A complex integral over two lines?

Question

Let C denote the union of segments $[-1-i,2]$ and $[2,3+2i]$. Find

$$\int_{c}(z^{2}-z+1)dz$$

Sorry about not showing any work. I know we are supposed to parameterize the lines but I'm have a lot of difficulty with that.

Answer 1

Let me show you how to parametrise the line from $-1-i$ to $2$. There are lots of ways of parametrising it actually, and it doesn't matter which one we choose. One simple example is: $$ z = (-1-i)(1-t) + 2t, \ \ \ \ \ 0 \leq t \leq 1.$$ Hopefully you can convince yourself that this works.

Having done this, you can write the integral in terms of the parameter $t$: $$z^2 - z + 1 = \left[(-1-i)(1-t) + 2t\right]^2 - \left[(-1-i)(1-t) + 2t\right] + 1$$ (Of course, you could simplify this if you wish...)

You should also write $dz$ in terms of $t$, using the chain rule. $$ dz = \left(-(-1-i)+2\right)dt$$

Finally, put everything together, taking care to put the correct limits on the integral: $$ \int_{[-1-i,2i]}(z^2 - z + 1)dz = \int_{0}^{1} \left( \left[(-1-i)(1-t) + 2t\right]^2 - \left[(-1-i)(1-t) + 2t\right] + 1 \right) \times \left(-(-1-i)+2\right)dt $$ (Again, this will look much nicer once you simplify everything!)