Evaluating the inverse $\mathcal Z$-transform on the unit circle
Hint:

With $z = e^{j\omega}$, you have:

\begin{align} 
\displaystyle \frac{dz}{d\omega} &amp;= je^{j\omega}\\
\Rightarrow dz &amp;= je^{j\omega} d\omega
\end{align}

With $-\pi \leq \omega \leq \pi$, your contour integral goes from $-\pi$ to $\pi$.


  Can somebody tell me intuitively how to simplify this equation to look
  like the inverse Fourier transform?


Plugging all the above should do the job.


  Is my confusion due to a lack of understanding of the contour
  integral?


It is more the change of variables.

EDIT:

Note that $\omega$ is taken in the fundamental interval $\left[-\pi, \pi\right]$ because $X(\omega)$ is periodic with period $2\pi$. So, $\displaystyle \int_c$ becomes $\displaystyle \int_{2\pi}$