Clearly the function $f(x)=\sin z+2\sin8z$ has many zeroes on the real line. Does it have any off the real line? I thought of inspecting its real and imaginary parts separately: $$f(x+iy) = (\sin x\cosh y+2\sin8x\cosh8y)+i(\cos x\sinh y+2\cos8x\sinh8y)$$ However, I didn't find this to be very helpful. In the case of just a single sine I'd have $$\sin(x+iy)=(\sin x\cosh y)+i(\cos x\sinh y)$$ and I could show that all zeroes are on the real line by observing that cosh is always positive (as a real function) so I must have $x=\pi k$, so for the imaginary part $0=\cos \pi k\sinh y=\sinh y$ implies $y=0$. However, it's not so simple for the case of $f(z)$ above. Can anyone help me solve this?

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    $\begingroup$ Try letting $q=e^{iz}$ and rewriting the equation as polynomial (of degree $8$) in $q$. $\endgroup$ – Barry Cipra Feb 4 at 22:50
  • $\begingroup$ Of course - thank you! Wouldn't that be a polynomial of degree 16, though, since we'd need to factor out $\frac{1}{q^8}$ from $\sin 8z=q^8+q^{-8}$? $\endgroup$ – BGreen Feb 4 at 23:42
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    $\begingroup$ Yes, sorry, for some reason I jotted it down as $\sin4z$. $\endgroup$ – Barry Cipra Feb 5 at 0:03
  • $\begingroup$ I thought I'd post the following here for anyone reading this in the future. Using your argument I can show that there are at most 16 "distinct" (meaning $z$ is considered the same as $z+2\pi$) zeroes, which I at first thought meant I had overlooked nonreal zeroes. However, I just realized that, on the contrary, it means that if I can find 16 distinct real zeroes then there cannot possibly be any imaginary zeroes. After that, it's easy to see that it has 16 distinct real zeroes, proving that there are no imaginary zeroes. $\endgroup$ – BGreen Feb 5 at 0:13
  • $\begingroup$ Excellent observation. You might consider posting it as a self-contained answer. $\endgroup$ – Barry Cipra Feb 5 at 2:42

To start, note that your question is analogous to asking whether or not


for some nonzero $y$. Of course, we may as well square this in order to simplify our question somewhat (for the sake of notation, call this $g(x,y)$). Therefore, let us expand this function out:


$$=2(\cos (x) \sinh (y)+2 \cos (8 x) \sinh (8 y))^2+2(\sin (x) \cosh (y)+2 \sin (8 x) \cosh (8 y))^2$$

$$=-4 \cos (9 x) \cosh (7 y)+4 \cos (7 x) \cosh (9 y)-\cos (2 x)-4 \cos (16 x)+\cosh (2 y)+4 \cosh (16 y).$$

Thus, if we can show that $g(x,y)$ is $0$ only if $y$ is $0$, then we are done. In order to do this, consider the partial derivative with respect to $y$:

$$\frac{\partial}{\partial y}\left[-4 \cos (9 x) \cosh (7 y)+4 \cos (7 x) \cosh (9 y)-\cos (2 x)-4 \cos (16 x)+\cosh (2 y)+4 \cosh (16 y)\right] $$

$$=2 (-14 \cos (9 x) \sinh (7 y)+18 \cos (7 x) \sinh (9 y)+\sinh (2 y)+32 \sinh (16 y)).$$

For the sake of notation, call this function $f(x,y)$. Now, if $y>0$, then

$$\frac{1}{2}f(x,y)=-14 \cos (9 x) \sinh (7 y)+18 \cos (7 x) \sinh (9 y)+\sinh (2 y)+32 \sinh (16 y)$$

$$\geq -14 \sinh (7 y)-18 \sinh (9 y)+\sinh (2 y)+32 \sinh (16 y)$$

$$>-14 \sinh (9 y)-18 \sinh (9 y)+\sinh (2 y)+32 \sinh (9 y)$$ $$=\sinh(2y)>0.$$

Now, note that $f(x,-y)=-f(x,y)$. Thus, for $y<0$, $f(x,y)<0$. Putting it all together, we have a function $g(x,y)$ which has the following properties:

$$g(x,y)\geq 0,$$

$$\frac{\partial}{\partial y}g(x,y)>0\text{ for } y>0,$$

$$\frac{\partial}{\partial y}g(x,y)<0\text{ for } y<0.$$

We conclude that if $g(x,y)=0$, then $y=0$.

  • $\begingroup$ Thank you! I must ask, did you have intuition that it would be increasing (decreasing) whenever $y>0$ ($y<0$) before you began? I had thought that there was a possibility of the two sines being out-of-phase and canceling somewhere, so I wasn't expecting an argument of that nature to work. $\endgroup$ – BGreen Feb 5 at 0:23
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    $\begingroup$ Not so much intuition, but the first thing I did was graph the absolute value as a function of $x$ and $y$ (just to check if your conjecture was true). From there, it was easy to see that this method would work. $\endgroup$ – Nick Guerrero Feb 5 at 1:17

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