How do I examine if the signal can be z-transformed?
You can use the basic properties of Z-transform to find the result:

Given, $$ x[n] = \sin(\frac{2 \pi}{N} m n) u[n] = \sin(\omega_m n) u[n]  $$

Decompose the sine wave into complex exponentials:

$$ x[n] = \frac{1}{2 j} \left( e^{j \omega_m n} - e^{-j \omega_m n} \right)   u[n] \tag{1}$$

And observe the following :


1 :
$$ U(z) = \mathcal{Z} \{ u[n] \} = \frac{1}{1-z^{-1}} ~~~~,~~~ |z| &gt; 1 $$
2 :


if  $x[n] \longleftrightarrow X(z)$ then 
$$ e^{j \omega_m n} ~ x[n]  \longleftrightarrow X( z ~e^{ -j \omega_m} ) ~~~,~~~\text{ ROC the same} $$

Then apply these two observations to Eq (1) :

$$ \begin{align}
X(z) &amp;= \mathcal{Z} \{ x[n] \} = \mathcal{Z} \{ \frac{1}{2 j} \left( e^{j \omega_m} - e^{-j \omega_m} \right)   u[n] \} \\
&amp;= \frac{1}{2 j} \left( \mathcal{Z} \{  e^{j \omega_m} u[n] \} - \mathcal{Z} \{ e^{-j \omega_m} u[n] \} \right)  \\
&amp;= \frac{1}{2 j} \left(  U( z e^{-j \omega_m} ) - U( z e^{j \omega_m} ) \right)  \\
&amp;= \frac{1}{2 j} \left(  \frac{1}{1-e^{j \omega_m} z^{-1}} - \frac{1}{1-e^{-j \omega_m} z^{-1}}  \right)  \\
\end{align}
$$