这是一份leeds利兹大学MATH5356M01作业代写的成功案例

Proof. Consider an arbitrary power series of the form
$$
\lambda+a_{n} w^{n}+a_{n+1} w^{n+1}+\ldots
$$
Let $n$ be even, and let $a_{n}>0$. Then this series is formally conjugate to the polynomial $\lambda+z^{n}$. The formula of the corresponding change is as follows:
$$
z=w\left(a_{n}+a_{n+1} w+\ldots\right)^{1 / n}
$$
Here, of course, we mean the formal Taylor expansion of the radical into power series in $w$. If $a_{n}<0$, then we set $z=-w\left(-a_{n}-a_{n+1} w-\ldots\right)^{1 / n}$.
If $n$ is odd, then the formula is analogous.

MATH5356M01 COURSE NOTES ：

where $\min$ and $\max$ are taken over the circle ${F=c}$ centered at $P_{0}$. Taking the limit as $c \rightarrow 0$, we obtain the equality
$$
\Pi(0)=2 \pi \omega(0,0) .
$$
On the other hand, the linearization of $w=\frac{\xi}{\omega(x, y)}$ at the equilibrium point $P_{0}$ has the form $\frac{\xi}{\omega(0,0)}=\frac{(-y, x)}{\omega(0,0)}$. Therefore, the eigenvalues of the linearized mapping $\sigma$ equal
$$
\nu=\exp \left(\frac{\pm i}{\omega(0,0)}\right)
$$
Comparing this expression with $\Pi(0)$, we get the desired for mula:
$$
\nu=\exp \left(\frac{\pm 2 \pi i}{\Pi(0)}\right) .
$$