Well, it looks like you can transform the equation into the form

e^z_1 + e^z_2 = 1

(for a different z_1 and z_2, of course),

and then use some of the reasoning that goes into the first statement

the imaginary parts of the two exponentials still have to add to 0,

and the real parts to 1 -

that should give you some useful constraints,

and those might be enough to answer your question

Well, it looks like you can transform the equation into the form

e^z_1 + e^z_2 = 1

(for a different z_1 and z_2, of course),

and then use some of the reasoning that goes into the first statement

the imaginary parts of the two exponentials still have to add to 0,

and the real parts to 1 -

that should give you some useful constraints,

and those might be enough to answer your question

The solution suggested above is naturally nice!

You can solve the equivalent equation e^(z_1-z_3)=-e^(z_2-z_3)-1 (in the unknown v_1= z_1-z_3 v_2=z_2-z_3) using the trigonometric representations for both the left and the right hand member of the equation. The relation is, obviously, trickier

Question. Let e^z_1+e^z_2=1,

where z_k=x_k+i y_k, k=1,2 are complex numbers. What about the real and imaginary parts x_1,x_2,y_1 and y_2?

Solution . We have the system of real numbers

e^x_1 cos[y_1]+e^x_2 cos[y_2]=1, e^x_1 sin[y_1]+e^x_2 sin[y_2]=0, which has

a solution if:

1. y_2= y_1 +k \pi , k\in Z. Then,

either y_1=2n\pi, y_2=2m\pi and e^x_1+e^x^2=1 or,

y_1=2n\pi, y_2=(2m+1)\pi and e^x_1-e^x^2=1 .

or

2. y_2 \neq y_1 +k \pi , k\in Z.

Then, either

sin[y_2-y_1]>0, sin[y_2]>0, sin[y_1]

I might be missing something here, but the original statement does not appear to be true. Under the assumptions given, e^z_1+e^z_2=-2e^(Re(z_1)) , which is never 0.

Right!

It looks like the first equality in the original problem statement should be a minus sign: Im(z_1) - Im(z_2) = (odd multiple of pi) -

this is easy to prove - first put one of the expression on the RHS -

then the magnitudes are equal, so Re(z_1) = Re(z_2),

then the arguments are equal with a phase shift (the minus sign you get when you put one of the expressions on the RHS), so the imaginary parts must differ by an odd multiple of pi

the magnitude argument will not easily apply to three terms, however

I think the second assumption should be Im(z_1) (+-) Im(z_2) = (2n+1)pi.

The prove is just use the term

e^(a+ib)= e^a(cos b+ i sin b)

and cos(x (+-) y) = cos x cos y (-+) sin x sin y.

Thank you, Gail for checking!

Thank you, Kus Krisnawan, I did the correction, that you mention.

Thank you, Christopher Landauer for the first step and discussion!

Thank you, Stepan Tersian (it is nice to hear from Armenian) for your solution.

I know that the original statement for sum of 2 exponents gives ALL SOLUTIONS.

Is it true?

Resuming solutions for the sum of 3 exponents we have the main equation:

e^{z_1-z_3}+e^{z_2-z_3}=-1=e^{i\pi(1+2k)} (Christopher's step, division by e^{z_3})

By taking imaginary parts we get

e^{Re[z_1-z_3]}sin[Im[z_2-z_3]]+e^{Re[z_1-z_3]}sin[Im[z_1-z_3]]=0

Assuming Im[z_1-z_3]=k\pi, Im[z_1-z_3]=n\pi, and by taking real part

of the main equation we get

e^{Re[z_1-z_3]}cos[Im[z_1-z_3]]+Re[z_2-z_3]}cos[Im[z_2-z_3]]=0

or first family of Stepan's solutions:

(-1)^ke^{Re[z_1-z_3]}+(-1)^n e^{Re[z_1-z_3]}=1,Im[z_1-z_3]=k\pi, Im[z_1-z_3]=n\pi.

Second family of Stepan's solutions is (under some restrictions that ln() is real valued):

Re[z_1-z_2]}=ln(-sin[Im[z_1-z_3]]/sin[Im[z_1-z_3]]),

Im[z_1-z_3]\neq k\pi,

e^{Re[z_1-z_3]}cos[Im[z_1-z_3]]+Re[z_2-z_3]}sin[Im[z_2-z_3]]=-1

Some other solution families could be found by permutations z_3 \to z_1 and z_3 \to z_2.

