I need some help manipulating a transfer function to a specific form
Firstly, it seems like you have miscalculated the value of Z:

\begin{eqnarray*}
\dfrac{1}{Z} &amp;=&amp; j \omega C_{1} + \dfrac{R_{2}}{1+j \omega R_{2} C_{2}} + \dfrac{1}{R_{3}} \\ &amp;=&amp; j \omega C_{1} + \dfrac{1 + R_{2}R_{3} + j \omega R_{2}C_{2}}{R_{3} + j\omega R_{2}R_{3} C_{2}} \\ &amp;=&amp; \dfrac{1 + R_{2}R_{3} + j\omega (R_{2}C_{2} + R_{3}C_{1}) - \omega^{2} R_{2}R_{3}C_{1}C_{2}}{R_{3} + j\omega R_{2}R_{3} C_{2}}
\end{eqnarray*}

and hence,

\begin{equation*}
Z = \dfrac{R_{3} + j\omega R_{2}R_{3} C_{2}}{1 + R_{2}R_{3} + j\omega (R_{2}C_{2} + R_{3}C_{1}) - \omega^{2} R_{2}R_{3}C_{1}C_{2}}
\end{equation*}

Given this, the Transfer Function turns out to be:

\begin{eqnarray*}
H(j\omega) &amp;=&amp; \dfrac{R_{1}(1 + R_{2}R_{3} + j\omega (R_{2}C_{2} + R_{3}C_{1}) - \omega^{2} R_{2}R_{3}C_{1}C_{2})}{R_{1}+R_{3}+R_{1}R_{2}R_{3} + j\omega (R_{1}R_{2}C_{2} + R_{1}R_{3}C_{1}+R_{2}R_{3}C_{2}) - \omega^{2} R_{1}R_{2}R_{3}C_{1}C_{2})} \\ 
&amp;=&amp; \dfrac{R_{1}(1 + R_{2}R_{3}) \left \{ 1 + j\omega \dfrac{(R_{2}C_{2} + R_{3}C_{1})}{(1 + R_{2}R_{3})} - \omega^{2} \dfrac{(R_{2}R_{3}C_{1}C_{2})}{(1 + R_{2}R_{3})} \right \}}   {(R_{3}+ R_{1}(1 + R_{2}R_{3})) \left \{ 1 + j\omega \dfrac{(R_{1}R_{2}C_{2} + R_{1}R_{3}C_{1}+R_{2}R_{3}C_{2})}{(R_{3}+ R_{1}(1 + R_{2}R_{3}))} - \omega^{2} \dfrac{(R_{1}R_{2}R_{3}C_{1}C_{2})}{(R_{3}+ R_{1}(1 + R_{2}R_{3}))} \right \}}
\end{eqnarray*}

You can now compare this equation to the form that you mentioned (the last equation in your question) and solve for all the T values.