We will show (1) if $z_1,z_2,z_3,z_4$ lie on a generalized circle, then $[z_1,z_2,z_3,z_4]\in\mathbb R$, and (2) if $[z_1,z_2,z_3,z_4]\in\mathbb R$, then $z_1,z_2,z_3,z_4$ lie on a generalized circle.
- First, suppose $z_1,z_2,z_3,z_4$ lie on a generalized circle. We know that, given three real numbers $x_1,x_2,x_3 \in \mathbb{R}$, there exists a Mobius transformation
\begin{equation*}
F=F_{x_1,x_2,x_3}^{-1} \circ F_{z_1,z_2,z_3}
\end{equation*}
such that $F(z_i)=x_i$, $i=1,2,3$. Also, since Mobius transformations map generalized circles to generalized circles, we know that $z_4$ is also mapped to some $x_4 \in \mathbb{R}$. Therefore, since $x_1,x_2,x_3,x_4 \in \mathbb{R}$, then their cross ratio $[x_1,x_2,x_3,x_4]=\frac{x_1-x_3}{x_2-x_3}\cdot\frac{x_2-x_4}{x_1-x_4} \in \mathbb{R}$. Therefore, since the cross ratio is invariant under Mobius transformation, we have
\begin{align*}
[z_1,z_2,z_3,z_4]
&=[F(z_1),F(z_2),F(z_3),F(z_4)] \\
&=[x_1,x_2,x_3,x_4] \\
&=\frac{x_1-x_3}{x_2-x_3}\cdot\frac{x_2-x_4}{x_1-x_4} \\
&\in \mathbb{R}
\end{align*}
- Next, suppose $[z_1,z_2,z_3,z_4] \in \mathbb{R}$. Let
\begin{equation*}
F=F_{x_1,x_2,x_3}^{-1} \circ F_{z_1,z_2,z_3}
\end{equation*}
be the Mobius transformation which sends $F(z_i)=x_i$, $i=1,2,3$, and let $z_4'=F(z_4)$. Since the cross ratio is invariant under Mobius transformation, we know that
\begin{equation*}
[z_1,z_2,z_3,z_4]=[x_1,x_2,x_3,z_4']
\end{equation*}
Therefore the cross ratio $[x_1,x_2,x_3,z_4']$ is also a real number, which implies that $z_4'$ is a real number as well. Now consider $F^{-1}$, which is also a Mobius transformation. Since $x_1,x_2,x_3,z_4'$ lie on a generalized circle (the real line), and since a Mobius transformation maps generalized circles to generalized circles, then $z_1,z_2,z_3,z_4 \in F^{-1}(\mathbb{R})$ belong to a generalized circle.
I cannot improve your approach, instead I'll use one I saw some time ago. I use it often in these "characterize the entire functions with this property"
questions, there were many good examples in Conway and online of this kind.
Let $f$ be a function as above. Define a new function via "double-conjugation inversion", a technique I've seen a few times:
$$
g(z) = \overline{f(\bar z^{-1})}^{-1}
$$
What double conjugation ensures is that $g$ is meromorphic (doesn't have any essential singularities) on $\mathbb C \backslash \{ 0\}$. We also have that $g(z)=f(z)$ on the circle $|z| = 1$. By analytic continuation, we have $g(z)=f(z)$ for non-zero $z$.
The key idea : furthermore, $g$ "carries" the pole of $f$ to $\infty$, in that it has a pole at infinity, of the same order $k$, as that of $f$ at $0$.
Indeed, this actually implies that $f$ is a polynomial of degree $k$, since an entire function with a pole at infinity is a polynomial (so $g$ is a polynomial).
Since $k$ is also the order of the pole at zero, it also follows that $f$ has all it's zeros situated at $0$, so $f(z) = az^k$. Substitute $z=1$, then $f(1)=a$, so $a$ has modulus $1$. Hence $f(z) = az^k$, where $|a| =1$ additionally.
Best Answer
Cauchy's Residue theorem states:
From Brown and Churchill, Complex Variables and applications.
From what you've written:
I understand that the Contour $C$ encloses $z=z_1$ and that all other points, $z_j$ with $j>1$ do not reside in $C$. Therefore we need to find the residue at $z=z_1$ and only at that point. Because it is the only point in the area enclosed by the Contour for which the function "blows up". For that we can use your method:
Let $g(z)=\frac{1}{f(z)}$ for which we have:
$$\oint_C g(z)\ \mathrm{d}z = 2\pi \textbf{i}\ \mathrm{Res}_{z=z_1}\frac{1}{f(z)}$$
To find the residue, apply the theorem:
In our case $z_0$ is $z_1$, which is isolated by virtue of the contour we have chosen to enclose it. So our isolated point is $z_1$ and the pole is simple, i.e. $m=1$
$$g(z) = \frac{1}{f(z)} = \frac{\phi(z)}{(z-z_1)}$$
From here we see that we can indeed express our function in the form indicated and that $\phi(z)$ is in fact $\frac{1}{(z-z_2)...(z-z_n)}$.
So the Residue is simply:
$$\mathrm{Res}_{z=z_1}g(z) = \phi(z_1) = \frac{1}{(z_1-z_2)...(z_1-z_n)}$$
So:
$$\oint_C f(z)\ \mathrm{d}z = 2\pi \textbf{i}\ \phi(z_1) = \frac{2\pi \textbf{i}}{{(z_1-z_2)...(z_1-z_n)}}$$
Hope this helps