Jtdylktuy

The problem states that if $f$ is nonzero and analytic on a region $A$, and if $gamma$ is closed and homotopic to a point in $A$, then $displaystyleint_{gamma} frac{f'(z)}{f(z)}dz = 0$

My approach is to use the fact that since $f$ is analytic on $A$ then so too is $f'$ on $A$, and since $f$ is nonzero, $g = frac{f'(z)}{f(z)}$ is continuous and analytic on $A$ and so we can use the antiderivate theorem and deduce that the integral of this function on a closed curve is zero. Is this the right approach? Does the fact that $gamma$ is homotopic to a point in $A$ help in any way?

Consider the annulus with inner radius $1/2$ and outer radius $2$, $D=operatorname{Ann}(1/2, 2)$. Consider $f(z)=z^{-1}$ then clearly $f$ is analytic and nonzero on $D$, but
begin{align}
int_{|z|=1} frac{f'(z)}{f(z)} dz =& -int_{|z|=1}frac{1}{z} dz \
=& -int^{2pi}_0 e^{-itheta} ie^{itheta} dtheta = -2pi i.
end{align}

Suppose $gamma$ is homotopic to a point $p in D$, i.e.
begin{align}
H(t, z) = (1-t)gamma(z)+tp
end{align}
is continuous on $[0, 1]times D$, then we see that
begin{align}
int_gamma frac{f'(z)}{f(z)} dz = int_{H(t)} frac{f'(z)}{f(z)} dz
end{align}
for all $t in [0, 1]$. In particular, we see that
begin{align}
left|int_{H(t)} frac{f'(z)}{f(z)} dzright| leq& int_{H(t)} frac{|f'(z)|}{|f(z)|} |dz| \
leq& left(left|frac{f'(p)}{f(p)}right|+varepsilonright)operatorname{Length}left(H(t) right)\
leq& C(1-t)left(left|frac{f'(p)}{f(p)}right|+varepsilonright)
end{align}
for $t$ sufficiently close to $1$.