Sampling H(z) to get DFT
When grabbing the first 10 samples of $h_1[n]$ you are essentially multiplying the whole sequence $h_1[n]$ with the rectangular window

$$
  w[n] = 
\begin{cases} 
  1 &amp; 0 \leq n &lt; 10 \\
  0 &amp; \mbox{otherwise}
\end{cases}
$$

whose DFT is given by:

$$
\begin{align}
  W\left[k\right]
    &amp;= \sum_{n=0}^{9} e^{-j 2\pi k n / N} \\
    &amp;= \frac{1 - e^{-j 2\pi k \cdot 10/15}}{1 - e^{-j 2\pi k / 15}}
\end{align}
$$

By the circular convolution theorem:


  Multiplication in the discrete-time domain becomes circular convolution in the discrete-frequency domain. Circular convolution in the discrete-time domain becomes multiplication in the discrete-frequency domain.


In other words, 

$$
  \mbox{DFT}\left\{h_1[n] \cdot w[n]\right\}
  = \frac{1}{N} H_1\left[k\right] \otimes W\left[k\right]
$$

where $\otimes$ is the circular convolution operator.
This can be seen with the following matlab script (you can play around with h1 to see what happens with other waveforms):

n = [0:14]; % discrete-time index
k = [0:14];
h1 = sin(pi*n/14); % some arbitrary function

% window function
w = [ones(1,10) zeros(1,5)];

% time domain operation
h2  = h1.*w;
% convert to frequency domain
H2t = fft(h2);

% frequency domain operation
W   = fft(w);
H1  = fft(h1);
H2f = cconv(W,H1) / 15;

% Compare H2t &amp; H2f
figure(1);
subplot(2,1,1);
hold off; plot(k, abs(H2t), 'b');
hold on;  plot(k, abs(H2t), 'bx');
hold on;  plot(k, abs(H2f), 'ro');
title('Magnitude');
subplot(2,1,2);
hold off; plot(k, angle(H2t), 'b');
hold on;  plot(k, angle(H2t), 'bx');
hold on;  plot(k, angle(H2f), 'ro');
title('Phase');




Note that if $h_1[n]$ is already no longer than 10 samples (i.e. if those truncated samples are already zero), then windowing in the time-domain will have no effect and similarly the circular convolution with $W[k]$ in the frequency domain will happen to produce $H_1[k] = \frac{1}{N} H_1[k] \otimes W[k]$ (but only as a special case).