How did you get $\hat{I}_n$ in the last Fourier analysis problem? Why was it negative in $e^{-j\phi_n}$?

The $\hat{I}_n$ for a particular Fourier component is found from $\hat{V}/\hat{Z}$. We wrote $\hat{Z}= R+ j \omega_n L + \frac{1}{j\omega_n C}$ in polar form, $\hat{Z} = \vert\hat{Z}\vert e^{j\phi_n}$, where the magnitude is $\vert\hat{Z}\vert = \sqrt{R^2+\left(\omega_n L - \frac{1}{\omega_n C}\right)^2}$ and the phase is $\phi_n = \tan^{-1}\left(\frac{\omega_n L - \frac{1}{\omega_n C}}{R}\right)$. The $e^{-j\phi_n}$ shows up with a negative exponent in the expression for $\hat{I}_n$ because $e^{j\phi_n}$ is in the denominator.