How to take the linear convolution of these two signals?
For $n = 1\ldots 8$
$$x[n] = e^{j\pi n}\{{u[n]}-u[n-8]\} = (-1)^{n}$$ 
and for $n = 0\ldots 3$
$$h[n] = (-1)^n$$
Else, if $n &gt; 8$ or $n &lt; 1$, then $x[n] = 0$. Similarly, if $n &lt; 0$ and $n &gt; 3$ then $h[n] = 0$. Using the definition of convolution, 
$$y[k] =(h * x)[k] = \sum\limits_{m = 0}^3 h[m]x[k-m] $$
For $k = 1$
$$y[1] =(h * x)[1] = h[0]x[1]  = 1$$

For $k = 2$
$$y[2] =(h * x)[2] = h[0]x[2] + h[1]x[1] = -1 -1 = -2$$
$\vdots$

For $k = 5$
$$y[5] =(h * x)[5] = h[0]x[5] + h[1]x[4] + h[2]x[3] + h[3]x[4]= 1 +1 + 1 +1 = 4$$
$\vdots$

For $k = 8$
$$y[8] =(h * x)[8] = h[0]x[8] + h[1]x[7] + h[2]x[6] + h[3]x[5]= -1 -1 - 1 -1 = -4$$

$\vdots$

For $k = 11$
$$y[11] =(h * x)[11] = h[0]x[11] + h[1]x[10] + h[2]x[9] + h[3]x[8]= 0 + 0 + 0 + 1 = 1$$