The level of difficulty of complex variables vs. Real analysis

1. Sep 6, 2008

Benzoate

Which course is more difficulty in terms of which subject contains more rigorous proofs, Complex variables or Real analysis. I don't know whether I should dropped Complex variables, but the only reason I am taken it is because of the useful physics applications found in this course. I my school, the only prereq for CV is calculus 3 , and the prereq for Real analysis is a transitional proof class and abstract algebra. I think complex variables is an interesting class and the content of the class is not as rigorous as say Real analysis to me, but my prof(not my advisors ) thinks I should dropped it. I can't dropped a 3 hour class because I am taken only 12 hours of classes right now, and therefore would become a part time student if i dropped it.

2. Sep 6, 2008

SCV

So your professor thinks Complex Variables will be too hard for you? If so did the professor say why?

The usual undergraduate course in Complex Variables has very little rigor.

This doesn't mean it can't be hard. In complex analysis there are different techniques to learn than in Real Analysis. There are different ideas.

3. Sep 6, 2008

lasm2000

Depends on the course. But usually are at the same level of rigor than Calculus, the usual book is the one by Churchill and Brown which is definitely only a small step ahead of the usual calculus rigor. Real analysis on the other side should be a rigorous course, if you use the text by Rudin then expect a really heavy but quite rewarding course, I remember using it at third semester and it was hard but quite worthy.

As far as applications go, well Real analysis leaves you in the doorstep of topology but unless you aspire to do some theoretical things it might not be a really obvious it has applications (series and convergence are used in all physics but not really in a quite rigorous way, which is fine to me). Complex variables is almost inmediately useful, you can use it in 2D potential problems for example, conformal method mappings are omnipresent and integration in the complex plane is something you MUST acquire at some moment.

4. Sep 6, 2008

Benzoate

Did you really need to take Real analysis and other rigorous math courses in order to take Complex variables. I don't think Complex variables has been hard so far. A prof will not always know whats best for the student.

5. Sep 6, 2008

Benzoate

Do you think a person who has only had a transitional proof math course, Linear Algebra, Abstract Algebra and Partial differential equations course can handle Complex variables? The complex variables class at my school doesn't require me to take an intro to proof course , but it is a grad level course but a lot of undergrads have taken this course.

6. Sep 6, 2008

Staff Emeritus
The content of courses called "complex variables" varies enormously from school to school. It's going to be difficult for someone on the outside to tell you that the professor is wrong and that you're completely qualified.

My experience has been that if the course instructor says you're underprepared, (s)he's usually right.

7. Sep 6, 2008

Benzoate

The professor has not really assessed my ability. He only look at my math background and told me i was underprepared. In the past, I had professors tell me I wasn't prepared for Calculus and ended up doing fairly decent in my calculus courses. Dropping Complex variables isn't an option for me because I will no longer be considered a full time student if I dropped this course.

8. Sep 6, 2008

SCV

Complex Variables was the first upper division class I took. I took it from the hardest professor that teaches that class. I took Real Analysis at the same time, the courses were very different. I had to work very hard in this course but I did good (A-)

I took the Graduate course in Complex Variables at the end of my second year. Now in many gradute courses they may assume knowledge of analysis and other topics. Graduate Complex Variables will usually be rigorous. However if yours isn't then it might actually be like an undergraduate course and then you could handle it.

Can you post the course description?

If this was an undergraduate course then I would say take it. But since a graduate course can be significantly harder. How many weeks into the class are you? The course may seem easier at the beginning because it may be review for the graduate students.

Is there another course which you could enroll in to drop this one and still be a full time student?

When I took the Graduate Complex Variables, I was one of two second year undergraduate students taking it. I did good (A-) but the other guy ended up dropping all three of his classes because he was not prepared to take the finals for any his classes. The complex variables course ate up about 90% of my time that quarter and has been the hardest course I have ever taken. You have to be careful.

9. Sep 6, 2008

will.c

You say that you are not able to drop it; you have no option, so what's the question here? As others have mentioned we have no idea what your complex variables class actually entails, but even if we did, we can't reassure your problems away!

If the material is harder than you're used to, well, that's good experience! Talk to the professor and study until you understand it. If your other classes have to suffer a bit, then this is simply the position you're in. I've never heard of someone failing a class that I know they actually worked at.