The next QUESTION is there any other solutions or not?

Is it possible to simplify these families (they are too complicated)?

Let e^zi=e^ai(cos(bi)+isin(bi)), i=1,2,3. The following solutions are included among the infinitely many solutions of e^z1+e^z2+e^z3=0:

Take any b2 with sin(b2) \ne cos(b2).

If sin(b2)>cos(b2), take b1=\pi 2n ( i.e. zero or a positive or negative even integer),

a1=ln(e^(a2)(sin(b2)-cos(b2))), and b3=\pi-arccos(exp(a2)sin(b2)/exp(a3))

where a2 and a3 are chosen so that the argument of arccos lies between -1 and 1

If sin(b2)

Gail your solution does not work since for your solution

Im[e^z1+e^z2+e^z3]= exp(a2) sin(b2)+exp(a3)sin(arccos(exp(a2)sin(b2)/exp(a3))) \neq 0

Assume that z_1=z_2. Then we have

2e^(z_1)=-e^(z_3).

This holds if and only if 2e^(x_1)=e^(x_3) and y_1-y_3=pi(2n+1), which is equivalent to

Re(z_3)=ln[2e^(Re(z_1))] and Im(z_1-z_3)=pi(2n+1).

Gro, you are correct. Sorry, I had a typo writing cos(b3) instead of sin(b3) in my calculation! In any case, there appear to be simple solutions to your problem.

Again, let e^zi=e^ai(cos(bi)+ I sin(bi)), i=1,2,3. The following are solutions:

Let b1 be any even multiple of pi,

and b2 and b3 both be any odd multiples of pi.

Take any a1>a3, and a2=ln(e^a1-e^a3)

OR

Let b1 and b2 be any even multiples of pi

and be any odd multiple of pi.

Take a1a3, and a2=ln(e^a1-e^a3)

OR

Let bi=(2n_i+1)pi/2, i=1,2 where n_i are any even integers

and b3=(2k+1)pi/1 where k is an odd integer.

Take a1

Dear Hassan,

I know, there are many famous theorems about oscillations of differential equations with REAL coefficients. For differential equations with complex coefficients there are no much results.

My questions pops up when you are trying to find oscillation conditions of the simplest linear differential equations with complex constant coefficients.

I think that there are many simple interesting questions from the complex analysis that are not considered (may be) before .

Gail, your solutions are correct, but they are contained in Stepan's set of solutions.

Could you find new ones?

Dear Oktay,

Your solutions formally contained in Stepan’s set when k=0,

but they are important, since they are simple and are similar to 2 dimensional case.

The general answer for e^w_1 + e^w_2 = 1is messy but doable

choose real y_1 and y_2 so that

sin y_1 and sin y_2 have different signs,

and sin(y_2 - y_1) and sin y_1 have the same sign

(we need sin(y_2 - y_1) != 0 also)

then choose real x_1 and x_2 so that

e^x_1 * sin(y_2 - y_1) = sin y_2,

e^x_2 * sin(y_2 - y_1) = - sin y_1

(the sign conditions above guarantee that these two equations are solvable,

since we know that e^x_1 and e^x_2 have to be positive)

then w_i = x_1 + i*y_1 and w_2 = x_2 + i*y_2 satisfy the equation (the converse is easy to show)

this is related to the original equations in a simple but ugly way

(it might be less ugly to solve the equation e^w_1 + e^w_2 + 1 = 0,

since the derivation from and to the original question is much simpler)

Thank you Christopher for killing my question! Since you killed that one I have another one: when e^(z1)+e^(z2)+e^(z3)+e^(z4)=0.

