transmission line - Derivation of Characteristic Impedance?

I start from the telegrapher's equation: $-\frac{dV(z)}{dz}=(R'+j\omega L')I(z)$, where $V(z)$ and $I(z)$ are the phasors of voltage and current respectively, in the transmission line model. $R'$ and $L'$ are resistance per unit length and inductance per unit length respectively.

The solution to the wave equation $\frac{d^2V(z)}{dz^2}-\gamma^2V(z)=0$ where $\gamma=\sqrt{(R'+j\omega L')(G'+j\omega C')} $ has the form $V(z)=V_o^+e^{-\gamma z}+V_o^-e^{\gamma z} $. $G'$ and $C'$ are respectively the conductance per unit length and capacitance per unit length of the transmission line.

From the telegrapher's equation we get: $-\frac{dV(z)}{dz}=\gamma V_o^+e^{-\gamma z}-\gamma V_o^-e^{\gamma z}=\gamma (V_o^+e^{-\gamma z}-V_o^-e^{\gamma z})=(R'+j\omega L')I(z)$

$I(z)=\frac{\gamma}{R'+j\omega L'}(V_o^+e^{-\gamma z}-V_o^-e^{\gamma z})$

...and I'm stuck here.

Given that characteristic impedance $\frac{V_o^+}{I_o^+}=Z_o=\frac{V_o^-}{I_o^-}$, how do I arrive at $Z_o=\frac{R'+j\omega L'}{\gamma}=\sqrt{\frac{R'+j\omega L'}{G'+j\omega C'}}$?

I'm not sure how to get $Z_o$ from $\frac{V_o^+e^{-\gamma z}+V_o^-e^{\gamma z}}{I(z)}=\frac{V_o^+e^{-\gamma z}-V_o^-e^{\gamma z}}{I_o^+e^{-\gamma z}+I_o^-e^{\gamma z}}$