When I was reading complex analysis by Joseph Bak Newman I came across the word one one $C^1$ mapping. What is it? Is it same as one one mapping? It came in the definition of smooth curves. The text does not explain the notation.
[Math] one-one $C^1$ mapping
complex-analysis
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The theorems are quite simple to understand if you realize where they came from (i.e. why they are natural questions to consider).
Morera's Theorem: The Cauchy-Goursat theorem tells us that if $f$ is holomorphic on a domain $D$, and $\gamma$ is a contractible loop in $D$ then
$$\int_\gamma f=0$$
for all holomorphic $f\in\mathcal{O}(D)$. This is a HUGE theorem, it is of the utmost importance in univariate (and multivariate, although suitably adapted) complex analysis. In fact, most of the big theorems concerning holomorphic functions are just corollaries of this theorem.
Because of this fact, becauseit seems like all of the important properties about holomorphic functions come from the Cauchy-Goursat theorem, it seems obvious to wonder whether we can extend our utilization of this fantastic theorem to a larger class of functions. Said differently, if we really only care about functions which have properties following from the Cauchy-Goursat theorem, what if, instead of holomorphic functions, we studied the set
$$X=\left\{f\in C(D,\mathbb{C}):f\text{ satisfies Gourat's theorem for every }\gamma\right\}$$
(where $C(D,\mathbb{C})$ is the set of continuous functions $D\to\mathbb{C}$--we always want continuity for well-behave functions). Clearly $\mathcal{O}(D)\subseteq X$, but do there exists more functions? Morera's theorem tells us the answer to this question: no.
In fact, Morera's theorem tells us something much stronger. It tells us that a function being analytic is actually equivalent to $f$ satisfying the Cauchy-Goursat theorem for every contractible triangle $\Delta\subseteq D$.
While this is the reason why it is natural to expect the statement of Morera's theorem to come up, it is not practically where it shows up. Practically it shows up in the following workhorse situation. We want to know if $f\in C(D,\mathbb{C})$ is holomorphic, and by Morera's we are reduced to just having to check that a set of integrals, over very nice curves, are zero.
For example, one is able to show using this idea the equivalence of holomorphic (i.e. $C^1$ and satisfies the Cauchy-Riemann equations) functions and complex differentiable (the limit $$\displaystyle \lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}$$ exists). This is a huge deal, since we really want to think about holomorphic functions as just being complex differentiable, but really all of the basic theorems require them to be $C^1$ and satisfy the Cauchy-Riemann equations.
Schwarz's Reflection Principle: This I really have less to say about, hopefully someone else can fill in some of the holes.
The basic idea of why one would want to consider something like the Schwarz's Reflection Principle (SRP) is the following fact: continuation theorems are king. A big part of complex analysis deals with when we can extend a given holomorphic function to larger domains (an unique to univariate complex analysis). I'll leave the reasons for this for you to look up, but they show up all over the place, especially in number theory.
Well, a very obvious extension problem is whether or not we can extend a holomorphic function $f:\mathfrak{h}\to\mathbb{C}$ (where $\mathfrak{h}$ is the upper half-plane) which has a continuous extension $f:\overline{\mathfrak{h}}\to\mathbb{C}$ which takes real values on $\mathbb{R}$, to an entire function. Why is this so obvious? because there is a very obvious way one might attempt this extension. Namely, the upper half-plane (or other domains in the upper half-plane lying on the real line) have a natural map taking them to subsets of the lower half-plane--conjugation. We might try to define our extension of $f$ on the lower half-plane by reflecting $z\in\mathfrak{h}$ to $\bar{z}$ in the lower half-plane by reflecting the image of $z$ under $f$ with conjugation. The content of the SRP is that this naive approach to an extension theorem actually works.
EDIT: The SRP also applies to harmonic functions, which may mean that thinking about it purely from the point of view of complex analysis may be misleading. There may be a more potential theoretic reasons to consider the SRP, and why it should be true.
I hope that this helps!
When $f$ has a pole at $z_0\in{\mathbb C}$ then $\lim_{z\to z_0} f(z)=\infty\in\bar{\mathbb C}$. It follows that $\lim_{n\to\infty }f(\beta_n)=\infty$ for all sequences $(\beta_n)_{n\geq1}$ converging to $z_0$. In particular there is no sequence $\alpha_n\to z_0$ with $\lim_{n\to\infty}f(\alpha_n)=0$.
When $f$ has an essential singularity at $z_0$ then by the Casorati-Weierstrass theorem for each $n\geq1$ there is an $\alpha_n$ with $|\alpha_n-z_0|<{1\over n}$ and $|f(\alpha_n)|<{1\over n}$. It follows that $\lim_{n\to\infty}\alpha_n=z_0$ and $\lim_{n\to\infty} f(\alpha_n)=0$. Similarly, for each $n\geq 1$ there is a $\beta_n$ with $|\beta_n-z_0|<{1\over n}$ and $|f(\beta_n)|>n$. It follows that $\lim_{n\to\infty}\beta_n=z_0$ and $\lim_{n\to\infty} f(\beta_n)=\infty$.
Best Answer
$C^1$ means continuously differentiable. In general $C^k$ means the function has $k$ derivatives and they are all continuous functions. A continuous function is then $C^0$ in this notation.
Usually for defining paths in complex analysis, people take a $C^1$ function $\gamma \colon [0,1] \to {\mathbb{C}}$. That means that $\gamma'$ exists and is a continuous function. This is useful for integration, as when you integrate along $\gamma$, with respect to $dz$, then the $dz$ becomes $\gamma'(t)dt$, so you want a continuous function under the integral. That is, you define $$ \int_\gamma f(z)\, dz = \int_0^1 f(\gamma(t)) \gamma'(t) dt $$ Notice that it is useful to have both $\gamma$ and $\gamma'$ to be continuous, since integrating continuous functions is no problem.