Let us consider $$f(z):=\sum\limits_{j=1}^{j=n}a_j\sin(\lambda_jz) $$ where all $a_j$ and $\lambda_j$ (of course, $\lambda_j$ are distinct) are real numbers and $n \in \mathbb{Z},\, n \ge 3$. The question is: are all the zeroes of $f(z)$ real?

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    $\begingroup$ Obviously not. Any 3 complex numbers are linearly dependent over reals. So, as long as $n\geq 3$ you can choose non-zero $a_j$ to make your expression zero at any given complex number. $\endgroup$ – Oleg Eroshkin Feb 18 '20 at 14:16
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    $\begingroup$ I think also for $n=2$ there are counter examples: $\sin z+3\sin(z/2)=0$ for $z=2\pi+2i\log(\tfrac{3}{2}+\tfrac{1}{2}\sqrt{5})$ $\endgroup$ – Carlo Beenakker Feb 18 '20 at 14:22
  • $\begingroup$ @Carlo Beenakker: Thank you. How about $n \ge 3$? I refined my question. $\endgroup$ – user64494 Feb 18 '20 at 14:43
  • $\begingroup$ @Oltg Eroshkin: Can you kindly elaborate your comment in details? $\endgroup$ – user64494 Feb 18 '20 at 14:44
  • $\begingroup$ there are complex roots for any $n$; for example, for $n=3$ try $\sin z+\tfrac{1}{2}\sin 2z+\sin(3z/2)$, which vanishes at $z=4.57903+0.785214 i$. $\endgroup$ – Carlo Beenakker Feb 18 '20 at 14:58

Note that $\sinh$ is a strictly increasing positive unbounded function on $(0,\infty)$ so for say $n \ge 3$, the equation $\sinh 1+ \sinh 2+...\sinh (n-1) =\sinh x$ has a unique positive solution $x_n> n-1$.

Using that $\sin {iy}=i\sinh y$ for real $y$, the above means that $i$ is a root of the equation $\sin z+\sin 2z +..\sin {(n-1)z}-\sin x_nz=0$

For $n=2$ we can adapt this and use that $2\sinh 1=\sinh x_2$ for a unique $x_2>1$ and get that $i$ is a root of $2\sin z-\sin x_2z$


Proof of the existence of a complex root in the general case, expanding Oleg's remark:
Think of the $n$ complex numbers $\sin(\lambda_j z)$ as vectors in the 2D plane; for $n\geq 3$ and fixed $z$ add the first $n-2$ of these vectors with coefficients $a_j\neq 0$ such that the sum is neither a real multiple of $\sin\lambda_n z$ nor of $\sin\lambda_{n-1}z$; then adjust $a_n\neq 0$ and $a_{n-1}\neq 0$ to cancel the full sum.

an example for $n=10$ (requested by the OP): $$\sum_{n=1}^8 \sin nz +\tfrac{1}{2}\sin 9z+\sin 10z=0$$ for $z=0.6142+0.0627\,i$.

To demonstrate that this zero is not a numerical artefact, here is a plot in the complex plane of contours where the real part of the sum vanishes (orange) and of contours where the imaginary part vanishes (blue); the contours intersect, demonstrating the existence of a complex root.

The Mathematica code (contour plot in the complex plane of the roots of the real and imaginary parts of sum of sines) :
ContourPlot[{Cosh[y] Sin[x] + Cosh[2 y] Sin[2 x] + Cosh[3 y] Sin[3 x] + Cosh[4 y] Sin[4 x] + Cosh[5 y] Sin[5 x] + Cosh[6 y] Sin[6 x] + Cosh[7 y] Sin[7 x] + Cosh[8 y] Sin[8 x] + 1/2 Cosh[9 y] Sin[9 x] + Cosh[10 y] Sin[10 x] == 0, Cos[x] Sinh[y] + Cos[2 x] Sinh[2 y] + Cos[3 x] Sinh[3 y] + Cos[4 x] Sinh[4 y] + Cos[5 x] Sinh[5 y] + Cos[6 x] Sinh[6 y] + Cos[7 x] Sinh[7 y] + Cos[8 x] Sinh[8 y] + 1/2 Cos[9 x] Sinh[9 y] + Cos[10 x] Sinh[10 y] == 0}, {x, 0.614, 0.6145}, {y, 0.06265, 0.06275}, WorkingPrecision -> 30]

the OP also requests an example for arbitrary $n$, let me take $z=\pi +i$ and $\lambda_k=k$, and all coefficients $a_k$ equal to unity except $a_n=a$, $$S_{n}=\sum_{k=1}^{n-1}\sin k(\pi +i)+a\sin[n(\pi+i)]=$$ $$\qquad\qquad=-\frac{i}{2 \cosh\left(\frac{1}{2}\right)} \left[(-1)^{n} \sinh \left(n-\tfrac{1}{2}\right)+\sinh \left(\tfrac{1}{2}\right)\right]+a(-1)^ni\sinh n,$$ which vanishes for a nonzero real $a$.

  • $\begingroup$ Thank you for you interest to the question. Hope you understand (these are nuts and bolts of numerical analysis), that $z=0.614+0.0627\,i$ for $\sum_{n=1}^8 \sin nz +\tfrac{1}{2}\sin 9z+\sin 10z=0 $ proves nothing even in view of its numerical verification with Maple: eval(add(sin(j*z), j = 1 .. 8) + 1/2*sin(9*z) + sin(10*z), z = 0.614 + 0.0627*I) which performs $- 0.0001542513- 0.0016096971\,i $. I will be waiting for a more solid answer. Thank you anyway. $\endgroup$ – user64494 Feb 18 '20 at 16:31
  • $\begingroup$ Also the command of Maple Digits := 15; RootFinding:-Analytic(add(sin(j*z), j = 1 .. 8) + 1/2*sin(9*z) + sin(10*z), z = -1 - I .. 1 + I); performs $ 0+0\,i$, not confirming your statement. $\endgroup$ – user64494 Feb 18 '20 at 16:34
  • $\begingroup$ Hope you understand that a plot without a command is not any argument. $\endgroup$ – user64494 Feb 18 '20 at 16:39
  • $\begingroup$ +1 for the idea, Thank you for the added code. I can reproduce it. The matter is interesting: two different CASes produce different results. We still havt no counterexample for $n=10$. $\endgroup$ – user64494 Feb 18 '20 at 17:10
  • $\begingroup$ Can you ground your statement " it's just not the only root."? I repeat numerical calculations do not confirm $z=0.6142+0.0627\,i$, producing $ - 0.0001542513- 0.0016096971\,i$ instead of zero. $\endgroup$ – user64494 Feb 18 '20 at 17:18

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