Contour Integral of $1/z dz$ with triangle

Question

I was asked to find the contour integral $dz/z$ of a triangle with vertices $1-2i$, $-2+i$ and $1+i$.

I am not sure how to begin this but my professor said that it would be:

$$\int_{-1}^{2} \frac{-i}{1-ti} dt$$

But this doesn't make sense to me. why would $z$ be equal to $1-ti$? I can see he may be working from the vertically connected vertices but that would be $1+iy$ right?

Answer 1

The total contour will be;

$$\oint\frac{dz}{z}=\int_{1+i}^{1-2i}\frac{dz}{z}+\int_{1-2i}^{-2+i}\frac{dz}{z}+\int_{-2+i}^{1+i}\frac{dz}{z}$$

For the first integral, we apply the substitution $z=1-ti$ which yields;

$$\int_{-1}^{2}\frac{-i}{1-ti}dt$$

The second integral will recieve the substitution $z=-(1+t)+ti$;

$$\int_{-2}^{1}\frac{-1+i}{-(1+t)+ti}dt=\int_{-2}^{1}\frac{-1+i}{-1+t(-1+i)}dt=\int_{-2}^{1}\frac{-1+i}{(-1+i)-i+t(-1+i)}dt=\int_{-2}^{1}\frac{1}{(1+t)-i}dt$$

The final integral will recieve the substitution $z=t+i$;

$$\int_{-2}^{1}\frac{1}{t+i}dt$$

Now, we can manipulate our second integral by applying another substitution of $m=-(1+t)$. This will yield;

$$\int_{-2}^{1}\frac{1}{(1+t)-i}dt\rightarrow\int_{1}^{-2}\frac{1}{m+i}dm$$

So, in total we will have;

$$\oint\frac{dz}{z}=\int_{-1}^{2}\frac{-i}{1-ti}dt+\int_{1}^{-2}\frac{1}{m+i}dm+\int_{-2}^{1}\frac{1}{t+i}dt$$

The last two integrals clearly cancel each other and we are left with;

$$\oint\frac{dz}{z}=\int_{-1}^{2}\frac{-i}{1-ti}dt$$

as your professor had claimed.

Answer 2

Considering the fact that the singularity of this function ($0$) lies inside the contour, we can use the residue theorem. So $$ \oint_C \frac{dz}{z} = 2i\pi \mathcal{Res}(0) = 2i\pi \lim_{\phi\to0}\frac{(\phi-0)}{\phi}=2i\pi $$ Where the contour $C$ is the triangle you were talking about

Answer 3

I'll assume that you want to do this directly, rather than appealing to the residue theorem.

Suppose we have a path $\gamma: [a, b] \to \mathbb{C}$. When we write down a contour integral $\int_{\gamma} f(z) \; dz$, this is the same as $$\int_{a}^b f(\gamma(t)) \gamma'(t) \; dt,$$ just like any other path/line/curve integral.

So, the integral that your prof told you to start with corresponds to the piece of the triangular contour going from $1 + i$ to $1 - 2i$. One way to parametrize this line segment is to define $\gamma(t) = 1 - ti$ with domain $t \in [-1, 2]$. Then $\gamma'(t) = -i$, so indeed the correct integral expression (using $f(z) = 1/z$ in my previous formula) is $$\int_{-1}^2 \frac{-i}{1 - it} \; dt.$$ You can do something completely similar with the other two pieces and add all of them up to get the full contour integral.

That being said, despite the fact that this integral looks simple, you need to be careful. It is true that the function $\log(1 - it)$ has this integrand as its derivative, but to have this make sense you need to choose a particular branch cut of the logarithm function. You can do the same for the other pieces, and you might think that at the end, everything cancels. But this turns out to be wrong: if you go in a circle around the origin, the branch cut of the logarithm that you start with is not the one that you end up with. In fact, these two branch cuts differ by exactly $2\pi i$, and indeed this is precisely the value of the integral. I will leave you to work out the subtleties.