This is the typical example where the Fast Analytical Circuits Techniques (FACTs) will help you get the answer without writing a line of algebra. Look at your circuit for s = 0: remove all caps. The dc transfer function is a simple resistive divider: H0 = R2/(R1+R2). What is the time constant of this circuit? Reduce the excitation source (Vin) to 0 V and look at the resistance driving the capacitors. As you can see, both capacitors come in // forming a single cap. equal to C1+C2. This is what we can a degenerate case (1 single independent state variable despite the 2 energy-storage elements). The time constant is Tau = (R1||R2)(C1+C2). So the pole is simply the inverse of Tau: Wp = 1 / ((R1||R2)(C1+C2)). Is there a zero in this circuit? Yes, if the impedance made Z by C1 and R1 becomes infinite a the zero frequency, the response disappears and this is your zero. What is the pole of this network (the value of s for which Z becomes infinite)? The time constant is R1C1, then the zero of this circuit is 1/R1C1. The complete transfer function in a clean and ordered form is thus:

H(s) = H0 * (1+s/Wz) / (1+s/Wp) with

H0 = R2/(R1+R2), Wp = 1 / ((R1||R2)(C1+C2)) and Wz = 1/(R1C1).

This where FACTs lead you to, no algebra, just inspection for these simple passive circuits.

The answer given by the gentleman before is valid, but factor R2 in the numerator R2*(1+sR1C1), then s in the denominator and then R1+R2. You obtain the same expression as in the above. This is a low-entropy form in which you see a dc gain (the leading term), a zero in the numerator and a pole in the denominator.

More details in the 2016 APEC presentation available here:

http://cbasso.pagesperso-orange.fr/Downloads/PPTs/Chris%20Basso%20APEC%20seminar%202016.pdf

This is the typical example where the Fast Analytical Circuits Techniques (FACTs) will help you get the answer without writing a line of algebra. Look at your circuit for \$s=0\$: remove all caps. The dc transfer function is a simple resistive divider: \$H_0=\frac{R2}{R1+R2}\$. What is the time constant of this circuit? Reduce the excitation source (\$V_{in}\$) to 0 V and "look" at the resistance driving the capacitors. As you can see, both capacitors come in // forming a single cap. equal to \$C_1+C_2\$. This is what we designate as a degenerate case (1 single independent state variable despite the 2 energy-storage elements). The time constant is \$\tau=(R_1||R_2)(C_1+C_2)\$. So the pole is simply the inverse of \$\tau\$: \$\omega_p=\frac{1}{(R_1||R_2)(C_1+C_2)}\$. Is there a zero in this circuit? Yes, if the impedance \$Z\$ made of \$C_1\$ and \$R_1\$ becomes infinite at the zero frequency, the response disappears and this is your zero. What is the pole of this network (the value of \$s\$ for which \$Z\$ becomes infinite)? The time constant is \$R_1C_1\$, then the zero of this circuit is \$\frac{1}{R_1C_1}\$. The complete transfer function in a clean and ordered form is thus:

\$H(s)=H_0\frac{1+s/\omega_z}{1+s/\omega_p}\$ with

\$H_0=\frac{R2}{R1+R2}\$, \$\omega_p=\frac{1}{(R1||R2)(C1+C2)}\$ and \$\omega_z=\frac{1}{R_1C_1}\$

This where FACTs lead you to, no algebra, just inspection for these simple passive circuits.

The answer given by the gentleman before is valid, but factor \$R_2\$ in the numerator \$R_2(1+sR_1C_1)\$, then \$s\$ in the denominator and then \$R_1+R_2\$. You obtain the same expression as in the above. This is a low-entropy form in which you see a dc gain (the leading term), a zero in the numerator and a pole in the denominator.

More details in the 2016 APEC presentation available here:

http://cbasso.pagesperso-orange.fr/Downloads/PPTs/Chris%20Basso%20APEC%20seminar%202016.pdf