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 $y=0$.
The left side has expressions of the form $\cos\theta + i \sin\theta,$ which can be written as $e^{i\theta}$ by Euler's formula:
$$e^{ix}-e^{inx}e^{ix} \over {1-e^{ix}}$$
$$={ e^{ix}(1-e^{inx}) \over {1-e^{ix}}}$$
$$={ e^{ix}e^{ {inx/2}}(e^{{-inx}/2}-e^{inx/2}) \over {e^{ix/2}(e^{-ix/2}-e^{ix/2})}}.$$
$$={{e^{ix/2}e^{inx/2}(-2i\sin{\frac {nx} 2)}}\over{-2i\sin{\frac x2}}}$$
$$=e^{i(n+1)x/2}{\frac{\sin(\frac {nx}2)}{\sin (\frac x2)}}$$
$$={\frac{\sin(\frac{nx}2)}{\sin(\frac x2)}}\biggl(\cos \biggl({\frac {(n+1)x} 2\biggr) +i\sin\biggl({\frac {(n+1)x} 2}}\biggr)\biggr)$$
That's basically it, right?
Best Answer
$$e^{iz_1}-e^{-iz_1}=e^{iz_2}-e^{-iz_2}$$
Denote by $u_l = e^{iz_l}$. Then $u_1 - \frac{1}{u_1} = u_2 - \frac{1}{u_2} = s$ and therefore $u_1$, $-\frac{1}{u_1}$ and $u_2$, $-\frac{1}{u_2}$ are roots of the equation $$\lambda ^2 - s \lambda -1=0$$ and therefore $$\{ u_1, -\frac{1}{u_1}\} = \{ u_2, -\frac{1}{u_2}\} $$
Therefore, $u_1 = u_2$ or $u_1 \cdot u_2 = -1$ which for $z_1$, $z_2$ means
$$z_1 - z_2 = 2 k \pi $$ or $$z_1 + z_2 = (2 k+1) \pi$$
which, given the restrictions on $\mathcal{Re}(z)$ implies $z_1 = z_2$.