The integral I have to evaluate is $\int_Czdz$, where $C$ is the line from 0 to $1+i$, and then from $1+i$ to 2.
My work:
$z_1(t)=(1+i)t$ and $z_2(t)=(t+1)+i(1-t)=t(i-1)+(1+i)$, $t\in[0,1]$. $z_1'(t)=1+i$ and $z_2'(t)=1-it$. $$\int_Czdz=\int_0^1z_1(t)\cdot z_1'(t)dt+\int_0^1z_2(t)\cdot z_2'(t)dt$$ $$=\int_0^1(1+i)t(1+i)+[t(1-i)+1+i](1-it)dt$$ $$=\frac{5}{3}+i\frac{2}{3}$$
Is this correct? Are integrals of complex numbers supposed to yield complex results?
Well, to follow the aproach you took. First of all, the usual parametrization of a line segment $[a,b]$ is
$$\gamma(t)=(1-t)\cdot a+t\cdot b$$
when $t\in [0,1]$, so at the time $t=0$, you get the point $a$, and at the time $t=1$, you get the point $b$. Now, in your particular case, you need two parametrizations, one for the line segment $[0,1+i]$ and the other one for the line segment $[1+i,2]$. For the first one, a parametrization would be
$$\gamma_1(t)=(1-t)\cdot0 \ +t\cdot(1+i)$$ with derivative
$$\gamma_1^{'}(t)=1+i$$ when $t\in [0,1]$. And for the other one
$$\gamma_2(t)=(1-t)(1+i)+2t=1+i+(1-i)t$$ with derivative
$$\gamma_2^{'}(t)=1-i$$ Then
$$\int_Czdz=\int_0^1\gamma_1(t)\cdot\gamma_1^{'}(t)dt\ +\ \int_0^1\gamma_2(t)\cdot\gamma_2^{'}(t)dt=\int_0^1(1+i)^2\cdot tdt\ +\ \int_0^1((1+i)(1-i)+t\cdot(1-i)^2)dt=2$$ In your answer, you didn't parametrized the line segment $[1+i,2]$ well, but you were in the right path. Another way to solve this is to add the line segment from $2$ to $0$. Doing this, we have created a triangle $\Delta$ with vertex in $0,1+i$ and $2$. Then, by the Cauchy's Integral Theorem, since the function $f(z)=z$ is holomorphic on $\Delta$
$$\int_\Delta f(z)dz=0$$
But what does it have to do with your integral? Well, your integral is just
$$\int_Cf(z)dz=\int_\Delta f(z)dz\ -\ \int_{[2,0]}f(z)dz$$ where by $[2,0]$, I mean the line segment from $2$ to $0$. Parametrizing the line segment by $\gamma_3(t)=(1-t)\cdot 2$, which has derivative $\gamma_3^{'}(t)=-2$, we get that
$$\int_{[2,0]}zdz=\int_0^1\gamma_3(t)\cdot \gamma_3^{'}(t)dt=-2$$ Then
$$\int_Cf(z)dz=0-(-2)=2$$ which gives again the desired result.