thank you Gro, i can see where this is going 8-)

note that in the background equation to the original question, an arbitrary choice of z_2 still allowed infinitely many choices for z_1 (all having the same real part)

similarly in my version of the original equation, we can see that there are still infinitely many pairs of choices for y_1 and y_2, which then determine x_1 and x_2

that translates back in the original question (in the ``less ugly'' version of my change to the original equation, we can have w_1 = z_1 - z_3 and w_2 = z_2 - z_3) to an arbitrary choice for z_3, and still infinitely many (but not arbitrary) choices for z_2 and z_1

the more terms you have in the single equation, the more degrees of freedom will be available, and the solution will get much more complicated

the key observation occurs in the background question: the expression e^z does not uniquely determine z in the complex plane, so no matter what particular solution you have for these sums, adding any integer multiples of 2*pi*i to the exponents (even different integer multiples for different exponents) leaves all of the terms in the sum the same

Problem to describe the solutions of the n-dimensional equation e^{z1}+…e^{zn}=0 is still alive.There should be a (nice?) solution involving Euler’s formula, linear algebra, geometry (mentioned by Stepan) and symmetry (mentioned by Christopher).

Excluding singular cases (sin(Im[z_k-z_m])=0, Stepan’s observations)

for 3 dimensional case dividing e^{z1}+ e^{z2}+ e^{z3}=0 by e^{z3} and solving for

e^{\Re(z1-z3)} ,e^{\Re(z2-z3)} (Christopher’s observations)

one can get the law of sine for two triangles

sin(Im[z2-z3])/e^{Re[z1-z3])}= sin(Im[z1-z2])/e^{Re[z2-z3])},

sin(Im[z3-z1])/e^{Re[z2-z3])}= sin(Im[z1-z2])/e^{Re[z1-z3])}.

if you consider the real and imaginary parts separately, and choose y_j = Im(z_j) arbitrarily for each j, then there is still the ambiguity of 2*k*pi for each y_j, but we know s_j = sin(y_j) and c_j = cos(y_j) for all j

in any case, the equation becomes two linear equations in the unknowns p_j = e^x_j, where as you should expect, x_j = Re(Z_j), and the condition is that we want a solution with all values p_j positive so we can solve for x_j

the equations are sum(j) p_j * c_j = 0 and sum(j) p_j * s_j = 0 (the equations are independent for most but not all choices of the y_j; for example, if each y_j is pi/4 (plus any multiple of pi), then the two equations are identical)

it is easy to see that there is not always a positive solution - for example, when all c_j and s_j are positive, or all negative), and that when at least one j has both c_j and s_j for that same j is negative and the rest positive, then there are positive solutions, just by making the corresponding p_j large enough to compensate for the rest

the remaining case i have not described is when some of the c_j and s_j are positive and some are negative, which may or may not have positive solutions (i think some do and some don't, and the rules for which is which are tricky)

however, it is easy to see that if there is a positive solution, then there are many, by multiplying all p_j by the same positive constant f

actually, we have much more than that - consider the space P of all n-vectors p with all positive coordinates that satisfy the two linear equations - we noted just above that if p is in P, then so is f*p for any f > 0, and we can easily see that if p and q are in P, then so is p+q

so P is sort of a half linear space or cone or something like that (i don't know the actual technical term), and i suspect that it has co-dimension 2 because there are the two equations (by analogy to linear equation solutions in a linear space), but i don't know whether such a space always has a basis of the appropriate size (i suspect it does), or any of the usual properties of subspaces defined by linear equations

interesting

It's weird, i was dreaming about this problem yesterday, and i think i have a general approach (uising the notation of my previous answer) - if you pick z_2, ..., z_n arbitrarily, then you can pick y_1 to have an appropriate sign for both cos and sin, depending on the sum of the other terms, and then compute x_1 - in retrospect, this is an obvious generalization of the original background question (the tricky part is choosing y_1 to adjust for the ratios in the sum of e^z_j from j=2 to n - that may make y_1 unique (up to a multiple of 2*pi, of course) - this is really about computing the complex logarithm of that sum

Here the desired statement for the sum of 3 exponents (any exceptions?):

Assume that 0 \le Im [z_1-z_3]

Hi,

I do not have a general analytical solution, but only two comments:

a) rewrite e^z_1=e^x_1*e^(i y_1) etc. Then the sum e^(z_j) = 0 may be interpreted as follows: put "masses" e^x_j > 0 on the unit circle at angles y_j mod 2 pi. Then, the center of mass (or centroid) of this mass distribution must be the center of the unit circle. This is a geometric meaning of the equation. Alternatively, you could put masses 1 at e^(z_j) in the complex plane and again, the center of mass is the origin. This does hold also for more than 4 summands. Compare p. 102ff of [Visual Complex Analysis,Tristan Needham, Clarendon Press, Oxford, 2000]. In a sense, this is a general geometric solution. In the book, you find also information on sum f(z_j), that might be helpful, especially since exp is a conformal mapping.

b) look for the symmetries of your equation, e.g. complex conjugation, rotating all your e^(z_j) by a common angle around the origin.

