integration of an holomorphic function on a loop

Question

Let $f: D(0,1) \rightarrow \mathbb{C}$ be continuous and holomorphic on $D(0,1) \setminus \left]0,1\right[$

I must prove that:

• for all $a,b,c \in D(0,1): \int_{[a,b,c,a]}f(z) \, dz=0$.

What I did:

$$\int_{[a,b,c,a]} f(z) \, dz= \int_{[a,b,c,a]} f(z) \, dz=\int_a^b f(z) \, dz + \int_b^c f(z)\,dz+\int_c^a f(z) \, dz = 0.$$

It must be valid in the real analysis, but I didn't use the holomorphic property of $f$, so something is missing in my work.

Any help is much appreciated.

Many thanks!

Answer 1

I do not see why the sum of these integrals should be $0$. You do not provide any argument for that (hence it is hard to see where that argument would fail for arbitrary $f$).

Probably, you know $$\int_{[a, b, c, d]} g(z) \,\mathrm{d}z = 0 $$ if $g: U \to \mathbb C$ is holomorphic and the triangle lies completely in $U$.

Hence, if the triangle does not intersect with $]0, 1[$, you do not need to show anything.

Assume that the intersection of $[a, b, c, d]$ and $]0, 1[$ is a single point, that is, a corner of the triangle, let‘s say $a$. You can split the triangle in three smaller triangles: One with corner $a$ and circumference $\le \varepsilon$ and two lying completely in $D(0, 1) \setminus ]0, 1[$. The former integral is at most $\varepsilon \sup_{z \in \Delta} |f(z)|$ while the latter two are $0$.

The cases where the intersections are two points or a line can be handled similarly.

Answer 2

Presumably in setting your sum of three integrals equal to zero, you have said something like $$\int_{a}^{b} f(z)dz + \int_{b}^{c} f(z)dz + \int_{c}^{a} f(z)dz = \int_{a}^{a} f(z)dz = 0$$ If $z$ were real, this would be perfectly legitimate. For complex integrals, the integral will in general depend on which path you choose to integrate along. For example, if $C$ is a circle of radius $1$ with centre $0$, then $$\int_{C} \frac{1}{z} dz = 2\pi i$$ although this could be written as $$\int_{1}^{1} \frac{1}{z} dz$$ if we assumed the integral only depended on the endpoints. If you then complain that $C$ is not made up of straight lines, that still doesn't fix it, because the same integral (with $C$ replaced by any triangle containing $0$) still evaluates to $2\pi i$.

It is probably better to not use the $\int_{a}^{b}$ notation when referring to complex integrals, at least until you know which theorems hold in the real case and which hold in the complex case. The holomorphicity of $f$ on a region $U$ tells you that the integral over any triangle in $U$ will vanish, but this fact is not really obvious.