If a function of the complex variable z is analytic in an open set if it has a derivative at each point in that set. This definition seems fairly straight-forward but some of the terminology has got me stumped. A holomorphic function; is that analytic or not? Could someone explain what a holomorphic function is and what a regular function (and perhaps an example). Thanks.

Secondly, wouldn't any polynomial be an analytic function since it has a derivative everywhere?

Thirdly, what if the function is analytic except at some points which do not obey the definition? What happens here?

Now when dealing with analytic functions is it true that we are dealing with complex variables f(z) and not real variables f(x)? If so can we apply the normal differentiation/integration rules to complex variables like chain rule, I.B.P, etc...?

Thankyou for any comment on these issues.

Ben.

Originally posted by oxymoron
If a function of the complex variable z is analytic in an open set if it has a derivative at each point in that set. This definition seems fairly straight-forward but some of the terminology has got me stumped. A holomorphic function; is that analytic or not? Could someone explain what a holomorphic function is and what a regular function (and perhaps an example). Thanks.


From what I understand, holomorphic is just another term for analytic (as is regular), the only difference being that people whose field of study is not Mathematics use tend to use analytic, while Mathematicians are more partial to holomorphic. Secondly, I am not sure what you mean by a regular function, but if you mean non-holomorphic, then just pick a function and a set on which it does not have a derivative everywhere, like f(z) = 1/z on the set where |z|<3.

Originally posted by oxymoron
Secondly, wouldn't any polynomial be an analytic function since it has a derivative everywhere?


Yes, and there is a special term for functions that are analytic everywhere, they are called entire functions.

Originally posted by oxymoron
Thirdly, what if the function is analytic except at some points which do not obey the definition? What happens here?


Well, you could either redefine your set to make the function analytic, or simply say that the function has singular points, (Note: for a point to be singular the function must be analytic in the neighborhood of the singularity) e.g., f(z) = 1/z is analytic everywhere except at the singularity z = 0 + 0i.

Originally posted by oxymoron
Now when dealing with analytic functions is it true that we are dealing with complex variables f(z) and not real variables f(x)? If so can we apply the normal differentiation/integration rules to complex variables like chain rule, I.B.P, etc...?


I believe that analytic functions refer to functions defined only on a complex domain, so to deal with real functions, one would have to consider them in the complex domain.

Concerning integration and differentiation: my knowledge here is currently limited (hopefully some of the other posters will enlighten you as to the details) but I know that integration of functions of a complex variable is dependent on whether one is integrating with respect to a real or complex variable. I am not sure about the differentiation of complex functions and their similarities and differences to differentiation of real functions at the present time.

If you are interested in studying this material more in depth, a course in Complex Variables (usually called Complex Analysis or something of the sort) would address all of these issues, including items such as Taylor and Laurent series, Residues, Conformal Mappings, and so on. Classes concerning this material are usually advanced undergraduate or first year graduate level courses.

Complex derivatives are defined in much the same way as real ones. The derivative of a function f(z) is:

df/dz = lim(h->0) [ f(z+h) - f(h) ]/h

where z and h are both complex numbers.

For a function to have a derivative, the requirement is very strict in complex analysis. Not only must the above limit exist, but it must also be the same no matter how the complex h goes to zero. Since a complex number h can approach zero from many different directions instead of just the two in the real case, if a complex derivative exists at all a lot more can be said about the function than can be said about a real function with a derivative.

This is the starting point for examining analytic functions.

It turns out that many, but not all, functions of a complex variable behave in very similar ways to their real counterparts. In a lot of cases, the complex function can be treated as an extension of the real function. For example, a lot of "short cuts" in differentiation apply equally to complex functions as to real ones.

"f is holomorphic" is the same as saying "f is analyic for all complex numbers". This is the same as "entire function".

"Analytic" is a point property- a function may be analytic at some points and at others (although it is different from "continuous" or "differentiable" [for functions of real numbers] in that one can show that if a function is analytic at a point it is analytic in some neighborhood of that point).

A function is "holomorphic" if and only if it is analytic for ALL real numbers. Normally just saying "f is analytic" without restrictions implies holomorphic. If you want to be more precise, say "f is holomorphic".

