pi}+1+i\right) -\ln\left|3+i\right|+iArg\left(2e^{i\cdot 0}+1+i\right)\right)=\\
3i\left(Arg\left(2e^{i2\pi}+1+i\right) -Arg\left(2e^{i\cdot 0}+1+i\right)\right)=3i(2\pi)$$

Or very easy way, use Cauchy's integral formula
$$f(a)=\frac{1}{2\pi i}\int\limits_{|z|=2}\frac{f(z)}{z-a}dz$$
where $f(z)=3$ and this
$$3=\frac{1}{2\pi i}\int\limits_{|z|=2}\frac{3}{z+1+i}dz$$
