I need to compute the winding number of $\alpha$ with respect to the point p=(1/2,0)
Where $\alpha: [0,2\pi] \to \mathbb R^2$ $\alpha(t)=((2 Cos[t] - 1)*Cos[t], (2 Cos[t] - 1)*Sin[t])$
The winding number with respecct to p is:
$i_p(\alpha)=1/2\pi i\int_{0}^{2\pi}\alpha'/ (\alpha-p) $
How can I compute this integral using complex analysis and not doing any calculation? I think I should use Cauchy's integral formula but I'm no sure...
In case is needed: $\alpha'(t)= (-2 Cos[t] Sin[t] - (-1 + 2 Cos[t]) Sin[t], Cos[t] (-1 + 2 Cos[t]) - 2 Sin[t]^2) $
I know the answer is 2.
Thank you for your help and time.
The trick is to rewrite the integral defining the winding number as a "complex integral" which then can be computed "by inspection".
Let $e^{it}=:z$. Then $\cos t={1\over2}\bigl(z+{1\over z}\bigr)$. Therefore $\alpha$ can be written as $$\alpha(t)=(2\cos t-1)e^{it}=z^2-z+1,\qquad z:=e^{it}\ .$$ Denoting differentiation with respect to $t$ with a dot we therefore have $$\dot\alpha(t)\>dt=\alpha'(z)\> iz\> dt=(2z-1)\>dz\ .$$ It follows that $${1\over 2\pi i}\int_0^{2\pi}{\dot\alpha(t)\over \alpha(t)-{1\over2}}\ dt={1\over 2\pi i}\int_{\partial D}{2z-1\over z^2-z+{1\over2}}\ dz={1\over 2\pi i}\int_{\partial D}\left({1\over z-{1+i\over2}}+{1\over z-{1-i\over2}}\right)\ dz\ .$$ Since the two points ${1\pm i\over2}$ are within $D$ each of them produces a residue of $1$. Therefore the winding number we are after is $2$.