Best regards

Herbert

Thank you, Herbert for nice comment.

you are welcome. BTW, it could well be that it might be an interesting topic to regard the equation as defining a seven-dimensional real differential manifold. Since there are relatively many symmetries here, it could be a candidate for a Lie analysis. Where are all the differential geometers? :) Especially since you came up with the problem in the context of differential equations ...

With respect to Herbert Homeier's last answer I would like to add a few remarks on the description of the sulution set:

The equation

w_1+w_2+w_3+w_4=0

defines a hyperplane H in \C^4, that is a 3-dimensional complex sumbmanifold of \C^4, which can as well be considered as a 6-dimensional (not 7!) real submanifold of \R^8. There are no secrets about this hyperplane as a linear submanifold, you can easily parametrize it, etc.

On the other hand we have a map:

F: \C^4 \to ( \C \setminus \{ 0\} )^4,

(z_1,z_2,z_3,z_4) \mapsto (e^{z_1}, ..., e^{z_4}).

Obviously the solution set of the original equation is F^{-1}(H), which is again a 3-dimensional complex submanifold of \C^4. The above answers describe different ways of describing or parametrizing this set. Unfortunately the usual (linear) parametrizations of H will yield parametrizations of F^{-1}(H) with somewhat strange domains of definition (excluding the case w_i=0).

For example, if we parametrize H as \lambda (1,0,0,-1)+\mu (0,1,0,-1)+\nu (0,0,1,-1)

and define for w\not=0 a set-valued Logarithm Log(w)=\{ z\in\C | e^z=w\}, then

F^{-1}(H) is the union of

Log(\lambda)\times Log(\mu)\times Log(\nu) \times Log(-\lambda-\mu-\nu)

for all \lambda, \mu,\nu \in \C \setminus \{0\} with \lambda+\mu+\nu \not=0.

In order to avoid the multivalued Logarithm, we may as well consider the solutions 'modulo 2\pi i' in every component, that is we consider the map:

G: \C^4 \to (\R\times S^1)^4

(z_1,...,z_4)\mapsto ([z_1],...,[z_4]),

where [z_k] denotes the equivalence class of z_k modulo 2\pi i. The set of equivalence classes can be identified with a complex infinite cylinder (we glue the line Im(z)=2\pi to the line Im(0). The set of solutions F^{-1}(H) is closed with respect to exchaning some z_k by z_k +2\pi i l, that is, it can be obtained as

G^{-1} (M)

for some three-dimensional complex submanifold M of (\R\times S^1)^4

(which is topologically T^4\times \R^4). I would consider this submanifold M as the most adequate description of the set of Solutions. Its parametrizations can be obtained similarly to the parametrizations of F^{-1}(H):

(\lambda,\mu,\nu)\mapsto ([Log(\lambda)],[Log(\mu)},[Log(\nu)],[Log(-\lambda-\mu-\nu)])

P.S. Is there no LaTeX support for Research Gate? (Needed!)

Thank you, Stefan.

One can deduce dimensions using Chris Landauer's suggestion

to go down from "homogeneous equation" or "projective geometry" by dividing by e^{z_4} and getting the lower dimension complex plane w_1+w_2+w_3=1. Other way to lower dimensions is Oktay Duman's suggestion to consider particular solutions in the case z_4=z_3

which deduce 4 dimensional case to 3-dimensional: e^{z_1}+e^{z_2}+e^{z_3+ln2}=0

Octav Olteanu, University Politehnica of Bucharest

In my opinion, the answer of Herbert Homeier solves the problem. Namely, multiplication of your equality by e^it, t real and using complex conjugate-operation, lead to: Sum (e^x_j*e^(i(y^j+t)=0, Sum (e^x_j*e^(-i(y^j+t)=0. Addition and subtracting the two equalities yield the following new equalities in real:

Sum (e^x_j*cos(y^j+t)=0, Sum (e^x_j*sin(y^j+t)=0, t real. A next idea could be the analytic continuation of the preceding relations. Because the sums are finite, we can write: Sum (e^x_j*cos(y^j+z)=0, Sum (e^x_j*sin(y^j+z)=0, z arbitrary complex number. In other words, 0 is the centroid for the polygon having vertices e^y_j if and only if 0 is the centroid of many other polygons, with the same weights (masses) e^x_j, but having different vertices.

