# $\int_{\alpha} \frac{f'(z)}{f(z)}\ dz = 0$ for closed contour when $|f(z)-1|<1$

Suppose $f$ an analytic function in a domain $\Omega$ and that $|f(z)-1|<1$ for all $z\in \Omega$. Show that:

$$\int_{\alpha} \frac{f'(z)}{f(z)}\ dz = 0$$

for every closed contour in $\Omega$.

There's two ways that I know about integration of complex variables. The first is to choose a contour $\alpha(t)$ from $a$ to $b$ and then $\int_{\alpha} f(z) \ dz = \int_a^b f(\alpha(t))\alpha'(t)\ dt$ but I don't think this will help in anything, because I'd get:

$$\int_{\alpha} \frac{f'(\alpha(t))}{f(\alpha(t))}\alpha'(t)\ dt$$

The other way is to take the antiderivative if the integrand is analytic and if the countour is included in the region where the integrand is analytic.

Since there's a supposition that $|f(z)-1|$ is analytic, I guess that I'll have to use the last one, and I guess that if I make the substitution $w = f(z)$, then the integral becomes

$$\int \frac{1}{w}dw = \ln w$$

but I don't see how $|f(z)-1|<1$ comes in the story and if I can make the substitution.

• Try using Rouche's Theorem?
– Moya
Nov 5, 2016 at 2:58

Since $\log(s)$ is holomorphic on $|s-1| < 1$,
$|f(z)-1|< 1$ on $\Omega$ means that $\log( f(z))$ is holomorphic on $\Omega$.
Hence $\frac{f'(z)}{f(z)} = \frac{d}{dz}\log( f(z))$ has an holomorphic anti-derivative on $\Omega$, and for any closed-contour $\alpha \subset \Omega$ : $$\int_{\alpha} \frac{f'(z)}{f(z)}dz = \int_a^b \frac{f'(\alpha(t))}{f(\alpha(t))}\alpha'(t)dt =\log (f(\alpha(t)))|_a^b = 0$$