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I want to compute $$ I = \int_C z \frac{f'(z)}{f(z)} dz $$, where $ C = \{z : z = e^{i\theta}, 0 \leq \theta \leq 2\pi \}$ and $f$ is analytic, with a simple unique root at $z_0$ inside the circle $C$.

So I say $f$ can be written in the form of $f(z) = (z-z_0) g(z)$, where $g(z)$ is analytic and $g(z) \neq 0$ for all $z \in \mathbb{C}$. Thus $f'(z) = (z- z_0)g'(z) + g(z)$ so

$$ \begin{align} I &= \int_C z\frac{(z- z_0)g'(z) + g(z)}{(z-z_0)g(z)} dz \\ &= \int_C z \frac{g'(z)}{g(z)}dz + \int_C \frac{z}{z-z_0}dz \\ &= I_1 + I_2 \end{align} $$

Now $z \frac{g'(z)}{g(z)}$ is analytic everywhere so $I_1 = 0$ and $I_2 = 2\pi i z_0$ so

$$ I = 2\pi i z_0. $$

Is my thought correct or am I missing something? Any comment would be appreciated.

Da Mike
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1 Answers1

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Your calculation is correct, it is a special case of this generalization of the argument principle.

In this simple case you can also use the residue theorem directly: $$ I = 2 \pi i \operatorname{Res}(z \frac{f'(z)}{f(z)}, z_0) $$ For the simple pole at $z_0$ $$ \operatorname{Res}(z \frac{f'(z)}{f(z)}, z_0) = \lim_{z \to z_0}z \frac{(z-z_0)f'(z)}{f(z)} = z_0 $$ because $$ \frac{(z-z_0)f'(z)}{f(z)} = \frac{f'(z)}{\frac{f(z)-f(z_0)}{z-z_0}} \to 1 $$ for $z \to z_0$.

Martin R
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