Essentially, all these new polygons are obtained from the initial one by means of rotations.

The symmetry with respect to the Ox axes is also used to obtain new "valid" polygons.

Dear Octav,

I think Herbert's answer gives a nice geometric interpretation of the problem:

0 must be the centroid of the corresponding polygons.

But it does not solve the problem, since to solve means to describe all possible polygons with that property. In 2 dimensional case (e^z1+e^z2=0) the description is simple.

Octav Olteanu, University Politehnica of Bucharest

Dear Gro,

by solving the problem I mean characterization in terms of real and imaginary parts, written above by me, based on Herbert's ideas.

The polygons are also written above. They have vertices at points cos(y^_j+z), and sin(y^_j+z), j=1,...,n, z arbitrary complex number. Other polygons have vertices

exp(i*(y^_j+t)), exp(- i*(y^_j+t)), j=1,...,n, t arbitrary real number. The polygons written firstly are obtained from the second ones, so that these last ones are basic. So we have an infinity of basic polygons, each of them having n vertices situated on the unit circle.

Octav Olteanu, University Politehnica of Bucharest

Dear Gro,

The basic polygons have the vertices at points exp(i*(y^_j+t)), exp(-i*(y^_j)+t)), j=1,...,n, t real number (it was a misprint in my first answer).

Dear Octav, Your construction is OK. But more details are needed.

I think parameter "t" does not help much, since it is just rotation of polygon and may be ignored.

Could you add more details in your construction?

For example,

if you draw straight lines that connect possible polygon vertices with the centroid what are possible angles between these lines? For 2 dimensional case possible angles are only 0 and pi. For 3 dimensional case not all angles are possible.

Octav Olteanu, University Politehnica of Bucharest

Dear Gro, I think that all these polygons of vertices exp(i*y^_j), j=1,...,n, and their symmetric polygons of vertices exp(- i*y^j), j=1,...,n are two dimensional. The argument is that all their vertices are on the unit circle. On the other hand, concerning the angles, if we take n=3 and an equilateral triangle with vertices on the unit circle, assuming also that x^_1=x^_2=x^_3, the centroid is the origin. Neighbor wedges of this triangle form angles equal to pi/3, and there are no opposite vertices with respect to the origin. Going back to the general case, one attach to the polygon with the vertices w^_j^1,2=exp(+- i*y^_j), j=1,...,n, the algebraic equation Product (z-w^_j^1,2)=0. This equation can be regarded as the characteristic equation attached to a linear differential equation with constant coefficients.

Octav Olteanu, University Politehnica of Bucharest

Dear Gro, The algebraic equation from above has real coefficients and pairwise complex conjugate roots. But you can take only w^_j=exp(i*y^_j), j=1,...,n and the equation Product (z-w^_j)=0. In this general case, the coefficients are complex numbers. In the particular case of equal masses x^_j, j=1,...,n, the coefficient of

z**(n-1) is vanishing, because of the initial relation, that becomes:

Sum (exp((i*y^_j))=Sum w^_j = 0.

Octav Olteanu, University Politehnica of Bucharest

Dear Gro, concerning the rotations introduced by the real t, I think that it might be important. To fix the ideas, consider the case n = 3 and equal masses exp(x^_j), j=1,2,3. By means of a rotation, we always may assume that one of the vertices of the equilateral triangle is equal to 1. Then the other two vertices are the complex conjugate roots of third degree of the unit. It is very probable that in the general case of n terms, with equal masses exp(x^_j), j=1,...,n, using the rotation and assuming that one of the vertices is the real number 1, the other vertices are the other n-ht roots of the unit. This might be helpful in solving some differential equations, especially when n is odd.

