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For $f\in Hol(\Bbb{C}\setminus \Bbb{R})\cap C(\Bbb{C})$ show that $\int_{\gamma}f(z)dz=0$ for all closed curve $\gamma$ piecewise-continuous.

I am completely clueless but this is primarily because I feel hopeless regarding the winding numbers. Can you give me an advice/guidance?

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    Use Morera's theorem to show that such an $f$ is actually holomorphic on the entire plane. – 2017-01-04
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    How can I? Doesn't the theorem require me to show what I have to prove to begin with? – 2017-01-04
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    Should I look at a contour? – 2017-01-04
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    Morera's theorem only requires you to look at the special case of triangles. That's much simpler than looking at the full zoo of piecewise continuously differentiable closed curves. Now if the triangle completely lies in the (open) upper or the lower half-plane, we're done by Goursat. Next step, the triangle lies in the closed upper (or lower) half-plane, and has just one vertex on the real axis. After that, consider triangles that have one side on the real axis. Finally, look at triangles having one vertex in the upper and one in the lower half-plane. – 2017-01-04
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    The real axis cuts these into two parts (either two triangles, or one triangle and one quadrilateral) each having one side on the real axis. If we have a quadrilateral from the splitting by the real axis, cut it into two triangles using a diagonal. Then you have two or three triangles of the type treated above. – 2017-01-04
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    Suppose I decided to split to little triangles with more and more triangles from one step to another, how can I be sure it becomes zero ? I mean, there are sums of decreasing sequences that don't converge to zero of course... – 2017-01-04

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We use Morera's theorem to conclude that such a function is actually holomorphic on the entire plane, and then Cauchy's integral theorem yields the vanishing of $\int_{\gamma} f(z)\,dz$ for all closed $\gamma$.

So let $\Delta \subset \mathbb{C}$ a triangle. We need to show

$$\int_{\partial \Delta} f(z)\,dz = 0.\tag{1}$$

Case $0$: $\Delta \cap \mathbb{R} = \varnothing$: In that case, Cauchy's integral theorem immediately yields $(1)$ since $f$ is holomorphic on a neighbourhood of $\Delta$.

Case $1$: $\Delta \subset \{ z \in \mathbb{Z} : \operatorname{Im} z \geqslant 0\}$. For $h \in (0,1]$, let $f_h(z) = f(z + ih)$. Since $f_h$ is holomorphic on $\{ z \in \mathbb{C} : \operatorname{Im} z > -h\}$, Cauchy's integral theorem yields

$$\int_{\partial \Delta} f_h(z)\,dz = 0,\tag{2}$$

and thus the standard estimate yields

$$\Biggl\lvert\int_{\partial\Delta} f(z)\,dz\Biggr\rvert = \Biggl\lvert \int_{\partial\Delta} f(z) - f_h(z)\,dz\Biggr\rvert \leqslant L(\partial\Delta)\cdot \max \{ \lvert f(z) - f_h(z)\rvert : z \in \partial \Delta\}.$$

Since $f$ is uniformly continuous on the compact set $\{ z\in\mathbb{C} : \operatorname{dist}(z,\Delta) \leqslant 1\}$, for every $\varepsilon > 0$ there is a $\delta > 0$ such that $\max \{ \lvert f(z) - f_h(z)\rvert : z \in \partial \Delta\} \leqslant \varepsilon$ for all $h \in (0,\delta]$, it follows that

$$\Biggl\lvert\int_{\partial\Delta} f(z)\,dz\Biggr\rvert \leqslant L(\partial\Delta)\cdot \varepsilon$$

for all $\varepsilon > 0$. But that means $(1)$ holds.

Case $2$: $\Delta \subset \{ z \in \mathbb{C} : \operatorname{Im} z \leqslant 0\}$. This case is analogous to case $1$, only we use $f_h(z) = f(z - ih)$ now.

Case $3$: $\Delta$ intersects both, the open upper and lower half-plane.

Then let $A = \Delta \cap \{ z \in \mathbb{C} : \operatorname{Im} z \geqslant 0\}$ and $B = \Delta \cap \{ z \in \mathbb{C} : \operatorname{Im} z \leqslant 0\}$. By the argument of cases $1$ and $2$, we have

$$\int_{\partial A} f(z)\,dz = 0 = \int_{\partial B} f(z)\,dz,$$

and since the segment $\Delta \cap \mathbb{R}$ occurs with opposite orientations in $\partial A$ and $\partial B$, while the remaining parts of the boundaries of $A$ and $B$ together make up the boundary of $\Delta$, we have

$$\int_{\partial\Delta} f(z)\,dz = \int_{\partial A} f(z)\,dz + \int_{\partial B} f(z)\,dz = 0 + 0 = 0.$$

So the premise of Morera's theorem is verified, and we conclude $\operatorname{Hol}(\mathbb{C}\setminus \mathbb{R}) \cap C(\mathbb{C}) = \operatorname{Hol}(\mathbb{C})$.

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    Which path do you choose to define $F(z)$? If you don't specify the path, you need the result of the exercise to know that $F$ is well-defined. If you take the straight line segment, you need the result for triangles to show that $F$ is differentiable (on $\mathbb{C}\setminus \mathbb{R}$ at least) with derivative $f$. – 2017-01-04
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    Yes, the path doesn't matter, but proving that is the point of the exercise. And if we take the axis-parallel paths, we get (for $\operatorname{Im} z \neq 1$) the partial derivative with respect to $y$ immediately, but to get $\partial_x F = f$, one needs the result of the exercise for some class of closed paths, to prove you can take the last path horizontal without changing $F$. – 2017-01-04
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    Ok, I'd start by showing $\int_\square f(s)ds = 0$ for any square contour crossing the real axis in 2 points, before defining $F(x+iy) = \int_i^{x+i} + \int_{x+i}^{x+iy} f(s)ds$ , which is equivalent to the answer you wrote. – 2017-01-04