How to find Winding number Let $r:[0, 2\pi]\to\Bbb C$ be the circle given by $r(t)=1+2e^{2it}$, what is the winding number of $r(t)$ around $z=2$? 


I struck at :

If $r(t)=1+2e^{it}$ then we can write as $|z-1|=2$, but
How can write $r(t)=1+2e^{2it}$,  in the circle form?
You can still write this as the circle described by $|z-1|=2$, the function just goes around the circle twice, since the exponent over $i$ ranges from $0$ to $4\pi$, which is twice $2\pi$
Just for fun, $\dfrac 1{2\pi i}\oint_r\dfrac {\operatorname dz}{z-2}=\dfrac 1{2\pi i}\int_0^{2\pi}\dfrac 1{1+2e^{2it}-2}4ie^{2it}\operatorname dt=\dfrac 1{2\pi i}\int_0^{2\pi}\dfrac 1{2e^{2it}-1}4ie^{2it}\operatorname dt=\dfrac  2{\pi}[-\dfrac 14\ln (2e^{2it}-1)]_0^{2\pi}=\dfrac{2\pi}{\pi}=2$, where at the end the principal branch of log was used.

This checks out, since $r(t)$ goes around $2$ twice as $t$ goes from $0$ to $2\pi$.  Note $r(t)$ is just the circle $2e^{2it}$ of radius $2$ centered at the origin, translated  $1$ unit to the right.