Parametrization of a path for line integral

Question

Compute the integral $$\int\limits_{\alpha} z \,\mathrm{d}z$$ over the path $\alpha$ from $i+1$ to $-i$ to $1-i$.

I sketch this path and it's a incomplete triangle. We've never done any examples what to do and in which way. How I get the boundaries for the integral? Thank you so much!

Answer 1

Assume straight line from $1+i$ to $-i$, and then straight line from $-i$ to $1-i$.

For the first straight path, $$x=1-t, y=1-2t$$ where $t=0\to 1.$

The integral is $$\int_{t=0}^{t=1} \left(x+iy\right)\left(dx+idy\right)=\int_0^1 \left(1-t+i(1-2t)\right)\left(-dt-2idt\right)$$ $$=-\int_0^1\left(1-t+i(1-2t)\right)(1+2i)dt$$ $$=-\int_0^1\left[(-1+3t)+i(3-4t)\right]dt$$ $$=-\left[-t+\frac{3t^2}{2}\right]_0^1-i\left[3t-2t^2\right]_0^1$$ $$=-\frac{1}{2}-i$$

For the second straight path, $$x=t, y=-i$$ The integral is $$\int_{t=0}^{t=1}(x-i)dx$$ $$=\int_0^1(t-i)dt$$ $$=\left[\frac{t^2}{2}-it\right]_0^1$$ $$=\frac{1}{2}-i$$

So the total integral is $$(-\frac{1}{2}-i)+(\frac{1}{2}-i)$$ $$=-2i$$

Answer 2

Observe that $\;F(z)=\cfrac12z^2\;$ is a primitive to $\;z\;$ , and thus for any two points $\;z_0,\,z_1\in\Bbb C\;$ ,we have that the integral from $\;z_0\;$ to $\;z_1\;$ is

$$\int_{z_0}^{z_1}z\,dz=F(z_1)-F(z_0)$$