Good luck!

10. Sep 7, 2008

Howers

Real analysis is MUCH harder. This is mainly because complex variables, as the name suggests, can be taught by using a strictly applied apporach. Real analysis can't (atleast I've never heard of an applied approach to r.analysis, isnt that calculus?). The only pre-reqs for complex variables would be calc 3, linear algebra, and differential equations. Its definately not something you should be afraid of.

Complex analysis, on the other hand, is another beast. I'm assuming complex variables is an applied approach, like something out of Fisher. A course a typical engineer would even be able to handle. In terms of rigour, complex analysis can be up there with real but again I doubt this is the honors class we're talking about. Sometimes, complex variables courses will teach you vector calculus as well in some schools. But if you have the 3 pre-reqs I listed, you should have no problems. I am interested in knowing why your advisor told you not to take it.

Edit: Ah, your talking grad complex. Well... this is a different ball game. So it would be on par with honors complex and be taught out of something lke Ahlorfs. I would say some experience with real analysis is essential. You also have to be good with proofs. If you haven't seen many proofs in your calculus and algebra courses I would steer clear. You mentioned you took Abstract Algebra. That may be a good sign if it was from a more rigorous book then Gallain. PDE is another good sign, but again I doubt it was that rigorous. If you are looking soley for applications to physics, get something like Dovers Complex Variables by Fisher and read it on your own time. In my opinion, you are not ready for graduate level complex analysis. But if you know the book you are using it would help us determine the appropriate level.

Last edited: Sep 7, 2008
11. Sep 7, 2008

mathwonk

in a sense real and complex analysis are polar opposites, because complex is about the worlds nicest functions: namely analytic functions given locally by power series, and reals is about the worlds worst functions: i.e measurable functions, often only determined up to changing their values on a set of measure zero, so even their values are not well defined at any specific point. to me this makes complex infinitely easier and more enjoyable.

12. Sep 7, 2008

Benzoate

As a math professor, which math classes do you think a typical student should have had before they take complex variables? the prereq at my school says I should have calc 3 but do you think I should know more math beyond the prereq liked topology , abstract algebra or Real analysis. I think complex variables is interesting and the material isn't too difficult. There no way I can dropped this class because I would be considered a part time student.

Here is a course description of the class:

Introduction To Complex Variables 3(3-0-0) F,S,Sum
Prereq: MA 242
Operations with complex numbers, derivatives, analytic functions, integrals, definitions and properties of elementary functions, multivalued functions, power series, residue theory and applications, conformal mapping.

13. Sep 7, 2008

mathwonk

there are very few prerequisites for complex, although elementary topology is useful. when i taught it in the past i included all the topology needed. elementary metric space topology is useful but can be taught in the course as in the book of ahlfors. indeed complex is often the course where people first see such concepts. ill see if i can find my old notes.

you do need to know a little about integrals and it helps if you have seen a powers series but again this can be taught in the course itself. look at the book (vol one) of hille for a pretty complete explanation of evrything.

14. Sep 7, 2008

mathwonk

here is a brief survey of complex variables:
MAT 814: Introduction to Complex Analysis

First Day: What is complex analysis , and why is it interesting?

Complex analysis is the study of the (differentiable) solutions of a single very important differential equation, the Cauchy-Riemann differential equation: ∂f/∂zbar = 0, where ∂f/∂zbar = (1/2)[∂f/∂x - i∂f/∂y]. We can explain this a bit more as follows: Recall that two functions f,g:U-->R2 are called tangent at the point p of the open set U in R2 if:
1) f(p) = g(p), and
2) (z -->p) implies ( |f(z)-g(z)|/|z-p| --> 0 ).
(It means their graphs are tangent at [p,f(p)] - think about it.) A function L:R2-->R2 is called R-linear if L(z+w)=L(z)+L(w) for all z,w in R2, and if L(tz)=tL(z) for all t in R and all z in R2. Then a function f:U-->R2 is called (R-) differentiable at a point p in the open set U, if there is some R-linear function L such that f(z)-f(p) is tangent at p to the function L(z-p), (as a function of z). L is called the (Frechet) derivative of f at p. [See Rudin's Principles of Mathematical Analysis defn.9.11, p.212, or Spivak's Calculus on Manifolds, chap.2, pp.15-16 if you don't remember this definition of derivative.]