By the way, it is more common to DEFINE analytic at a point, p, as meaning "the Taylor's series for f(z), at p, exists and converges to f(z) for z in some neighborhood of p". Of course, this turns out to be equivalent to "differentiable".

Some texts use the term "real analytic" to mean a function on the real numbers whose Taylor's series exists and converges to the function.

Thankyou Dapthar and HallsofIvy for your quick and concise answers and to James R for his explanation of differentiation. This was very helpful and easy to read. :)

Now can someone tell me how the Cauchy-Riemann Equations fit into this. I can do the questions but don't understand why we need to verify if the C-R equations are satisfied and how we go about this.

Originally posted by oxymoron
Thankyou Dapthar and HallsofIvy for your quick and concise answers and to James R for his explanation of differentiation. This was very helpful and easy to read. :)


No problem, glad to help. I have a question for you though, and if you don't care to answer it just ignore it (and please don't take it in a derogatory manner), but what was the last Math class you took? Just wondering so I can get a frame of reference.

Originally posted by oxymoron

Now can someone tell me how the Cauchy-Riemann Equations fit into this. I can do the questions but don't understand why we need to verify if the C-R equations are satisfied and how we go about this.

The purpose of Cauchy-Riemann Equations is two fold. From a computational standpoint, they describe f'(z0) (the 0 is supposed to be a subscript) in terms of the first order partial derivatives of u and v, (where f(z) = u(x,y)+i&bull;v(x,y)). From a more theoretical standpoint, they constitute a necessary (but not sufficient) condition for the existence of f'(z0). Technically, verification is not necessary, if you know that f"(z0) exists, but you are probably just having to do this as an exercise.

Okay, so just say you have f(z) = z^2 and f'(z) = 2z. This function is differentiable everywhere but you still need to verify that the C-R equations are satisfied everywhere?

If this is correct then I know z^2 = x^2 - y^2 + i2xy then taking partial derivatives I would obtain...
ux = 2x and uy = -2y then substituting into C-R...

f(z) = u(x,y)+i•v(x,y))
f'(z) = 2x + i2y = 2z = f(z)

Well this came out nice! but can someone supply a worked example where C-R equ. don't fit (and if it isn't too much trouble, perhaps a harder example where C-R equ Do fit!)

Dapthar, I took MATH1220 (Advanced Mathematics).

If a function of a complex variable is differentiable at a point, then the C-R equations MUST be satisfied. That comes from considering the limits as you approach the point along a horizontal line and a vertical line.

For example, z= x+ iy and f(z)= u(z)+ iv(z)= u(x,y)+ iv(x,y),
Then for z approaching (x₀,y₀) "horizontally", z= x₀+h+ iy₀ so that f(z)= f(x₀+h,y₀) so forming the difference quotient and taking the limit gives df/dz= du/dx+ i dv/dx (the derivatives wrt x are partial derivatives).
For z approaching (x₀,y₀) "vertically", we have z= x₀+ i(y₀+h) and forming the difference quotient and taking the limit gives df/dz= - i du/dy+ dv/dy (the -i is from the "ih" in the denominator- also that cancels the i in the dv/dy term).

Since the limit must be the same no matter how the point is approached, we have du/dx+ idv/dx= -i du/dy+ dv/dy. Equating real and imaginary parts, du/dx= dv/dy and dv/dx= -i du/dy, the Cauchy-Riemann equations.



[Moderator note to HallsofIvy: to get subscripts you must use angled brackets in the tags rather than square ones. i.e. write &lt;sub&gt; not [sub]]

what exactly is an <i>anti</i>holomorphic function?

Originally posted by lethe
what exactly is an <i>anti</i>holomorphic function?
If one flips the sign on only one side of the equalities provided by HallsofIvy, one gets the conditions for an antiholomorphic function, namely: u_x = -v_y andu_y = v_x.

Originally posted by oxymoron
Okay, so just say you have f(z) = z^2 and f'(z) = 2z. This function is differentiable everywhere but you still need to verify that the C-R equations are satisfied everywhere?

Like HallsofIvy said, if the function is differentiable at a point, the Cauchy-Riemann equations must be satisfied there, since the Cauchy-Riemann equations are necessary (but not sufficient) conditions for differentiability.

