How can I reduce this circuit to a source in series with a single impedance?

Question

How can I Thevenize the following circuit looking into the out terminal:

I want to deduce this circuit to a source and a single series impedance.

Here what I tried:

\$\frac{R1\cdot R2}{R1+R2}\$ is in series with C1

This becomes \$\sqrt{\frac{R1\cdot R2}{R1+R2}^2 + \frac{1}{\omega C1}^2 }\$

And now \$\sqrt{\frac{R1\cdot R2}{R1+R2}^2 + \frac{1}{\omega C1}^2 }\$ is in parallel with R5

I cannot proceed more. How could this be achieved if mine is wrong as an alternative way?

is the following correct?: \$\frac{R1\cdot R2}{R1+R2}\$ is in series with C1

\$Z_{thevenin} = \frac {\sqrt{\frac{R1\cdot R2}{R1+R2}^2 + \frac{1}{\omega C1}^2 } \cdot R5} {\sqrt{\frac{R1\cdot R2}{R1+R2}^2 + \frac{1}{\omega C1}^2 } + R5}\$

And the Thevenin voltage needed to be found as well.

Vth = V1 * ((|R5+(1/jwc)| // R2)/ (R1 + (|R5+(1/jwc)| // R2))) * R5/|R5+(1/jwc)|

Answer 1

If I do not make any mistake in the math the correct equation for \$Z_{th}\$ will look like this:

$$Z_{th} = \frac {\sqrt{R5^2 + (\omega\ C_1\ R_T\ R_5)^2}}{\sqrt{1+(\omega\ C_1 (R_T+R_5))^2}}$$

Where:

$$R_T = \frac{R_1 \cdot R_2}{R_1+R_2}$$

And the Thevenin voltage

$$Vth = V_{IN}\cdot\frac{R_2||R_5}{R_1 + R_2||R_5}\cdot \frac{\omega\ (R_T +R_5)C_1}{\sqrt{1 +(\omega\ (R_T + R_5)C_1)^2} }$$