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Integer-valuedness of the winding number:

I assume that the function $f$ has an isolated root at the origin and I integrate along a circle centered at the origin, $x+iy=\varepsilon\cos 2\pi t+i\varepsilon\sin 2\pi t$. The radius $\varepsilon$ is small enough that no other root is inside the circle, and moreover $f(t)$ does not vanish on the circle.

I decompose $f(t)$ on the circle into modulus and phase, $$f(t)=|f(t)|e^{i\phi_0}e^{i\phi(t)},\;\;|f(t)|>0,\;\;f(0)=f(1).$$ I define $z(t)\equiv-i\ln e^{i\phi(t)}\in(-\pi,\pi]$. The logarithm has a branch cut on the imaginary real axis. The $t$-independent offset $\phi_0$ is chosen such that $z(0)=0=z(1)$, in order to avoid the branch cut at the initial and final times.

We wish to show that the integral $$I=-i\int_0^1 \frac{d\ln f(t)}{dt}\,dt=-i\int_0^1 \frac{d\ln |f(t)|}{dt}\,dt+\int_0^1 \frac{dz(t)}{dt}\,dt$$ is an integer multiple of $2\pi$.

The modulus $|f(t)|$ does not cross the branch cut of the logarithm, hence $$\int_0^1 \frac{d\ln |f(t)|}{dt}\,dt=\ln|f(1)|-\ln|f(0)|=0.$$ The function $z(t)$ is a piecewise continuous function, consisting of $N$ continuous segments $z_1(t),z_2(t),\ldots z_N(t)$, connected by $\pm 2\pi$ jumps at each $t=t_1,t_2,\ldots t_{N-1}$ where the branch cut of the logarithm is crossed: $$z_p(t_{p})-z_{p+1}(t_p)=\sigma_p 2\pi,\;\;p=1,\ldots N-1,\;\;\sigma_p\in\{+1,-1\}.$$ We define $t_0=0$, $t_N=1$, hence $z(t_0)=0=z(t_N)$. The integral then evaluates to $$\int_0^1 \frac{dz(t)}{dt}\,dt=\sum_{p=1}^N \int_{t_{p-1}}^{t_p}\frac{dz_p(t)}{dt}\,dt=\sum_{p=1}^N \bigl(z_p(t_p)-z_p(t_{p-1})\bigr)=\sum_{p=1}^{N-1}\sigma_p 2\pi.$$ The net sum of the $\pm 2\pi$ jumps is the winding number $W=\sum_{p=1}^{N-1}\sigma_p$, which indeed is an integer.

Comment: I did not use analyticity of $f$ in the complex plane, the winding number $W$ is an integer for any continuous function which does not vanish on the integration contour. The connection with the multiplicity of the root inside the contour requires analyticity, so that near a root $f\propto(x+iy)^W$.

Example:

In these plots I compare $z(t)$ for the two functions mentioned by the OP, $f_1=x+iy$ (left plot) and $f_2=x+iy^2$ (right plot). So you see that the first function has one $2\pi$ jump, winding number 1, while the second function has no jump at all, winding number 0.

Integer-valuedness of the winding number:

I assume that the function $f$ has an isolated root at the origin and I integrate along a circle centered at the origin, $x+iy=\varepsilon\cos 2\pi t+i\varepsilon\sin 2\pi t$. The radius $\varepsilon$ is small enough that no other root is inside the circle, and moreover $f(t)$ does not vanish on the circle.

I decompose $f(t)$ on the circle into modulus and phase, $$f(t)=|f(t)|e^{i\phi_0}e^{i\phi(t)},\;\;|f(t)|>0,\;\;f(0)=f(1).$$ I define $z(t)\equiv-i\ln e^{i\phi(t)}\in(-\pi,\pi]$. The logarithm has a branch cut on the imaginary real axis. The $t$-independent offset $\phi_0$ is chosen such that $z(0)=0=z(1)$, in order to avoid the branch cut at the initial and final times.

We wish to show that the integral $$I=-i\int_0^1 \frac{d\ln f(t)}{dt}\,dt=-i\int_0^1 \frac{d\ln |f(t)|}{dt}\,dt+\int_0^1 \frac{dz(t)}{dt}\,dt$$ is an integer multiple of $2\pi$.

