Why the map $z→z+ \overline z^m $ has fixed point with local Lefschetz number $m$ at the the origin of C (m≥0)? My professor went through an example in class and making following claim


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*The map $z→z+z^m$ has a fixed point with local Letfschetz number $m$ at the origin of $C$ $(m>0)$ 

*For any $c≠0$, the homotopic map $z→z+z^m+c$ is Lefschetz with $m$ fixed point that are all close to zero if $c$ is small

*The map $z→z+ \overline z^m $ has fixed point with local Lefschetz number $-m$ at the the origin of C (m≥0) 
He says it's easy to recognize these claim, so he didn't justify it. But clearly, it's not easy for me to see it at all.
I know that $m$ can't be bigger than  to $1$, so $0 < m \leq 1$. So we have a map $z\to z+ z^{x/y}$ for $0< x \leq y$ and $m=\frac x y$. How can I show that this map has   a fixed point with local Letfschetz number $m$ at the origin of $C$

The local Lefschetz number of a fixed point can be defined in two ways:


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*if $x$ is a Lefschetz fixed point (i.e. if $df_x$ has no nonzero fixed vectors), we just consider the sign of $\det(df_x - I)$

*if $x$ is not a Lefschetz fixed point, but just a fixed point, but is nonetheless isolated, then the local Lefschetz number at $x$ is the degree of the application
$$z\mapsto \dfrac{f(z)-z}{|f(z)-z|}$$
on the boundary of a small ball around $x$ containing no other fixed points.


In the case of the map $z\mapsto z+z^m$, if $m=1$ I would say that it is obvious that $L_0(f)=1$: $f(0)=0$ and $df_0=2I$ so $df_0-I=I$ which simultaneously implies that $0$ is a Lefschetz fixed point and that the local Lefschetz number is $1$.
If $m>1$, $df_0=I$, so we have to use the second definition:
$f(z)-z=z^m$ so we consider $z\mapsto \dfrac{z^m}{|z|^m}$ which is the map $\theta\mapsto m\theta$ on $S^1$, so it is of degree $m$ (no balls contains other fixed points).
The second statement should not present any difficulty: those maps have $m$ fixed points which are all Lefschetz with local Lefschetz index $1$ ($df_{z_i}-I=m|z_i|^{2(m-1)}>0$).
As for the third statement, I would say that the Lefschetz index in $0$ is $-m$ and not $m$: the map $z\mapsto z+ \bar{z}^m$ has an isolated fixed point in $0$, which is not Lefschetz, so using the second definition we get to calculate the degree of the mapping 
$$z\mapsto \frac{\bar{z}^m}{|z|^m}$$
on $S^1$, i.e. of $\theta\mapsto -m\theta$, which has degree $-m$.
NOTE If $m$ is not in $\mathbb{N}$ (non negative integer) then the map $z\mapsto z+z^m$ is not well-defined in a neighbourhood of $0$.
