I'm trying to evaluate $\int_c(z^3+2z) dz$ where C is the union of the three line segments $l(1,i),l(i,1-2i)$ and $l(1-2i,4)$.
What I've done so far is :
$$\int_C(z^3+2z)dz=\int_Cz^3dz+\int_C2zdz$$
then taking this piece by piece ;
$$\int_cz^3dz=\int_{l(1,i)}z^3dz+\int_{l(i,1-2i)}z^3dz+\int_{l(1-2i,4)}z^3dz$$
and then taking this part piece by piece too
$$\int_{l(1,i)}z^3dz$$ : this line segment can be parameterised as $\gamma(t)=t+i-it$
We can the use the fact that $\int_{\gamma}f(z)dz=\int^{b}_{a}f(\gamma(t))\gamma'(t)dt$, and keep going about it like this for all the separate parts.
However this is a very long and tedious way to calculate this integral. Is there any more expeditious methods using Cauchy's theorems ?
@gt6989b
SO is the following answer correct ?
We have the contour integral $\int_C(z^3+2z)dz$ where $C:=l(1,i) \bigcup l(i,1-2i, \bigcup l(1-2i,4)$
Now define $C*:=l(1,i) \bigcup l(i,1-2i, \bigcup l(1-2i,4) \bigcup \gamma$ where $\gamma:=l(4,1)$, i.e. the line closing the curve.
now we can note that $f(z)=(z^3+2z)$ is holomorphic as it is in fact entire, and $C*$ is certainly rectifiable. $\therefore $ by Cauchy's Integral theorem :
$\int_{C*}f(z)=0.$
That parts easy though, this next part I'm a little more unsure if I'm doing correctly.
This time we calculate just:
$\int_{\gamma}(z^3+2z)dz$
the line segement $\gamma:=l(4,1)$ has parametrisation $\gamma(t)=1+3t$, and so $\gamma'(t)=3$, where $t \in [0,1]$.
Then splitting this integral in two and applying $\int_{\gamma}f(z)dz=\int^{b}_{a}f(\gamma(t))\gamma'(t)dt$. We get:
$\int_{\gamma}z^3=\int_{0}^1(1+3t)^33dt=63.5$
$\int_{\gamma}2zdz=\int_{0}^12(1+3t)dt=5$
$\Rightarrow$
$\int_C(z^3+2z)dz=\int_{C*}(z^3+2z)dz-\int_{\gamma}(z^3+2z)dz=0-68.5=-68.5$
HINT
Add a 4th leg to close the contour, note the integrand is holomorphic and apply Cauchy's integral theorem. You will only have to evaluate that 4th leg...