The modulus $|f(t)|$ does not cross the branch cut of the logarithm, hence $$\int_0^1 \frac{d\ln |f(t)|}{dt}\,dt=\ln|f(1)|-\ln|f(0)|=0.$$ The function $z(t)$ is a piecewise continuous function, consisting of $N$ continuous segments $z_1(t),z_2(t),\ldots z_N(t)$, connected by $\pm 2\pi$ jumps at each $t=t_1,t_2,\ldots t_{N-1}$ where the branch cut of the logarithm is crossed: $$z_p(t_{p})-z_{p+1}(t_p)=\sigma_p 2\pi,\;\;p=1,\ldots N-1,\;\;\sigma_p\in\{+1,-1\}.$$ We define $t_0=0$, $t_N=1$, hence $z(t_0)=0=z(t_N)$. The integral then evaluates to $$\int_0^1 \frac{dz(t)}{dt}\,dt=\sum_{p=1}^N \int_{t_{p-1}}^{t_p}\frac{dz_p(t)}{dt}\,dt=\sum_{p=1}^N \bigl(z_p(t_p)-z_p(t_{p-1})\bigr)=\sum_{p=1}^{N-1}\sigma_p 2\pi.$$ The net sum of the $\pm 2\pi$ jumps is the winding number $W=\sum_{p=1}^{N-1}\sigma_p$, which indeed is an integer.

Comment: I did not use analyticity of $f$ in the complex plane, the winding number $W$ is an integer for any continuous function which does not vanish on the integration contour. The connection with the multiplicity of the root inside the contour requires analyticity, so that near a root $f\propto(x+iy)^W$.

Example: In these plots I compare $z(t)$ for the two functions mentioned by the OP, $f_1=x+iy$ (left plot) and $f_2=x+iy^2$ (right plot). Both have an isolated root at the origin, but only $f_1$ is analytic. The first function has one $2\pi$ jump in $z(t)$, winding number 1, while the second function has no jump at all, winding number 0.

Integer-valuedness of the winding number:

I assume that the function $f$ has an isolated root at the origin and I integrate along a circle centered at the origin, $x+iy=\varepsilon\cos 2\pi t+i\varepsilon\sin 2\pi t$. The radius $\varepsilon$ is small enough that no other root is inside the circle, and moreover $f(t)$ does not vanish on the circle.

I decompose $f(t)$ on the circle into modulus and phase, $$f(t)=|f(t)|e^{i\phi_0}e^{i\phi(t)},\;\;|f(t)|>0,\;\;f(0)=f(1).$$ I define $z(t)\equiv-i\ln e^{i\phi(t)}\in(-\pi,\pi]$. The logarithm has a branch cut on the imaginary real axis. The $t$-independent offset $\phi_0$ is chosen such that $z(0)=0=z(1)$, in order to avoid the branch cut at the initial and final times.

We wish to show that the integral $$I=-i\int_0^1 \frac{d\ln f(t)}{dt}\,dt=-i\int_0^1 \frac{d\ln |f(t)|}{dt}\,dt+\int_0^1 \frac{dz(t)}{dt}\,dt$$ is an integer multiple of $2\pi$.

The modulus $|f(t)|$ does not cross the branch cut of the logarithm, hence $$\int_0^1 \frac{d\ln |f(t)|}{dt}\,dt=\ln|f(1)|-\ln|f(0)|=0.$$ The function $z(t)$ is a piecewise continuous function, consisting of $N$ continuous segments $z_1(t),z_2(t),\ldots z_N(t)$, connected by $\pm 2\pi$ jumps at each $t=t_1,t_2,\ldots t_{N-1}$ where the branch cut of the logarithm is crossed: $$z_p(t_{p})-z_{p+1}(t_p)=\sigma_p 2\pi,\;\;p=1,\ldots N-1,\;\;\sigma_p\in\{+1,-1\}.$$ We define $t_0=0$, $t_N=1$, hence $z(t_0)=0=z(t_N)$. The integral then evaluates to $$\int_0^1 \frac{dz(t)}{dt}\,dt=\sum_{p=1}^N \int_{t_{p-1}}^{t_p}\frac{dz_p(t)}{dt}\,dt=\sum_{p=1}^N \bigl(z_p(t_p)-z_p(t_{p-1})\bigr)=\sum_{p=1}^{N-1}\sigma_p 2\pi.$$ The net sum of the $\pm 2\pi$ jumps is the winding number $W=\sum_{p=1}^{N-1}\sigma_p 2\pi$, which indeed is an integer multiple of $2\pi$.

