Steady state value of a complex convolution I am trying few problems on the introductory part of DSP. One of the problem asks to calculate the steady state response of a system with impulse response $h[n] = (\frac{j}{2})^{n} u[n] $ to an input signal $x[n]=\cos(\pi n) u[n]$. However, I have no idea how to handle this complex $j$ in the convolution. I want someone to confirm if my solution is correct or not. If it is incorrect please tell me where I am getting wrong.

I start with calculating the Z transform of $h[n]$.
$$\begin{align}
H(z) &amp;= \sum_{n=0}^{\infty}h[n] \, z^{-n} \\
 &amp;= \sum_{n=0}^{\infty}\left(\frac{j}{2}\right)^{n} \, z^{-n} \\
 &amp;= \frac{1}{1-\frac{j}{2}z^{-1}} \\
\end{align}$$ 
such that $|z| &gt; \frac{1}{2}$.

Now the z transform of $x[n]$ is ,
$$\begin{align}
X(z) &amp;= \sum_{n=0}^{\infty}\cos(\pi n)z^{-n} \\
 &amp;= \sum_{n=0}^{\infty}\Big[\frac{e^{j\pi n}+e^{-j\pi n}}{2}\Big]z^{-n} \\
 &amp;= \frac{1}{2} \left(\frac{1}{1-e^{j\pi}z^{-1}}+\frac{1}{1-e^{-j\pi}z^{-1}}\right) \\
 &amp;=\frac{1}{1+z^{-1}} \\
\end{align}$$

The convolution output in the z domain is $Y(z)$.
$$Y(z) = H(z)X(z) = \frac{2}{(2-jz^{-1})(1+z^{-1})}$$
Applying the final value theorem, I get, steady state output $y_{ss}$ as 
$$y_{ss} = \lim_{z \to 0} \, zY(z) = 0$$.

I definitely think that I am doing something wrong over here. Please help. I don't know how to handle this $j$ in the convolution operation like this, please help me in that too.
You can't apply the final value theorem here. First of all, the correct form of the final value theorem for the $\mathcal{Z}$-transform is

$$\lim_{n\to\infty}y[n]=\lim_{z\to 1}(z-1)Y(z)\tag{1}$$

Eq. $(1)$ assumes that the poles of $(z-1)Y(z)$ are inside the unit circle, which is not the case in your example. In your case, the limit $(1)$ does not exist.

What you have to do is find the frequency response of the system, which is just the transfer function evaluated on the unit circle $z=e^{j\omega}$ (if it converges on the unit circle). Note that the region of convergence (ROC) you've found does not make sense because $|z|$ must be a real number. The requirement for the sum to converge is

$$\left|\frac{j}{2z}\right|&lt;1\quad\Longrightarrow\quad |z|&gt;\frac12\tag{2}$$

Since $H(z)$ converges on the unit circle $|z|=1$, the frequency response is given by

$$H(e^{j\omega})=\frac{1}{1-\frac{j}{2}e^{-j\omega}}\tag{3}$$

An LTI system's response to a sinusoidal input $x[n]=\cos(\omega_0n)$ is

$$y[n]=|H(e^{j\omega_0})|\,\cos[\omega_0n+\phi(\omega_0)]\tag{4}$$

where $\phi(\omega)$ is the system's phase response:

$$H(e^{j\omega})=|H(e^{j\omega})|e^{j\phi(\omega)}\tag{5}$$

So in the limit, after all transients have died out, the system's (steady-state) response to a switched sinusoidal input $x[n]=\cos(\omega_0n)u[n]$ is given by $(4)$. What is left now is to determine the magnitude and phase of $H(e^{j\omega})$ given by $(3)$ and plug these values into Eq. $(4)$.