(I am not sure what part is throwing you, please ask for elaboration in the comments and I will do my best to explain.)
The usual convention is to use $j = \sqrt{-1}$ to avoid confusion with the symbol for current, $i$.
If the supply voltage is $v(t) = \sqrt{2} \operatorname{Re} V_{\text{sup}} e^{j \omega t}$ and the steady state current is $i(t) = \sqrt{2} \operatorname{Re} I e^{j \omega t}$, then the instantaneous power delivered to the circuit is (as is usual convention, $V_{\text{sup}}, I \in \mathbb{C}$ are RMS quantities, also note that $\operatorname{Re} Z = \frac{1}{2}(Z+\overline{Z})$):
\begin{eqnarray}
p(t) &=& v(t)i(t) \\
&=& 2(\operatorname{Re} V_{\text{sup}} e^{j \omega t})(\operatorname{Re} I e^{j \omega t}) \\
&=& 2 \frac{1}{2}(V_{\text{sup}} e^{j \omega t}+ \overline{V_{\text{sup}}} e^{-j \omega t}) \frac{1}{2}(I e^{j \omega t}+ \overline{I} e^{-j \omega t}) \\
&=& \frac{1}{2}( V_{\text{sup}} I e^{j 2 \omega t} +
\overline{V_{\text{sup}}} I +
V_{\text{sup}} \overline{I} +
\overline{V_{\text{sup}}} \overline{I} e^{-j 2 \omega t}) \\
&=& (\operatorname{Re} V_{\text{sup}} I e^{j 2 \omega t}) + (\operatorname{Re} \overline{V_{\text{sup}}} I)
\end{eqnarray}
from which we see that the average power (integrating $t \mapsto e^{j 2 \omega t}$ over a cycle results in zero) is given by $P = \operatorname{Re} \overline{V_{\text{sup}}} I$.
Since the circuit is linear (lumped, time invariant, etc.), the voltage and current are related by $V_{\text{sup}} = I Z(j\omega)$.
In this case, $Z(j\omega) = R + \frac{1}{j \omega C}$, so we obtain the average power delivered to the circuit is
\begin{eqnarray}
P &=& \operatorname{Re} \overline{V_{\text{sup}}} I \\
&=& \operatorname{Re} \frac{|V_{\text{sup}}|^2 }{Z(j\omega)} \\
&=& |V_{\text{sup}}|^2 \operatorname{Re}( \frac{1}{R} \frac{(\omega R C)^2 + j \omega R C}{1+(\omega R C)^2}) \\
&=& \frac{|V_{\text{sup}}|^2}{R} \frac{(\omega R C)^2}{1+(\omega R C)^2}
\end{eqnarray}
Now consider the average power delivered to the resistor: $P_R = \operatorname{Re} \overline{V_{\text{R}}} I$. Since the capacitor and resistor form a voltage divider, we have $V_{\text{R}} = V_{\text{sup}} \frac{R}{Z(j \omega)}$, from which we get $P_R = \operatorname{Re} \overline{V_{\text{sup}} \frac{R}{Z(j \omega)}} \frac{V_{\text{sup}}}{Z(j \omega)} = R \frac{|V_{\text{sup}}|^2}{|Z(j \omega)|^2} = P $. Hence the average power delivered to the circuit is the same as the average power delivered to the resistor.
For the computational part, we use $P= \frac{|V_{\text{sup}}|^2}{R} \frac{(\omega R C)^2}{1+(\omega R C)^2} = \frac{|V_{\text{sup}}|^2}{R} \frac{1}{1+\frac{1}{(\omega R C)^2}}$, with $|V_{\text{sup}}| = 110$, $\omega = 120 \pi$, $C=1 \mu F$, $R=1k$ to get $P \approx 1.5057 W$.
(Contrast this with the $> 12 W$ power that would be delivered if the capacitor was removed.)