Hi Octav, I like your additional points on unit circle that shows the angles between neighbor edges, that are equal to y_k-y_j=Im(z_k-x_j). From the solution for 3 dimensional case that I mention before when the points e^z_j j=1,2,3 are in general position possible polygons should satisfy the laws of sin:

sin(y_1-y_2)/e^x_3=sin(y_2-y_3)/e^x_1 =sin(y_1-y_3)/e^x_2

and (y_1-y_2)+(y_2-y_3)+(y_1-y_3)=0 mod 2\pi

I am still not sure about domain of angles: geometry says

(y_1-y_2)+(y_2-y_3)+(y_1-y_3)=2\pi

(y_1-y_2)>0,(y_2-y_3)>0,(y_3-y_1)>0

Is it correct domain for the angles between the edges?

What is the geometry for 4 dimensional case?

Octav Olteanu, University Politehnica of Bucharest

Hi Gro, Your idea concerning several dimensions seems interesting, although just now I cannot understand it very well. In my opinion, all the n points exp(z^_j), j=1,...,n are complex numbers, so that all of them are in the same plane. In the next week I will be not in town and the connection to Internet will be unsure. Let us leave our discussion for a while. I will think about when I come back.

As regards the question of dimensionality raised by Stefan, I think he is correct. There are 8 real parameters (all the real and imaginary parts of the four complex z_j), and two real equations, Re(sum)=0 and Im(sum)=0 - I was thinking of only one equation :)... Thus, the dimension of the manifold seems indeed to be 6. Also, the complex mappings are interesting. But the main point that I wanted to make is the possible use of symmetries for the classification of the solutions. Maybe that via the mappings that Stefan described one can obtain the symmetries of the solution manifold by those of C^4 and vice versa. E.g., translations z_j->z_j+w in C^4 lead to a common factor exp(w) which does not vanish, and thus, is irrelevant since the RHS is zero.

One might also think of flows Z_j(t) such that sum_j e^{Z_j(t)} = 0 holds for all t if it holds for t=0, thus demanding Z_j(0)=z_j. By differentiation wrt t, one obtains

sum_j Z'_j exp(Z_j) =0 ( with '=d/dt ) .

If we put d exp(Z_j) = K_j dt with complex constants K_j, with solution

exp(Z_j(t)) = K_j t + exp(z_j)

then sum_j K_j = 0 entails sum_j exp(Z_j(t)) = 0 if sum_j exp(z_j)=0 holds. Thus, for any solution {z_j} and constants K_j with centroid 0, a one-parameter family of solutions, i.e., a flow of solutions is obtained. Normally, flows depend on a real parameter t. But it seems to me that here, also complex t lead to solutions {Z_j(t)}.

Particular solutions {z_j} may be obtained from the roots of unity:

Put exp(z_j-z_1) = w^{j-1}. Then

sum_{j=1..n} exp(z_j - z_1 ) = sum_{m=0..n-1} w^m = (1-w^n)/(1-w) = 0

if w is an n-th root of unity. Factoring exp(-z_1) on the LHS, we see that there is a solution {z_j=z_1 + (j-1) ln w}. Lets put zeta_j(a,w) = a + (j-1) ln w.

An interesting question is, whether one can rewrite every solution in the form

exp(z_j) = K_j t + exp(zeta_j(a,w))

...................................... ...................................................... (1)

with complex t, a, K_j with sum_j K_j = 0 and where w is an n-th root of unity.

Perhaps this question could be answered by dimensionality plus connectivity arguments for the manifolds (all solutions vs. solutions of the form (1))

At least, such solutions of the form (1) yield a nontrivial submanifold of all solutions.

Herbert, I don't understand how you get e^Z_j=t K_j+e^z_j. Maybe you mean e^Z_j=e^(t K_j)e^z_j? Solutions with roots of unity are nice but already mentioned by Octav.