Example:

In these plots I compare $z(t)$ for the two functions mentioned by the OP, $f_1=x+iy$ (left plot) and $f_2=x+iy^2$ (right plot). So you see that the first function has one $2\pi$ jump, winding number 1, while the second function has no jump at all, winding number 0.

Integer-valuedness of the winding number:

I assume that the function $f$ has an isolated root at the origin and I integrate along a circle centered at the origin, $x+iy=\varepsilon\cos 2\pi t+i\varepsilon\sin 2\pi t$. The radius $\varepsilon$ is small enough that no other root is inside the circle, and moreover $f(t)$ does not vanish on the circle.

I decompose $f(t)$ on the circle into modulus and phase, $$f(t)=|f(t)|e^{i\phi_0}e^{i\phi(t)},\;\;|f(t)|>0,\;\;f(0)=f(1).$$ I define $z(t)\equiv-i\ln e^{i\phi(t)}\in(-\pi,\pi]$. The logarithm has a branch cut on the imaginary real axis. The $t$-independent offset $\phi_0$ is chosen such that $z(0)=0=z(1)$, in order to avoid the branch cut at the initial and final times.

We wish to show that the integral $$I=-i\int_0^1 \frac{d\ln f(t)}{dt}\,dt=-i\int_0^1 \frac{d\ln |f(t)|}{dt}\,dt+\int_0^1 \frac{dz(t)}{dt}\,dt$$ is an integer multiple of $2\pi$.

The modulus $|f(t)|$ does not cross the branch cut of the logarithm, hence $$\int_0^1 \frac{d\ln |f(t)|}{dt}\,dt=\ln|f(1)|-\ln|f(0)|=0.$$ The function $z(t)$ is a piecewise continuous function, consisting of $N$ continuous segments $z_1(t),z_2(t),\ldots z_N(t)$, connected by $\pm 2\pi$ jumps at each $t=t_1,t_2,\ldots t_{N-1}$ where the branch cut of the logarithm is crossed: $$z_p(t_{p})-z_{p+1}(t_p)=\sigma_p 2\pi,\;\;p=1,\ldots N-1,\;\;\sigma_p\in\{+1,-1\}.$$ We define $t_0=0$, $t_N=1$, hence $z(t_0)=0=z(t_N)$. The integral then evaluates to $$\int_0^1 \frac{dz(t)}{dt}\,dt=\sum_{p=1}^N \int_{t_{p-1}}^{t_p}\frac{dz_p(t)}{dt}\,dt=\sum_{p=1}^N \bigl(z_p(t_p)-z_p(t_{p-1})\bigr)=\sum_{p=1}^{N-1}\sigma_p 2\pi.$$ The net sum of the $\pm 2\pi$ jumps is the winding number $W=\sum_{p=1}^{N-1}\sigma_p$, which indeed is an integer.

Comment: I did not use analyticity of $f$ in the complex plane, the winding number $W$ is an integer for any continuous function which does not vanish on the integration contour. The connection with the multiplicity of the root inside the contour requires analyticity, so that near a root $f\propto(x+iy)^W$.

Example:

In these plots I compare $z(t)$ for the two functions mentioned by the OP, $f_1=x+iy$ (left plot) and $f_2=x+iy^2$ (right plot). So you see that the first function has one $2\pi$ jump, winding number 1, while the second function has no jump at all, winding number 0.

Let me work this out, since the comments do not seem to have resolved the issue.

