continuous image of simple closed curve

Let $C$ be a simple closed curve, and let $f(z)$ be analytic inside and on $C$, $f(z) \neq 0$ on $C$. Then how can I prove the followings?

If $f(z)$ has no zeros inside $C$, then $f(C)$ does not surround the origin.

If $f(z)$ does have zeros inside $C$, then $f(C)$ must wrap around the origin.

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You may assume that $C$ is the unit circle (why?). If $f$ has no zeroes, you can calculate the winding number for shrinking radii and see what happens – Hagen von Eitzen Sep 16 '12 at 8:46

From the argument principle, we can write $$\frac{1}{2\pi i}\oint_C \frac{f^\prime(z)}{f(z)}\ dz = n - p$$ where $n$ denotes the number of zeroes inside $C$ and $p$ denotes the number of poles inside $C$. Since $f$ is analytic in $C$, it follows that $p=0$. We can then see that $$\frac{1}{2\pi i}\oint_C \frac{f^\prime(z)}{f(z)}\ dz = 0$$ if and only if $n=0$. But we also have $$\frac{1}{2\pi i}\oint_C \frac{f^\prime(z)}{f(z)}\ dz = \frac{1}{2\pi i}\oint_{f(C)}\frac{1}{z}\ dz = \mathrm{Wnd}(f(C),\ 0)$$ so that the integral gives the winding number of the curve $f(C)$ around the origin. Therefore the curve winds about $0$ if and only if there exists zeroes inside $C$.