You need to evaluate $H(z)$ on the unit circle $z=e^{iw}$ in order to get the frequency response (assuming that the system is stable, i.e. the region of convergence contains the unit circle). But your $\mathcal{Z}$-transform looks a bit unusual because for causal signals (or filter impulse responses) you should get negative powers of $z$:
$$H(z)=\sum_{n=0}^{\infty}x(n)z^{-n}$$
In your case $x(n)=c^n$, $n\ge 0$, so you get
$$H(z)=\frac{1}{1-cz^{-1}}=\frac{z}{z-c}\tag{1}$$
From (1) you see that the pole is at $z_{\infty}=c$. Note that you need to determine the poles of $H(z)$ for general $z$, not on the unit circle $z=e^{iw}$ because stable systems cannot have poles on the unit circle (this is also why you got a complex frequency, which does not make sense). The frequency response is
$$H(e^{iw})=\frac{1}{1-ce^{-iw}}$$
It is true that for real filter coefficients you get complex conjugate pairs of poles and zeros. But this is not the whole truth. You can also get real-valued poles and zeros, and for a first-order system like yours, this must be the case. You need at least a second-order system to get a complex pole pair. In your case you simply got one real pole (assuming that $c$ is a real-valued constant).