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Let $g(t) = \left ( \frac{dz}{dt} \right )$

Therefore, $m_k = \left ( \frac {1}{k\pi} \sin ( \frac {k\pi}{2} ) \right) , n_k = \left( \frac {-1}{k\pi} \sin ( \frac {k\pi}{2} ) e^\frac{-j2k\pi2}{4} \right)$

But $g_k = m_k + n_k = \left ( \frac {1}{k\pi} \sin ( \frac {k\pi}{2} ) \right) + \left( \frac {-1}{k\pi} \sin ( \frac {k\pi}{2} ) e^\frac{-j2k\pi2}{4} \right)$

$\therefore g_k = \left ( \frac {1}{k\pi} \sin ( \frac {k\pi}{2} ) \right) + \left( \frac {-1}{k\pi} \sin ( \frac {k\pi}{2} ) (-1)^k \right)$

But we had taken the derivative of z(t) to get g(t) (and hence $g_k$). $\therefore z_k = \left ( \frac{g_k}{jk\omega_o} \right )$

$z_k = \left( \frac { \frac {1}{k\pi} \sin ( \frac {k\pi}{2} ) * (1 - (-1)^k) }{jk\pi/2} \right)$

$z_k = \frac {2}{j} \left( \frac {1}{(k\pi)^2} \sin ( \frac {k\pi}{2} ) \right) * (1 - (-1)^k) ~~\forall ~k ~\ne ~0$

$g_o = \frac {2t_{1m}}{T_m} + \frac {2t_{1n}}{T_n}$

$\therefore g_o = 0.5 - 0.5 ~~~and \therefore z_o = 0$

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