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Упростим многочлен <tex>H(z)</tex>:
: <tex>H(z)=\left( \dfrac{1-z^{10}}{1-z} \right) ^3\left(\dfrac{1-z^{-10}}{1-z^{-1}}\right)^3=\left(\dfrac{2-z^{10}-z^{-10}}{2-z-z^{-1}}\right)^3</tex> и применим замену <tex>z=e^{i\phi}</tex>:
: <tex>p_0=\dfrac{1}{2\pi}\displaystyle\int\limits_0^{2\pi}\left(\dfrac{2-e^{10i\phi}-e^{-10i\phi}}{2-e^{i\phi}-e^{-i\phi}}\right)^3d\phi: <tex>\dfrac{1}{2\pi}\displaystyle\int\limits_0^{2\pi}\left(\dfrac{2-e^{10i\phi}-e^{-10i\phi}}{2-e^{i\phi}-e^{-i\phi}}\right)^3d\phi=\dfrac{1}{2\pi}\displaystyle\int\limits_0^{2\pi}\left(\dfrac{2-2\cos(10\phi)}{2-2cos\phi}\right)^3d\phi: \dfrac{1}{2\pi}\displaystyle\int\limits_0^{2\pi}\left(\dfrac{2-2\cos(10\phi)}{2-2cos\phi}\right)^3d\phi=\dfrac{1}{2\pi}\displaystyle\int\limits_{0}^{2\pi}\left(\dfrac{\sin^2(5\phi)}{\sin^2(\frac{\phi}{2})}\right)^3d\phi: \dfrac{1}{2\pi}\displaystyle\int\limits_{0}^{2\pi}\left(\dfrac{\sin^2(5\phi)}{\sin^2(\frac{\phi}{2})}\right)^3d\phi=\dfrac{1}{\pi}\displaystyle\int\limits_0^{\pi}\left(\dfrac{\sin(10\phi)}{\sin\phi}\right)^6d\phi: \dfrac{1}{\pi}\displaystyle\int\limits_0^{\pi}\left(\dfrac{\sin(10\phi)}{\sin\phi}\right)^6d\phi=\dfrac{1}{\pi}\displaystyle\int\limits_{-\pi/2}^{\pi/2}\left(\dfrac{\sin(10\phi)}{\sin\phi}\right)^6d\phi</tex>
Рассмотрим функцию <tex>f(\phi)=\dfrac{\sin(10\phi)}{\sin\phi}</tex> на <tex>\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]</tex>. Вне отрезка <tex>\left[\dfrac{-\pi}{10},\dfrac{\pi}{10}\right]
