From my limited knowledge of contour integrals I believe I understand that if $$\int_{C_{1}+C_{2}}f(z)\:dz=\int_{C_{1}}f(z)\:dz+\int_{C_{2}}f(z)\:dt$$ where $$\int_{C}f(z)\:dz=\int_{C}f(\gamma(t))\gamma^\prime(t)\:dt$$ and $\gamma(t)$ is the parametrisation of the contour.
Now I have a question that asks to compute the integral of $f(z)=1+i-2\bar{z}$ from $z_0=0$ to $z_1=1+i$ using the definition of the countour integral along the following contours:
$1)$ Straight line segment $2)$ Parabola $y=x^2$ $3)$ Broken line $z_{0},z_{2},z_{1}$ with $z_{2}=1$
I am only interested in solving $3)$ but I made an attempt at first solving $1)$ as I expected that all three contours would yield the same answer (please correct me if I am wrong). I also made an attempt at $2)$ after realising that none of my answers matched up.
My attempt at $1)$:
First we parametrise the curve, so
$[0,1]\rightarrow \mathbb{C}: $$\gamma(t)=t(0)+(1-t)(1+i)=(1-t)+i(1-t) $ where $0\leq t \leq1$
I am a bit shaky on parametrisation (it has been a long time since I did Calc III) but I used the fact that $\gamma(t)=tW^\prime+(1-t)W $ is the parametrisation of the straight line that connects $W$ to $W^\prime$
Then the integral is simple, we have $f(\gamma(t))=(1+i-2((1-t)-i(1-t))$ which gives $$\int_{C}f(z)\:dz=\int_{0}^{1}(1+i-2t+2ti)(1+i)\:dt=\int_{0}^{1}(2i-4t)\:dt=[2ti-2t^2]_{0}^{1}=2i-2$$
My attempt at $3)$:
I split the integral up so that $\int_{C_{1}+C_{2}}f(z)\:dz=\int_{0}^{1}f(z)\:dz+\int_{1}^{1+i}f(z)\:dt$
First parametrisation from $0$ to $1$: $\gamma(t)=t(1)+(1-t)(0)=t$
This integration simply becomes $\int_{0}^{1}(1+i-2(t))(1)\:dt=[t+it-t^2]_{0}^{1}=i$
Second parametrisation from $1$ to $1+i$: $\gamma(t)=t(1+i)+(1-t)(1)=1+it$
And so the integral becomes $\int_{0}^{1}(1+i-2(1-it)(i)\:dt=\int_{0}^{1}(i-1-2i-2t)\:dt=[it-t-2it-t^2]_{0}^{1}=-2-i$
Now strangely the sum of these does not give the same as $1)$ but subtracting the first integral from the second gives the conjugate of $1)$, namely $-2-2i$.
Can anyone please tell me what I have done wrong here and guide me in the right direction.
Also my poor attempt at $2)$ looked something like this:
$\gamma[0,2]\rightarrow\mathbb{C}$ : $\gamma(t)=t+it$ , $0\leq t\leq2$
Then our integral is $\int{0}^{2}(1+i-2(t-it))(1+t)\:dt=\int{0}^{2}(1+i-2t+2it+i+i^2-2it+2i^2t)\:dt=\int{0}^{2}(2i-4t)\:dt=[2it-2t^2]_{0}^{2}=4i-8$
I am aware that strangely if I take $0\leq t\leq1$ then $2)$ gives the same answer as $1)$ (namely $2i-2$). Again, I think my parametrisation is rusty because I thought a parabola was parametrised as $\gamma(t)=t+it$ , $0\leq t\leq2$
Thanks!!