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Let $c_{L},c_{T},\omega$ be positive constants with $c_{L}>c_{T}$. Define

$p=\sqrt{\frac{\omega^{2}}{c_{L}^{2}}-\xi^{2}}\qquad q=\sqrt{\frac{\omega^{2}}{c_{T}^{2}}-\xi^{2}}$

Consider the function $D_{S}\left(\xi\right)$ defined as follows:

$D_{S}=4\xi^{2}pq\sin p\cos q+\left(\xi^{2}-q^{2}\right)^{2}\cos p\sin q$

The function $D_{S}(\xi)$ has some zeros in the real axis. I need to know if all the zeros are on the real axis. My question is: Is it true that $D_{S}(\xi)$ has no zeros on the upper-half of the complex plane? I.e., that $D_{S}(\xi)\neq0$ whenever $\text{Im }\xi>0$? Thanks.

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    See also http://math.stackexchange.com/questions/67842/proving-that-zeros-of-a-function-are-first-order and http://math.stackexchange.com/questions/55872/numerical-computation-of-the-rayleigh-lamb-curves for related questions.2011-10-03

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Assuming you're using the principal branch of the square root, the answer appears to be that there can be many such zeros. Take $\omega = c_L = 1$, $c_T = 1/2$. Then, for example, there is a zero at approximately $\xi = 1.129125082+1.323674424 i$.