Now since R2 has a natural structure of multiplication, making it into the complex numbers C, just by setting [1,0] = 1 and [0,1] = i, and putting i2 = -1, we may generalize the last definition as follows: a function f:U-->C is said to be complex differentiable, or holomorphic, at the point p of the open set U in C, if f(z) is tangent at p to L(z-p)+f(p) for some C-linear function L:C-->C. Of course L is C-linear if L(z+w)= L(z)+L(w) for all z,w in C, and if L(tz)=tL(z) for all t,z in C. The only difference here is that the scalar t in the second condition is allowed to be any complex number and is not restricted to be a real number as in the definition of R-differentiable, but what a difference it turns out to make! Complex analysis is the study of holomorphic functions.

Exercise: If f:U-->R2 is a differentiable function at p, with (R-linear) derivative L, where z=x+iy, then ∂f/∂x and ∂f/∂y are both defined at p, and L is the linear function with matrix [ ∂f/∂x ∂f/∂y ], where both entries are regarded as column vectors. Moreover L is actually C-linear (i.e. f is holomorphic) if and only if ∂f/∂zbar = 0. Thus our two conditions for holomorphicity are equivalent. [At least they are equivalent for differentiable functions. It is conceivable however that a function f exists which has partial derivatives ∂f/∂x and ∂f/∂y which satisfy the equation ∂f/∂zbar = 0, but without f even being continuous, much less differentiable, and thus not holomorphic. Recall that a function can have partial derivatives without having a (total) derivative. There do exist functions having these properties at least at one point, but I don't know whether any f exists which has partials that satisfy the Cauchy-Riemann differential equation on a whole open set without f being continous on that set. The Looman-Menchoff theorem, p. 43, Narasimhan's Complex Analysis in One Variable, states that if a continuous function on an open set has partials everywhere on that open set which satisfy the Cauchy-Riemann equation then the function is holomorphic.]

Now why would anyone want to study holomorphic functions? [Life is too short to study them simply because they may occur on the Ph.D preliminary exams.] The study of complex numbers arose in the 19th century out of the study of real analysis and algebra, when it was found that the purely formal use of complex numbers actually facilitated the solution of some problems which appeared to concern only real numbers. There is a nice discussion of this in the book of Osgood. Once complex numbers were accepted as inevitable, which they must be if one ever wishes to solve simple algebraic equations like x2+1= 0, it becomes natural to ask about the calculus of complex numbers. From the point of view of differential equations holomorphic functions turn out also to be closely linked to real harmonic functions, and these had been discovered to be important in physics. That is, the function u(x,y) defining the temperature at a variable point inside a metal disc when that temperature is constant in time, can be shown from simple physical assumptions, to satisfy the Laplace equation ∆2u/∆x2 + ∆2u/∆y2=0. Now it is not hard to check that whenever f is holomorphic and has continuous partials of first and second order, then both the real and imaginary parts u,v of f=u+iv satisfy the Laplace equation, i.e. are harmonic. Moreover the relation between u and v is one which has physical significance. So our interest in physics, heat and electricity and magnetism, could lead us to study complex analysis. Complex analysis is also related to the science of map making! That is, the functions f:R2-->R2 which preserve angles locally (i.e. in very small regions), the so-called conformal mappings, are very closely akin to holomorphic functions, indeed they are almost the same thing.
In mathematics, holomorphic functions help us understand questions concerning the radius of convergence of a power series. For example, the real differentiable function f(t)=1/(1+t2), has infinitely many derivatives at every point of the real line, but its Taylor series centered at the point t=0 only converges on an interval of radius 1. Why? It turns out that the radius of convergence is determined by the phenomenon of absolute convergence, and hence the series cannot converge at any real point whose absolute value is greater than the absolute value of a complex number for which it does not converge. In this example the series fails to converge at z=i, and hence cannot converge at any real number of absolute value greater than | i | = 1. I am told that such questions have applications, within numerical analysis, to giving limitations on the domain of validity of certain numerical approximations. As another example, one not only needs complex numbers to solve for the eigenvalues of matrices in finite dimensions, but one uses complex path integration to study the analogue of the set of eigenvalues, the spectrum, for complex operators in infinite dimensions, [see Lorch, Spectral Theory, chap.4]. Further, the practice of substituting linear transformations into polynomials as in the Cayley-Hamilton theorem, is mirrored by the study of holomorphic functions of complex linear operators in functional analysis. Complex analysis and the associated study of the topology of Riemann surfaces also sheds light on the historically interesting question of studying the "non-elementary" transcendental functions defined by integrating differentials like dt/(p(t))1/2, where p is a polynomial of degree at least three.