The function $f$ has a root at $x=0=y$ so I integrate along a circle centered at the origin, $x+iy=\varepsilon\cos 2\pi t+i\varepsilon\sin 2\pi t$. I decompose into modulus and phase, $$f(x,y)=|f(t)|e^{i\phi(t)},$$ with $|f(0)|=|f(1)|$ and $\phi(0)=\phi(1)$. The logarithm has a branch cut along the negative real axis, which is not crossed by $|f(t)$, hence $$\int_0^1 \frac{d\ln |f(t)|}{dt}\,dt=\ln|f(1)|-\ln|f(0)|=0.$$ The function $z(t)=-i\ln e^{i\phi(t)}\in(-\pi,\pi]$ is a piecewise continuous function, consisting of $N$ continuous segments $z_1(t),z_2(t),\ldots z_N(t)$, connected by $\pm 2\pi$ jumps at each $t=t_1,t_2,\ldots t_{N-1}$ where the branch cut of the logarithm is crossed: $$z_p(t_{p})-z_{p+1}(t_p)=\sigma_p 2\pi,\;\;p=1,\ldots N-1,\;\;\sigma_p\in\{+1,-1\}.$$ We define $t_0=0$, $t_N=1$. The integral then evaluates to $$\int_0^1 \frac{dz(t)}{dt}\,dt=\sum_{p=1}^N \int_{t_{p-1}}^{t_p}\frac{dz_p(t)}{dt}\,dt=\sum_{p=1}^N \bigl(z_p(t_p)-z_p(t_{p-1})\bigr)$$ $$\qquad=z_N(t_N)-z_1(t_0)+\sum_{p=1}^{N-1}\sigma_p 2\pi=\sum_{p=1}^{N-1}\sigma_p 2\pi,$$ because $z_N(t_N)-z_1(t_0)=z(1)-z(0)=0$. The net sum of the $\pm 2\pi$ jumps is the winding number $W=\sum_{p=1}^{N-1}\sigma_p 2\pi$.

In these plots I compare $z(t)$ for the two functions mentioned by the OP, $f_1=x+iy$ (left plot) and $f_2=x+iy^2$ (right plot). So you see that the first function has one $2\pi$ jump, winding number 1, while the second function has no jump at all, winding number 0.

Integer-valuedness of the winding number:

I assume that the function $f$ has an isolated root at the origin and I integrate along a circle centered at the origin, $x+iy=\varepsilon\cos 2\pi t+i\varepsilon\sin 2\pi t$. The radius $\varepsilon$ is small enough that no other root is inside the circle, and moreover $f(t)$ does not vanish on the circle.

I decompose $f(t)$ on the circle into modulus and phase, $$f(t)=|f(t)|e^{i\phi_0}e^{i\phi(t)},\;\;|f(t)|>0,\;\;f(0)=f(1).$$ I define $z(t)\equiv-i\ln e^{i\phi(t)}\in(-\pi,\pi]$. The logarithm has a branch cut on the imaginary real axis. The $t$-independent offset $\phi_0$ is chosen such that $z(0)=0=z(1)$, in order to avoid the branch cut at the initial and final times.

We wish to show that the integral $$I=-i\int_0^1 \frac{d\ln f(t)}{dt}\,dt=-i\int_0^1 \frac{d\ln |f(t)|}{dt}\,dt+\int_0^1 \frac{dz(t)}{dt}\,dt$$ is an integer multiple of $2\pi$.

The modulus $|f(t)|$ does not cross the branch cut of the logarithm, hence $$\int_0^1 \frac{d\ln |f(t)|}{dt}\,dt=\ln|f(1)|-\ln|f(0)|=0.$$ The function $z(t)$ is a piecewise continuous function, consisting of $N$ continuous segments $z_1(t),z_2(t),\ldots z_N(t)$, connected by $\pm 2\pi$ jumps at each $t=t_1,t_2,\ldots t_{N-1}$ where the branch cut of the logarithm is crossed: $$z_p(t_{p})-z_{p+1}(t_p)=\sigma_p 2\pi,\;\;p=1,\ldots N-1,\;\;\sigma_p\in\{+1,-1\}.$$ We define $t_0=0$, $t_N=1$, hence $z(t_0)=0=z(t_N)$. The integral then evaluates to $$\int_0^1 \frac{dz(t)}{dt}\,dt=\sum_{p=1}^N \int_{t_{p-1}}^{t_p}\frac{dz_p(t)}{dt}\,dt=\sum_{p=1}^N \bigl(z_p(t_p)-z_p(t_{p-1})\bigr)=\sum_{p=1}^{N-1}\sigma_p 2\pi.$$ The net sum of the $\pm 2\pi$ jumps is the winding number $W=\sum_{p=1}^{N-1}\sigma_p 2\pi$, which indeed is an integer multiple of $2\pi$.

Example:

In these plots I compare $z(t)$ for the two functions mentioned by the OP, $f_1=x+iy$ (left plot) and $f_2=x+iy^2$ (right plot). So you see that the first function has one $2\pi$ jump, winding number 1, while the second function has no jump at all, winding number 0.