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?
Zeroes of $\sin(z)+2\sin(8z)$
complex-analysisroots
Best Answer
To start, note that your question is analogous to asking whether or not
$$|\sin(x+iy)+2\sin(8x+8iy)|=0$$
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:
$$2g(x,y)=2|\sin(x+iy)+2\sin(8x+8iy)|^2$$
$$=2(\cos (x) \sinh (y)+2 \cos (8 x) \sinh (8 y))^2+2(\sin (x) \cosh (y)+2 \sin (8 x) \cosh (8 y))^2$$
$$=-4 \cos (9 x) \cosh (7 y)+4 \cos (7 x) \cosh (9 y)-\cos (2 x)-4 \cos (16 x)+\cosh (2 y)+4 \cosh (16 y).$$
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 $y=0$.