Solution:

To demonstrate the eigenfunction property of complex exponentials for discrete-time systems, consider an input sequence $x[n]=e^{j \omega n}$ for $-\infty\ltn\lt\infty$, i.e., a complex exponential of radian frequency $\omega$.

The corresponding output of a linear time-invariant system with impulse response h[n] is,

$$ \begin{aligned}\\ y[n] & =\sum_{k=-\infty}^{\infty} h[k] e^{j \omega(n-k)} \\\\ & =e^{j \omega n}\left(\sum_{k=-\infty}^{\infty} h[k] e^{-j \omega k}\right) ...(1)\\ \end{aligned} $$

If we define,

$$ H\left(e^{j \omega}\right)=\sum_{k=-\infty}^{\infty} h[k] e^{-j \omega k}...(2), $$

Eq. (2) becomes,

$$ y[n]=H\left(e^{j \omega}\right) e^{j \omega n} ...(3). $$

Consequently, $e^{j \omega n}$ is an eigenfunction of the system, and the associated eigenvalue is $H\left(e^{j \omega}\right)$. From Eq. (3), we see that $H\left(e^{j \omega}\right)$ describes the change in complex amplitude of a complex exponential input signal as a function of the frequency $\omega$.

The eigenvalue $H\left(e^{j \omega}\right)$ is called the frequency response of the system. In general, $H\left(e^{j \omega}\right)$ is complex and can be expressed in terms of its real and imaginary parts as,

$$ H\left(e^{j \omega}\right)=H_R\left(e^{j \omega}\right)+j H_I\left(e^{j \omega}\right)...(4) $$

or in terms of magnitude and phase as,

$$ H\left(e^{j \omega}\right)=\left|H\left(e^{j \omega}\right)\right| e^{j H\left(e^{j \omega}\right)} ..(5) $$