# Is following good way to define analytic function? What other better ways there are to define Analyticity of complex function?

Is following good way to define analytic function? Complex function is analytic on a region $\Omega$ if it is complex differentiable at every point in $\mathbb{\Omega}$(Look at this)

What other ways there are to define analyticity of complex function?

Finally is for example $f(z)=z^2$ analytic on the region $D$, where $D=\{z \in \mathbb{C}| 1 \leq |z|< 2\}$? What about if $f(z)=\frac{z}{\bar{z}}$?

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Intuitively a function is analytic iff it is either expressible as a polynomial or as an "infinite polynomial". –  Shahab Sep 13 '12 at 17:58

The prime way to define analyticity of a function $f:\ \Omega\to{\mathbb C}$ is the one you have given: $\Omega$ is a region (i.e., a connected open set) in ${\mathbb C}$, and for each $z_0\in{\Omega}$ the limit $$\lim_{z\to z_0}{f(z)-f(z_0)\over z-z_0}=:f'(z_0)\ \in{\mathbb C}$$ exists.

There are several theorems describing or guaranteeing analyticity of functions $f$ expressed in particular ways, among them the following:

1. When the vector-valued function $$(x,y)\mapsto {\bf f}(x,y)=\bigl(u(x,y),v(x,y)\bigr)$$ is continuously differentiable on $\Omega$, and $u_x=v_y$, $u_y=-v_x$ for all $(x,y)\in\Omega$, then the function $$f:\quad \Omega\to{\mathbb C}\ ,\qquad z=x+iy\ \mapsto w:= u(x,y)+iv(x,y)$$ is analytic on $\Omega$.

2. When $\gamma\subset{\mathbb C}$ is a smooth arc and $\phi:\ \gamma\to {\mathbb C}$ is an arbitrary continuous function then the function $$f(z):=\int_\gamma{\phi(\zeta)\over z-\zeta}\ d\zeta$$ is analytic on $\Omega:={\mathbb C}\setminus\gamma$.

3. When an arbitrary complex sequence $(a_n)_{n\geq0}$ is given, such that $$\rho:={1\over\limsup_{n\to\infty}\root n\of{|a_n|}}>0$$ then the function $$f(z):=\sum_{k=0}^\infty a_k\ z^k$$ is analytic for $|z|<\rho$.

4. If $f(z)$ is an "analytic expression" using symbols like $z$ (i.e., ${\rm id}_{\mathbb C}$), $+$,$-$, $*$, $/$, $\circ$, principal value of $\sqrt{\cdot }$ or $\log$, $\exp$, $\cos$, ${\rm artanh}$, etc., then $f(z)$ is an analytic function wherever bona fide defined.

The function $f(z):=z^2$ is a polynomial in $z$, so it is analytic in all of ${\mathbb C}$. You also can prove this directly: $${f(z)-f(z_0)\over z-z_0}={z^2-z_0^2\over z-z_0}=z+z_0\to 2z_0\qquad(z\to z_0)\ ,$$ and as this holds for any fixed $z_0\in{\mathbb C}$ we deduce that the derivative $f':\ {\mathbb C}\to{\mathbb C}$ is given by $f'(z)=2z$.

When the function $f(z):=z/\bar z$ where analytic in the given annulus then by the principle 4. above the function $g(z):=z / f(z)=\bar z$ were analytic there also. But you know very well that this is not the case; or you may check the CR equations for $g$.

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Sorry, but is $f(z)=z^2$ analytic on the region $D$, where $D=\{z \in \mathbb{C}| 1 \leq |z|< 2\}$? What about if $f(z)=\frac{z}{\bar{z}}$ is it analytic in that region? –  alvoutila Sep 14 '12 at 8:26
@avoutila: See my edit. –  Christian Blatter Sep 14 '12 at 8:38
So is following true? $g(z)=\frac{z}{f(z)}=\frac{z}{\frac{z}{\bar{z}}}= \bar{z}=x-iy \Rightarrow u(x,y)=x, v(x,y)=-y \Rightarrow$ $u_x=1 \neq -1=v_y$ and $u_y=0=-v_x=-0 \Rightarrow$ $g(z)$ is not analytic? But if this true how do you use principle 4 to conclude that $f(z)$ is also analytic? Is $f(z)$ analytic? –  alvoutila Sep 15 '12 at 11:18
Excuse me, but how do you get that $f(z)$ is analytic? Or is it analytic? –  alvoutila Sep 15 '12 at 11:39
@alvoutila: (When I typed in my last answer your question was not displayed completely. I have therefore deleted it.) The logic is as follows: If $f(z):=z/\bar z$ were analytic, then $g(z):=z/f(z)=\bar z$ were analytic as well. Since this is not the case, it follows that this $f$ is not analytic. –  Christian Blatter Sep 15 '12 at 15:47

Perhaps to be a bit technical, it is not an immediately obvious observation (although it certainly is a well known one) that a complex differentiable function is "analytic" in the traditional sense of the word. As I was taught, a complex differentiable function is called holomorphic while functions given by a power series representation are termed analytic. It just so happens through a rather happy coincidence that the two concepts are equivalent in the complex plane, that is, holomorphic functions are automatically $C^\infty$ and have a convergent Taylor series. However, I believe there is merit in distinguishing the two concepts, for several important reasons.

1. It is important not to forget that nature of the two definitions are very different. Holomorphic talks about being differentiable. Analytic talks about being power series representable. It is not at all clear that there should be a connection between the two. To me at least, it is much more satisfying to define complex derivative function in terms of the traditional limit quotient and then show that holomorphic functions are analytic than to simply begin with power series right away. Although it certainly is more restrictive and tedious for the former, I feel it's a necessary burden.

2. It is not true that differentiability and analytic are equivalent. As I mentioned previously, the notion just happens to coincide for complex analysis which is perhaps the most natural setting for power series. In real analysis for example, differentiability and analytic are like-wise defined, but vastly different.

In the end, the terms complex differentiable, holomorphic and analytic are used more or less interchangeably. Ultimately, I feel that this sort of indifference is fine, provided you keep in mind the intrinsic differences.

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