Q:

Explanation of the argument principle.

https://www.youtube.com/watch?v=RRDmCC8gKpY
At 22:35 he's trying to explain how $ \arg h(z) $ changes.
I am not understanding this though why is $ f(z_0 + Re^{it})$ it not a circle? 
And also why is the change $4\pi$? As Z traversed through the curve, isn't 
$z_0$ fixed and all that is changing is $ t$ where $0 < t < 2\pi
$.
Please help me understand this principle I have been looking at this for past 2 hours 
without any luck.

A:

If you write $z = z_0 + R e^{it}$ then indeed the points $z$ fall on a circle
and you traverse the circle exactly once as $t$ increases from $0$ to $2\pi.$
But you cannot generally say that $f(z) = z,$ and there is no reason to think
that the points $f(z) = f(z_0 + R e^{it})$ will lie on a circle nor that if they do,
that $f(z)$ will traverse the circle just once as $t$ increases from $0$ to $2\pi.$
For example, let $f(z) = z^2,$ let $z_0 = 2,$ let $R = 3,$ and then plot
$f(z) = f(z_0 + R e^{it})$ for
$0 \leq t \leq 2\pi.$ The plot is not a circle, and it wraps around the origin twice
(argument increases by $4\pi$)
as $t$ increases from $0$ to $2\pi.$

