# Is there a more expeditious way to solve this contour integral?

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:

$\int_{\gamma}z^3=\int_{0}^1(1+3t)^33dt=63.5$

$\int_{\gamma}2zdz=\int_{0}^12(1+3t)dt=5$

$\Rightarrow$

$\int_C(z^3+2z)dz=\int_{C*}(z^3+2z)dz-\int_{\gamma}(z^3+2z)dz=0-68.5=-68.5$

Since $\frac 14 z^4 + z^2$ is a primitive of $(z^3+2z)dz$ on all of $\Bbb C$, and because the contour is a path from $1$ to $4$, the integral is $\frac 1 4 4^4 + 4^2 - \frac 14 - 1 = 80- \frac 54 = \frac {315}4$.