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If these two equations have imaginary roots, then,.
(2.6) We assume that (P_{n}(lambda; tau)) and (Q_{m}(lambda; tau)) cannot have imaginary roots.
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If the first equation has imaginary roots, then, which contradicts with or.
If the second equation has imaginary roots, then, which also contradicts with or.
Have imaginary friends.
After a month or two, your imaginary cat can have imaginary kittens.
Since, that is, this leads to We firstly claim that the equation has no imaginary root, that is, equations and both have no imaginary roots, where.
The necessary condition for a change in stability of the interior equilibrium point (E^ ) is that equation (23) should have purely imaginary roots, that is, stability switches for increasing τ in (I=[0,tau^)) may occur only with a pair of roots (lambda=pm iomega tau)). Let (tau=hat{tau}) be the particular magnitude of τ for which (mu(hat{tau})=0 ) and (omega(hat{tau })=hat{omega} ).
You have two imaginary roots as complex conjugate to each other.
Further, equation (7) has the unique positive root (omega_{n}), and the characteristic equation (4) has purely imaginary roots (pm iomega_{n}), where omega_{n}=sqrt{frac{-P_{n}+sqrt{P_{n}^{2}-4Q_{n}}{{2}}, quad n=0,1,2,ldots, N_{2}.
In a similar way to the one considered above, we get (this time Equation 2.5 has no imaginary roots, so the contour C will only bypass the origin on the right half-plane): (3.19).
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