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The winding number of $\gamma$ about $0$ is given as: $w(\gamma, 0)=\frac{arg(\gamma(1))-arg(\gamma(0))}{2\pi}$

$\gamma : [0,1] \rightarrow \mathbb{C}-\{0\}$ is a loop in $\mathbb{C}$ not passing through the origin (so $\gamma(0)=\gamma(1)$)

As $\gamma$ is a loop $arg(\gamma(1))-arg(\gamma(0))$ is an integer multiple of $2\pi$

Question 1: Compute the winding number of the loop $\alpha(z)=4z^4+2z^2+1: S^1 \rightarrow \mathbb{C} -$ {$0$} about the origin $0$

Question 2: Compute the winding number of the polynomial loop $\alpha(z)=6z^2+7z+2: S^1 \rightarrow \mathbb{C} -$ {$0$} about the origin $0$

The only way I can think of doing these is to prove the loops are homotopic to a simple loop, such as $4z^4$ and $6z^2$ respectively . Is this the right idea?

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I will mainly give something that is sometimes underrepresented in lectures on complex analysis: intuition.

In itself, "phase unfolding", the name given when dealing with angles in a range larger than $2 \pi$, is rather puzzling.

It is why I give a graphical hint (see figure below) under the form of a "Western" analogy for the winding number (see figure) you are attached to the torture pole in $0$: how many circles do the indians around you (Hollywood image of course...)? The answer here is 4 times, which is the winding number. The unfolding of the curve is represented as a "lifting" (classical term) of the 2D curve into a 3D curve with ordinates the angle (range $[0,2\pi]$) as can be understood on the program I have given in the Appendix below (which, as well, computes the winding number).

The definition of the winding number you may have already met is:

$$\dfrac{1}{2i\pi}\int_{C}\dfrac{f'(z)}{f(z)}dz=\dfrac{1}{2i\pi}\int_{f(C)}\dfrac{1}{Z}dZ \ \ \ (1)$$

(where (C) is the unit circle, but could be any simple closed loop around the origin). See mathexchange question (definition of winding number, have doubt in definition.) for the details of computation.

Once again, what intuitive grasp on the fact that this integral (1) is connected to the winding number?

A reason is that a primitive of the function $f'(z)/f(z)$ is

$$log(f(z))=log(Z)= log(R e^{i \Theta})=log R +i \Theta$$

(I should say: having agreed about a starting reference for $\Theta$, because for example $log R +i(\Theta+2\pi)$ would be another candidate).

And $log(f(z))$ is there for counting $+2 \pi$ each time there is a crossing of its "cut" along the negative real axis, i.e., counting the number of turns around the origin; the argument of $log(f(z))$ proceeds by "$2 \pi$ jumps" each time the current point crosses the negative part of the $x$-axis.

enter image description here

Appendix: here is the Matlab program that has produced the figure:

clear all;close all;hold on;
t=0:0.01:2*pi;
z=exp(i*t);
Z=4*z.^4+2*z.^2+1;
view([-6,28]);
plot(real(Z),imag(Z),'color','b')
plot3(real(Z),imag(Z),t,'color','r')
plot3([0,0],[0,0],[0,2*pi],'color','k');

logdev=@(z)((16*z.^3+4*z)./(4*z.^4+2*z.^2+1));
windnbr=(1/(2*i*pi))*integral(logdev,1,1,'waypoints',[i,-1,-i])
%integration on the square [1,i,-1,-i,1] instead of the unit circle
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