From: rusin@washington.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: How do you solve ln z = z ? Date: 1 Nov 1995 19:18:46 GMT In article <1995Oct31.171620.22793@lamont.ldgo.columbia.edu>, VADIUM FELDBAUM wrote: >The subject line is self-explanatory, I hope. Of course z is complex. ... >Also, another equation: > > 1+z >ln --- = 8z > 1-z The poster then goes on to exponentiate these equations -- a good move, since ln isn't single-valued for complex inputs, making the original question a little unclear. What remains then is an equation of the form exp(f(z))=g(z) for nice functions f and g. Then general question to answer, then, is >Is this in any way related to solvability in complex? Of interest here is a collection of theorems on the range of analytic functions. I will quote one useful results from classical complex analysis. "Little Picard Theorem": Every complex analytic function defined on the whole complex plane is either (a) onto C (i.e., for every w there is a z with f(z)=w) (b) onto C-{one point} (i.e., f misses at most one value) (c) or constant. In fact, in all cases, the inverse image of every point is either empty or infinite. So for example if f is the function f(z)=exp(z)-z, then since (as another poster has pointed out) there is one value of z with f(z)=0, there must be infinitely many. Moreover, the equations exp(z)=z+c are also all solvable, except possibly for one value of c. Likewise the equation (1+z)/(1-z) = exp(8z) may be recast as asking for the zeros of f(z)=(1-z)exp(8z)-(1+z). Without looking at f any more than checking it is entire, we know it would be nearly miraculous that it not have any zeros, and if it had one, it would have infinitely many. Incidentally, the Great Picard Theorem assures that f attains all values (with, possibly, one exception) infinitely often in any neighborhood of an essential singularity, so functions which appear to be poor candidates for the Little Picard Theorem are still easy functions for which to demonstrate the existence of many zeros. You will no doubt have noticed that I only "solved" the equation in a mathematician's sense: solutions exist, but I don't know where they are :-) If I had to, I suppose I would use Newton's method. Once I had one zero, I might look for others by applying Newton's method, not again to f(z) but rather to f(z)/(z-z0), so as to avoid converging to the same root again. dave ============================================================================== From: rusin@washington.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: How do you solve ln z = z ? Date: 2 Nov 1995 21:45:53 GMT In a recent article, VADIUM FELDBAUM asked for clarifications of a response I had made to his initial post: >> "Little Picard Theorem": Every complex analytic function defined on the whole >> complex plane is either >> (a) onto C (i.e., for every w there is a z with f(z)=w) >> (b) onto C-{one point} (i.e., f misses at most one value) >> (c) or constant. > >Do you mean to say that each and every equation would have a solution? That is correct: all of the equations f(z)=0, f(z)=1, f(z)=pi, etc. are solvable. The example f(z)=exp(z) shows that there can be just one equation (in this case, f(z)=0) which has no solution, but this theorem asserts that for each nonconstant entire function there is at most one such unsolvable equation. >> In fact, in all cases, the inverse image of every point is either empty or >> infinite. > >Is this a part of the above theorem? Does it mean that each and every >equation would have infinitely many solutions? As presented in Conway's "Functions of One Complex Variable", the preceding line is actually a corollary of the Great Picard Theorem. I don't know if historically they were presented as a unit or not; you may think of them as all one theorem, I suppose. >> Incidentally, the Great Picard Theorem assures that f attains all >> values (with, possibly, one exception) infinitely often in any neighborhood >> of an essential singularity, so functions which appear to be poor candidates >> for the Little Picard Theorem are still easy functions for which to >> demonstrate the existence of many zeros. > >Well, if I undertood correctly what you said here, I'll have to change >the way I understood the idea of singularities. Careful -- the phrase "essential singularity" has a technical meaning. There are "removeable singularities" at a point a (meaning that if a [new] value is assigned at f(a) then f is analytic at a -- for example, f(z)=(z^2-1)/(z-1) has a removeable singularity at 1 ) and there are "poles" (meaning (z-a)^n f(z) has a removeable singularity at a); it's the other singularities which are "essential". The usual example is f(z)=exp(1/z), which has an essential singularity at 0. > Say if you take 1/x then it does take on every value near zero. This is a simple pole, not an essential singularity. It does not attain the value 0 (nor indeed any value of small absolute value) in a neighborhood of zero. >If you take >|1/x|=ABS(1/x) then it take on every positive value, but not negative >ones. Are you saying that this cannot be achieved for complex? ABS >never takes on negative values, reals or complex! But then ABS is >not an analytic f-n. Still I can't quite beleive same thing cannot >be achieved with analytical f-ns only. This is very much a theorem of analytic functions. This function f(z)=1/sqrt(z*zbar) is not analytic at any point. (At 0 it fails on three counts, so to speak: f is built up from the complex functions "conjugate", "sqrt", and "reciprocal", none of which is analytic here. At nonzero points, it fails to be analytic only because of the first of these three: the function f(x+iy)=x-iy is often the example used to show the difference between complex-analytic and real-analytic functions f: C-->C.) It is a fundamental feature of complex analysis that the inoccuous definition of analyticity is actually quite restrictive, so that functions which do meet this condition can be counted on to be extremely well-behaved (e.g. they are _infinitely_-differentiable, uniquely determined by values on small sets, etc.) Take a good course on complex analysis (not just "Complex Variables"). It is an experience not to be missed in a student's mathematical training. dave (not a complex analyst) ============================================================================== From: rusin@washington.