# Argument Principle, Complex Analysis

• May 25th 2010, 12:42 PM
cribby
Argument Principle, Complex Analysis
I'm given that $\frac{1}{2 \pi i} \int_{\partial D(0,3)} \frac{f'(z)}{f(z)}\,dz = 2$ and also that $\frac{1}{2 \pi i} \int_{\partial D(0,3)} z \frac{f'(z)}{f(z)}\,dz = 2$. I know that $f$ is analytic on the curve and on the interior region as well, and that $f(z)\neq 0, \forall z \in \partial D(0,3)$.

The actual question of the problem aside, what does the second equation really tell me? I know that the first says that the winding number of $f(D(0,3))$ about the origin is 2, and that because $f$ is holomorphic in the disc there are no poles and so $f$ must have two zeroes in $D(0,3)$.

Eventually I'm trying to locate those zeroes, and there's a third equation as well that I imagine is supposed to help me along, but what more information can I glean from that second equation?

Thanks.
• May 25th 2010, 02:30 PM
Laurent
Quote:

Originally Posted by cribby
I'm given that $\frac{1}{2 \pi i} \int_{\partial D(0,3)} \frac{f'(z)}{f(z)}\,dz = 2$ and also that $\frac{1}{2 \pi i} \int_{\partial D(0,3)} z \frac{f'(z)}{f(z)}\,dz = 2$. I know that $f$ is analytic on the curve and on the interior region as well, and that $f(z)\neq 0, \forall z \in \partial D(0,3)$.

The actual question of the problem aside, what does the second equation really tell me?

If $z_0$ is a zero of multiplicity $k$, then $f(z)\sim C(z-z_0)^k$ and $f'(z)\sim Ck(z-z_0)^{k-1}$ as $z\to z_0$, hence $\frac{f'(z)}{f(z)}\sim\frac{k}{z-z_0}$ and $\frac{z f'(z)}{f(z)}\sim\frac{kz_0}{z-z_0}$. Thus, the residue of $z\mapsto\frac{f'(z)}{f(z)}$ at $z_0$ is the multiplicity of $z_0$ (as a zero), i.e. the "number of zeroes at $z_0$", while the residue of $z\mapsto\frac{zf'(z)}{f(z)}$ is $kz_0$, i.e. the "sum of zeroes at $z_0$" (counting multiplicity).

Thus, by the theorem of residues, the first integral is the number of zeroes inside the curve (counting multiplicity) and the second integral is the sum of these zeroes (counting multiplicity). With $z^2$ you would get the sum of their squares, and so on.