Originally posted by oxymoron
Well this came out nice! but can someone supply a worked example where C-R equ. don't fit (and if it isn't too much trouble, perhaps a harder example where C-R equ Do fit!)

Here's an example where the Cauchy-Riemann equations are used to find points where the function is not differentiable.

f(z)=|z|²
u(x,y)=x²+y² and v(x,y)=0
Thus u_x=2x, and u_y = 2y.
By the Cauchy-Riemann Equations, u_x=v_y and u_y=-v_x.
Thus, 2x=0, and 2y=0. Thus f'(z) does not exist at any point where (x,y) is not equal to (0,0). However, this alone does not ensure that f'(0) actually exists.
(In this case, f'(0) actually does exist though, since the Cauchy-Riemann equations plus continuity of the partial derivatives at the point of interest (and their existence in some neighborhood of the point) is a necessary and sufficient condition for differentiability)

Originally posted by oxymoron

Dapthar, I took MATH1220 (Advanced Mathematics).
I don't really have any idea what that entails, perhaps you could supply a short list of the topics covered?

Originally posted by Dapthar
If one flips the sign on only one side of the equalities provided by HallsofIvy, one gets the conditions for an antiholomorphic function, namely: u_x = -v_y andu_y = v_x.
thats what i thought. but:
Originally posted by Dapthar

Like HallsofIvy said, if the function is differentiable at a point, the Cauchy-Riemann equations must be satisfied there, since the Cauchy-Riemann equations are necessary (but not sufficient) conditions for differentiability.
does this then imply that an antiholomorphic function is not differentiable?

Originally posted by lethe
thats what i thought. but:

does this then imply that an antiholomorphic function is not differentiable?

Yes. This sort logic was used in the example in my previous post, since the function did not satisfy Cauchy-Riemann conditions except at (x,y)=(0,0), it implied that f'(z) was not differentiable at any non-zero point.

would it be correct to say that &fnof;(z) is antiholomorphic iff &fnof;(z*) is holomorphic?

Okay, now that I have differentiability under my belt, can analyticity of functions be applied anywhere else, or is it just a property of complex functions?

Dapthar,

Advanced Mathematics
Complex Numbers, Matricies, Determinants, Vectors (2 and 3 space) stuff like dot and cross product, lines and planes, Euclidean vector spaces and linear transformations, subspaces, basis & dimension, row space, rank and nullity, inner product spaces (inc. Gram-Schmidt, and QR Decomp.), change of basis and eigenvectors.

This is the linear algebra side. There is also the calculus side which deals with applied mathematics.

Originally posted by lethe
would it be correct to say that &fnof;(z) is antiholomorphic iff &fnof;(z*) is holomorphic?

I will have to think over that for a little bit, but I should have it worked out by the end of tomorrow.

Originally posted by oxymoron
Okay, now that I have differentiability under my belt, can analyticity of functions be applied anywhere else, or is it just a property of complex functions?


Just skimming through my Complex Variables book , I see that analytic functions are used in quite a few other places, primarily because they are "nice" functions. They seemed to be used about as often as the term "continuous, differentiable function" was used in Calculus I and II, and other material is built upon the concept of analytic functions in a similar manner.Originally posted by lethe
would it be correct to say that &fnof;(z) is antiholomorphic iff &fnof;(z*) is holomorphic? Yes it is true, and here's the proof of why.

Consider an antiholomorphic complex function, f(z)=u(x,y)+i&bull;v(x,y). Recall that antiholomorphic means that u_x = -v_y and u_y=v_x.

Now consider the complex conjugate of f(z), denoted by f(z*).

f(z*)=u(x,y)-i&bull;v(x,y), thus, u_x = -v_y&bull;-1 = v_y (the factor of -1 in front of v(x,y) in f(z*) introduces a -1 in the partial derivative. Similarly u_y=-v_x, as desired.

Thus if f(z) is antiholomorphic, f(z*) is holomorphic.

A similar proof (essentially the above one in reverse) proves that if f(z*) holomorphic, f(z) is antiholomorphic.

thanks dapthar. at first i was thinking that it is weird that an antiholomorphic function is not differentiable. but if i say it this way: a holomorphic function is differentiable with respect to z, while an antiholomorphic function is differentiable with respect to z*, it seems rather natural.