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: How do you solve ln z = z ? Date: 2 Nov 1995 22:00:19 GMT Sorry, forgot one little thing: > "Little Picard Theorem": Every complex analytic function defined on the whole > complex plane is either > (a) onto C (i.e., for every w there is a z with f(z)=w) > (b) onto C-{one point} (i.e., f misses at most one value) > (c) or constant. ... > In fact, in all cases, the inverse image of every point is either empty or > infinite. This last line only applies if f is not a polynomial. (The trick is just to observe f(1/z) has an essential singularity at 0 iff f is not a polynomial, so that the Great Picard Theorem can be used.) Sorry about that. dave ============================================================================== From: rusin@washington.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: How do you solve ln z = z ? Date: 3 Nov 1995 18:33:56 GMT Oh, dear: I've had to correct my own followups before, but I can't recall having to correct a correction! It seems I have long labored under a misconception which several in this group have now called to my attention. Let me try this one more time. This much, posted a while ago, is correct: > "Little Picard Theorem": Every complex analytic function defined on the whole > complex plane is either > (a) onto C (i.e., for every w there is a z with f(z)=w) > (b) onto C-{one point} (i.e., f misses at most one value) > (c) or constant. Here's where I blew it: > In fact, in all cases, the inverse image of every point is either empty or > infinite. I forgot to add that f is to be non-polynomial, but I hadn't realized that even with that condition, the statement is incorrect. Rather than risk botching it again, I will simply quote passages from Conway's book. "Great Picard Theorem. Suppose an analytic function f has an essential singularity at z=a. Then in each neighborhood of a, f assumes each complex number, with one possible exception, an infinite number of times. ... In the preceding chapter it was shown that an entire function of order \lambda, where \lambda is not an integer, assumes each value infinitely often. Functions of the form e^g, for g a polynomial, assume each value infinitely often, although there is one excepted value -- namely, zero. The Great Picard Theorem yields a general result along these lines (although an exceptional value is possible, so that the following result is not comparable with [that result from the preceding chapter]. Corollary. If f is an entire function which is not a polynomial then f assume every complex number, with one exception, an infinite number of times. Proof. Consider the function g(z) = f(1/z). Since f is not a polynomial, g has an essential singularity at z=0. The result now follows from the Great Picard Theorem. [qed] " (The main result of the preceding chapter is Hadamard's factorization theorem.) ============================================================================== From: Toke Lindegaard Knudsen Newsgroups: sci.math Subject: Complex analysis question. Date: 3 Nov 1995 16:17:14 GMT Hello. I was wondering about the following: Suppose f is a mapping from the set of complex numbers into the set of complex numbers and f is entire and that f does not take the value 0. Will there then exist an entire function g, so that f(x) = exp(g(x)) for all complex x. ? Can anybody help me? Thanks. Toke. [sig deleted] ============================================================================== From: rusin@washington.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: Complex analysis question. Date: 3 Nov 1995 23:03:10 GMT In article <47dfaa$nkv@danmat.math.ku.dk>, Toke Lindegaard Knudsen wrote: > >Suppose f is a mapping from the set of complex numbers into the set of >complex numbers and f is entire and that f does not take the value 0. > >Will there then exist an entire function g, so that > > f(x) = exp(g(x)) for all complex x. Since another respondent has posted something which I think misses the heart of the issue, I thought I'd follow up. Of course there ought to be such a g, namely g(x)=ln(f(x)), but there are two problems: how do we know that a continuous choice of branch of the logarithm may be made, and how do we know the resulting g is analytic? (This g will be well-defined on the whole plane since f is defined everywhere and assumed nonzero.) Actually the analyticity is immediate from the chain rule since the logarithm -- any branch -- is analytic wherever defined, and since f was assumed to be analytic. The remaining issue is the ability to choose a single branch of the logarithm over the entire plane. This requires nontrivial use of the hypotheses. Indeed, if f(x)=1/(x-a), then all the conditions are met except for a single point where f is not defined; and sure enough, there is no continuous function g with the same domain as f such that f = exp o g. The resolution of this problem is topological. Since f is assumed to be a continuous function from the _simply-connected_ domain C to the non-simply-connected region C-{0}, f factors through the universal cover of C-{0}. But the universal covering map p: C -> C-{0} is, you guessed it, the exponential map. (The lifting g: C-> C is uniquely defined by the condition f = p o g after choice of a single value g(x_0), which may be selected arbitrarily in the inverse image p^(-1)(f(x_0)).) The question of the existence of continuous liftings of this sort is often treated in first courses in algebraic topology. See, for example, Sieradski's "Introduction to Topology and Homotopy", Chap.14, although this specific example is somewhat more transparent. dave ============================================================================== From: ags@seaman.cc.purdue.edu (Dave Seaman) Newsgroups: sci.math Subject: Re: Complex analysis question. Date: 4 Nov 1995 11:59:05 -0500 >In article <47dfaa$nkv@danmat.math.ku.dk>, >Toke Lindegaard Knudsen wrote: [A good question, and not the one I answered.] I misread your question, but fortunately it has been addressed elsewhere. I had the Little Picard Theorem on the brain, having just read about it in another thread, and I was reminded of an old paradox. Those who are familiar with the theorem may as well stop reading, because this will be old hat. You have been warned. The question I was attempting to address here, not the one you asked, was the following: Consider the exponential function exp: C -> C. This is an entire function whose image is C - {0}. If we now form the composition g(z) = exp(exp(z)) we have an entire function that is obviously non-constant. By the Little Picard Theorem, the image of g must be the entire complex plane, with the possible exception of a single point. Obviously, 0 is not in the image of g. But since exp(0) = 1, and 0 is not in the image of exp, we find that 1 is also not in the image of g. Therefore, 0 = 1. (groan) Dave Seaman ==============================================================================