This gives some idea of some problems that might push one to study complex analysis, and which indeed did so to the great mathematicians of the previous century, but with the benefit of hindsight we can now look back on their successes and find further motivation for us to study such functions, even after the original motivating problems which led to their investigation may have been settled. That is to say, today the hypothesis of holomorphicity is one which is known to have tremendous mathematical consequences, and thus to take advantage of them in solving our own problems, it is useful to us to be able to recognize holomorphic functions when we meet them, and to know what properties they are guaranteed to have.

15. Sep 7, 2008

mathwonk

part 2: intro to complex vbls:

In many ways the property of holomorphicity is analogous to that of being a polynomial, but more subtle. For example one is always interested, in mathematics, in the problem of existence and uniqueness of solutions to equations. For polynomials the fundamental theorem of algebra tells us that if f(z) is a complex polynomial, and if z0 is a complex number such that f(z) = z0 has no solutions, then f is a constant. The analogous existence theorem is this: if f:C-->C is a holomorphic function (defined and holomorphic everywhere), and if there are two complex numbers z0, and z1, such that neither of the equations f(z)= z0, nor f(z)= z1 has a solution, then f is a constant. The corresponding uniqueness theorems are these: if f,g are complex polynomials with exactly the same zeroes (including multiplicities), then the quotient f/g is a constant; and if f,g:C-->C are holomorphic functions with exactly the same zeroes, then f/g=eh , for some holomorphic function h:C-->C. Since ez is never zero, even for complex z, this is the best we could hope for.
There are other analogies with polynomials, as follows: if one is given a finite set {a1,...,an} of complex numbers, not necessarily distinct, then there exists a polynomial f(z) of degree n, whose set of zeroes is precisely this set, and such an f can be expressed as a (finite) product, one factor for each ai. Moreover, if f is a complex polynomial of degree n, and if f vanishes (i.e. has the value zero) at a set of n+1 distinct points, then f is the zero polynomial. Hence a polynomial of degree n is entirely known if its values are known on a given set of n+1 points. Moreover there is a formula, called Lagrange's interpolation formula, which expresses the value at any point in terms of a finite sum involving the values at the given n+1 points. [The generalization to holomorphic functions is perhaps not so obvious - can you guess it? For the second part you have to figure out what we would mean by a set of infinity plus one points!] In fact if {a1,...,an,......} is an infinite set of complex numbers, such that in any circle centered at the origin there are only a finite number of them, then there is a holomorphic function f:C-->C whose zeroes occur precisely at the given points (and with given multiplicities), and this holomorphic function can be expressed as an infinite product, with essentially one factor for each point. Moreover, if {a1,...,an,......} is a infinite sequence of distinct complex numbers which converges to a (finite) complex number a0, and if f:C-->C is an entire holomorphic function, (holomorphic functions defined on all of C are called entire), which vanishes on all the points ai in the sequence and hence also at a0, then f is identically zero. Consequently an entire holomorphic function is completely determined by its values on the points of a convergent sequence. So the notion of "infinity plus one" points turns out to mean an infinite sequence plus a limit point of that sequence. In particular, an entire holomorphic function is determined by its values on any circle, and indeed there is a formula, Cauchy's integral formula, which gives the values of the function at points inside the circle in terms of an integral over the values on the circumference of the circle!

