I have no clue on how $V_o$ ends up looking like $\frac{V_{in}}{10}\frac{(iw)}{1 + iw(\frac{1}{2})}$ and would like to know how.
My attempt:
I plug in $Z_{iw} = \frac{1}{10}\frac{1}{1 + iw\frac{2}{5}}$ and get $$V_o = V_{in}\frac{\frac{1}{10}\frac{1}{1 + iw\frac{2}{5}}}{\frac{1}{iw}+\frac{1}{10}\frac{1}{1 + iw\frac{2}{5}}}$$ Multiply $\frac{1}{iw}$ in the denominator by $\frac{10}{10}$ to get $$V_o = V_{in}\frac{\frac{1}{10}\frac{1}{1 + iw\frac{2}{5}}}{\frac{10}{10iw}+\frac{1}{10}\frac{1}{1 + iw\frac{2}{5}}}$$ Factor out $\frac{1}{10}$ to get $$V_o = \frac{V_{in}}{10}\frac{\frac{1}{1 + iw\frac{2}{5}}}{\frac{10}{iw}+\frac{1}{1 + iw\frac{2}{5}}}$$ and from there I have no clue what the next step would be to get $$V_o = \frac{V_{in}}{10}\frac{(iw)}{1 + iw(\frac{1}{2})}$$ a step in the right direction would be greatly appreciated
The impedance seen from the output is
$$Z_{out}=\frac1{10+i4\omega}$$
and that from the input
$$Z_{in}=\frac1{i\omega}+Z_{out}.$$
Then the transmittance
$$\frac{Z_{out}}{Z_{in}}=\frac{\dfrac1{10+i4\omega}}{\dfrac1{i\omega}+\dfrac1{10+i4\omega}}=\frac{i\omega}{10+i4\omega+i\omega}=\frac1{10}\frac{i\omega}{1+i\dfrac\omega2}.$$