I was reading a proof for a proposition about winding numbers under perturbation, and ultimately it came down to saying $$n(\beta,0)=n(\alpha,0)+n(\gamma,0)$$ where $\beta(t)=\alpha(t)\gamma(t)$, and $\alpha,\gamma,\beta: [a,b]\rightarrow \mathbb{C}$ are curves in the complex plane.
Its `proof' of this was that "the argument of a product is the sum of the arguments." Taking Ahlfors viewpoint that the winding number not be defined with regards to the argument, I wanted to prove this result more generally using the definition: $$n(\gamma,w)=\frac{1}{2\pi i}\int_\gamma{\frac{dz}{z-w}}$$ where $\gamma$ is a piecewise continuously-differentiable curve not containing $w$ in its image.
I first proved it for when $w=0$, which is quite easy. See the end for this proof.
- Does this result actually hold true in the general case where $w$ is arbitrary? That is:
For $\alpha,\beta$ piecewise continuously differentiable closed curves $\alpha,\beta:[a,b]\rightarrow \mathbb{C}$ and $w$ not in the images of either $\alpha,\beta$, we define $\gamma:[a,b]\rightarrow \mathbb{C}$ by $\gamma(t)=\alpha(t)\beta(t)$. Then $$n(\gamma,w)=n(\alpha,w)+n(\beta,w).$$
- Is there a way to reduce it to the case where $w=0$?
Proof for $w=0$: Given $\alpha, \beta:[a,b]\rightarrow \mathbb{C}$ with 0 not in their images and piecewise continuously-differentiable, I define a curve $\gamma$ defined by $\gamma(t)=\alpha(t)\beta(t)$ for $t\in [a,b]$. Now, since this is product of piecewise continuously-differentiable closed, $\gamma$ is also piecewise continuously-differentiable closed, with derivative $\alpha'\beta+\alpha\beta'$.
Now, $n(\gamma,0)$ is defined since if there exists a $t\in [a,b]$ such that $\gamma(t)=0$, then $0=\alpha(t)\beta(t)$, and it follows that 0 is an element of the image of at least one of $\alpha,\beta$, contrary to assumption.
Next, we compute the following: $$\int_\gamma{\frac{dz}{z}}=\int_a^b{\frac{\gamma'}{\gamma}dt}=\int_a^b{\frac{\alpha'\beta+\alpha\beta'}{\alpha\beta}dt}=\int_a^b{\frac{\alpha'}{\alpha}dt}+\int_a^b{\frac{\beta'}{\beta}dt}$$ from which it follows that $$n(\gamma,0)=n(\alpha,0)+n(\beta,0)$$ by dividing by $2\pi i$.
Now, my hope was to reduce the general case when $w$ is arbitrary to the above case when $w=0$. The 'obvious' approach is to translate everything by $-w$, moving $w$ to the origin and $\alpha(t),\beta(t)$ to $\alpha(t)-w,\beta(t)-w$. However, this clearly does not preserve products, since then $(\alpha-w)(\beta-w)=\alpha\beta-w(\alpha+\beta)+w^2$, which is clearly not $\alpha\beta-w$. Moreover, $(\alpha-w)(\beta-w)$ has derivative equal to $\alpha'\beta+\alpha\beta'-w(\alpha'+\beta')$, and at least it doesn't seem to me that it is true that $$\int_a^b{\frac{\alpha'\beta+\alpha\beta'-w(\alpha'+beta')}{\alpha\beta-w(\alpha+\beta)+w^2}dt}=\int_a^b{\frac{\alpha'\beta +\alpha\beta'}{\alpha\beta-w}dt}$$
No. The fact that we use multiplication makes $0$ a special element of the space. For a counterexample, take $\alpha(t)=2e^{it}$, $\beta(t)=3e^{it}$ and $w=5$. Then $\gamma(t) = 6e^{2it}$, hence $n(\gamma, w)=2$. But $n(\alpha,w)=n(\beta,w)=0$.