The uniqueness property just stated tells us in particular that a differentiable function f:R-->R has at most one extension to a holomorphic function F:C-->C, whereas f always has infinitely many different extensions to a (R-) differentiable function G:R2-->R2. For example, sin(t), et, and a0+a1t+a2t2+....+antn, each has a unique extension to a holomorphic function on C, but 1/(1+t2), has no extension to a holomorphic function on all of C, and f(t)=exp(-1/t2) for t≠0 and f(0)=0, has no extension which is holomorphic on any neighborhood of 0, even though f(t) is infinitely differentiable on R. Now the function 1/(1+t2) does of course have a holomorphic extension to any open set in C that does not contain either of the points i or -i, and the extension is given by the same formula, 1/(1+z2). On the other hand there exist functions f:R-->R which have holomorphic extensions to certain open subsets of C, but such that the extensions may take several different values at the same point of C! That is, given a point z0 in C, there may be one holomorphic extension of f to one open set U containing z0 which has one value at z0, and there may be a second holomorphic extension of f to a second open set V which also contains z0 such that the second extension has a different value at z0 from the first one! For example, the real function f(t)=ln(t), defined and differentiable for t in R can be extended to a holomorphic function on the complement of any ray emanating from the origin in the complex plane. But if f1 denotes the holomorphic extension to the complement of the negative imaginary (y-) axis, and if f2 denotes the holomorphic extension to the complement of the positive imaginary axis, then f1(-1)=iπ, while f2(-1)=-iπ. So in cases of functions which extend only into part but not all of C, the values of the extension at a specific point are not determined by the point but rather by the whole open set on which the extension exists. It is true that for each connected open subset of C which meets R, there is at most one holomorphic extension of f to that set. So the holomorphic extension of a function is often inevitably multiple valued, and the attempt to work out the proper domain on which a full single-valued holomorphic extension can be defined led Riemann to the beautiful theory now called "Riemann surfaces". He analyzed the geometry of the resulting domains and thereby constructed the origins of the subject of topology. This problem was also studied abstractly by Abel and Galois in connection with the problem of integrating algebraic functions. [An algebraic function is one g(z) such that there exists a polynomial F(z,w) in two variables for which F(z,g(z))=0, for all z. g(z)=z1/2 is such a function since it satisfies this condition for F(z,w)=z-w2.] Indeed, Galois wrote down some of his very advanced insights on this question in the second part of the same famous letter which is so much better known for its statement of the criterion for a polynomial equation to be solvable in terms of radicals.

So much for motivation. Don't be dismayed if you don't follow all of this. Read it and think about it and refer back to it when the course has progressed further, and see how much more you will understand of it. Some of these results will not be proved until the second or third quarter. We will next take up a rapid review of the fundamental ideas of limits, continuity, completeness, compactness, and connectedness in the setting of arbitrary metric spaces. This will come in handy in this and every other course in which you need to construct solutions to problems by taking limits. Then we will go on to the geometry of the complex numbers and the Riemann sphere, elementary examples and properties of holomorphic functions, properties of power series, and then into the heart of complex analysis, the Cauchy theory of complex path integration. There is no logical necessity of waiting so long to go to the Cauchy theory, as you can see from the rapid introduction to it in the book of Knopp, but I think we will feel more comfortable with the calculus of complex functions if we get acquainted a bit with complex numbers and their topology first. Please let me know how the course is going for you, whether it is either too slow or too fast, too easy or too hard. I may not be able to solve all your problems but I can try if I know how things are going. And do lots of problems from Hille. To start with I want you to try all of them. We will see if this is feasible. In my opinion you will be very glad you did try them all when it comes time to be tested on the material.

Having said all this, I want to admit that I myself, and many others, think that complex analysis is worth studying simply because the results are so beautiful!

16. Sep 7, 2008

Benzoate

Can someone with a mathematical background of only calculus, linear algebra differential equations, can handle a course like intro to complex variables?

17. Sep 7, 2008

SCV

Why did you have to quote that long post to ask this?

Anyways, the answer is it depends on how the professor chooses to present the course and what he/she chooses to assume. The course description you posted seems like that of an undergraduate course. If, so far it doesn't seem to rigorous and that's the course description you will probably be fine. When I took complex variables (the undergrad one) I had only taken linear algrbra, diff eqs and the calculus classes, mine was rigorous though because of the professor I took it from.

18. Sep 7, 2008

Benzoate

what were the prereq's for the grad version of complex variables at your school?

19. Sep 7, 2008

SCV

For that one the prerequisites were 2 quarters of undergraduate real analysis.

20. Sep 7, 2008

Benzoate

What book did the professor used?