Winding number of closed curves

Question

Let $c_1,c_2$ be closed curves in $\mathbb C^{\times}$ and we define $c(t):=\frac{c_1(t)}{c_2(t)}$. Proof the following for the winding number $win(c,0)=win(c_1,0)-win(c_2,0)$.

I have no idea to solve this problem. I need some hints.

Answer 1

Answer (following Martin R. hint):

$$win(c,0)=\frac{1}{2\pi i}\int_c\frac{dz}{z}=\frac{1}{2\pi i}\int_0^1\frac{c'(t)}{c(t)}dt$$ Since $c(t)=\frac{c_1(t)}{c_2(t)}$ the derivative is $c'(t)=\frac{c_1'(t)}{c_2(t)}-\frac{c_1(t)\cdot c_2'(t)}{c_2^2(t)}$. We get then:$$\frac{1}{2\pi i}\int_0^1\frac{c'(t)}{c(t)}dt=\frac{1}{2\pi i}\left(\int_0^1\frac{c_1'(t)}{c_2(t)\cdot c(t)}-\frac{c_1(t)\cdot c_2'(t)}{c_2^2(t)\cdot c(t)}dt\right)$$$$=\frac{1}{2\pi i}\left(\int_0^1\frac{c_1'(t)\cdot c_2(t)}{c_2(t)\cdot c_1(t)}-\frac{c_1(t)\cdot c_2'(t)\cdot c_2(t)}{c_2^2(t)\cdot c_1(t)}dt\right)$$$$=\frac{1}{2\pi i}\left(\int_0^1\frac{c_1'(t)}{c_1(t)}-\frac{c_2'(t)}{c_2(t)}dt\right)=\frac{1}{2\pi i}\int_0^1 \frac{c_1'(t)}{c_1(t)}dt-\frac{1}{2\pi i}\int_0^1 \frac{c_2'(t)}{c_2(t)}dt$$$$=\frac{1}{2\pi i}\int_{c_1} \frac{dz}{z}-\frac{1}{2\pi i}\int_{c_2} \frac{dz}{z}$$$$=win(c_1,0)-win(c_2,0)$$

Answer 2

You have $$ win(c, 0) = \frac{1}{2 \pi i} \int_c \frac{dz}{z} = \frac{1}{2 \pi i} \int_0^1 \frac{c'(t)}{c(t)} \, dt $$ and analogously for $c_1$ and $c_2$.

From $c(t) = \frac{c_1(t)}{c_2(t)}$ one can easily compute a relationship between the "logarithmic derivatives" $$ \frac{c'(t)}{c(t)} \, , \, \frac{c_1'(t)}{c_1(t)} \, , \, \frac{c_2'(t)}{c_2(t)} $$ which leads to the intended formula.