Gro, I mean indeed + t K_j and sum K_j = 0, without exponents. The point is that looking at the flows, it is possible to look at centroids of complex numbers (which is somewhat easier), not at centroids of exponentials of complex numbers. And a single solution generates via the flow for each t a new solution. Due to the additional parameter a in z_j(a,w)= a+(j-1) ln w (there was a little error there in my original post, the ln was missing, I have corrected that), one may choose the starting angle for the roots of unity at will, and also the radius of the circle: for a=a_1 + i a_2, and w = exp(i 2\Pi * (m/n)), m in {0,1,...,n-1)} we have

exp(a+(j-1) ln w)= exp(a_1) exp(i a_2+ i (j-1)m/n)

whence exp(a_1) is the radius, and a_2 gives rise to a phase. Thus, there is somewhat more besides "root of unity".

Of course, one may obtain the flows also by noting that if sum L_j = 0 and sum K_j =0, then also (u sum L_j + v sum K_j) =0 for any complex u, v. Here, L_j corresponds for example to exp(z_j).

One could also say that two solutions Z_j and z_j are equivalent if there are k_j with sum k_j = 0 such that exp(Z_j) - exp(z_j) = k_j for j=1..n.

Of course, this implies sum exp(Z_j) - sum exp(z_j) = sum k_j.

Then, one has to find only a representative for each equivalence class. exp(z_j(a,w)) are representatives of certain classes. Rephrased in this way, one can ask whether there are other classes besides the classes represented by exp(z_j(a,w)). Here, k_j corresponds to t K_j .

Actually, this equivalence relation stuff also works for searching solutions of

sum_{j=1..n} exp(u_j ) = -1 (or any other nonvanishing constant on the RHS). Note, that you can rewrite this as

sum_{j=1..n+1} exp(Zeta_j) = 0 for Zeta_j=u_j for j=1,...,n, Zeta_(n+1) = 0

But this is nothing but an (n+1)-problem with a certain normalization. This could help to increase n inductively, i.e., to relate (n+1)-problems with n-problems, both with RHS=0.

Introducing an auxiliary complex number z_5 such that e^{z_5}=e^{z_3}+e^{z_4} one can reduce 4D equation to two 3D equations: e^{z_1}+e^{z_2}+e^{z_5}=0 and

e^{z_3}+e^{z_4}-e^{z_5}=0.

one comment the question

The zero set X of exp(z_1)+...+exp(z_n$ is a complex analytyc set in C^n.

The complex dim = n-1, the real dim = 2n-2.

This set is pullback of the hyperplain x_1+...x_n=0 in complex torus (C\0)^n

on the exp-map C^n \to (C\0)^n

This is a rather complicated geometric object.

For example, the intersection of X with a complex 1-dim subspace of C^n

is a very old mathematical object

(Ritt G.F. On the zeros of exponential polynomials. -- Trans. Amer. Math. Soc., 31, 1929, 680-686)

See also

On zeros of exponential sums

BY Kazarnovskiı

Dokl. Akad. Nauk SSSR (Russian) 257 (4), 804–808

Newton polyhedra and zeros of systems of exponential sums

BY Kazarnovskii

Functional Analysis and Its Applications 18 (4), 299-307

M. Voorhoeve. A generalization of Descartes' rule. -- J. London Math. Society (2 Series) 20, 1979, 446-456

A. Khovanskii and S. Yakovenko. Generalized Rolle theorem in $\R^n$ and $\C$, Journal of dynamical

and control systems. -- V.2, N.1, 103–123, 1996.

Dear Boris Kazarnovskii,

Thank you for your interesting answer. Could you download your papers that are mentioned in your answer on researchgate.net so I can access them.

It is interesting that in the original statement some solutions are missed!

So the original statement should be corrected:

e^(z_1)+e^(z_2)=0 if and only if Re(z_1-z_2)=0 and Im(z_2-z_1)=pi(2n+1),

or (missed part:)

Re(z_1-z_2)=0 and Im(z_2+z_1)=2k pi, where z_1, z_2 are complex numbers.

Ignore my previous statement. After double checking I found that the original statement for 2D equation is correct.

Dear Boris, I found that your paper deal with much more general object:

namely with the systems of linear combinations of exponents. Since my question is much simpler and has many symmetries the description of this specific set could be simpler that the one considered by you and others. For example the 3 dimensional equation has a simple description.

Hey guys, recently I published a paper (you may find it here) about oscillations of n-th order equations, where I gave thanks to everybody for a good discussion of my questions.