Well, notice that the input impedance of your circuit is given by:
\begin{equation} \begin{split} \underline{\text{Z}}_{\space\text{i}}\left(\omega\right)&=\underline{\text{Z}}_{\space\text{R}_1}+\left[\left(\underline{\text{Z}}_{\space\text{C}}\space\text{||}\space\underline{\text{Z}}_{\space\text{R}_2}\right)\space\text{||}\space\left(\underline{\text{Z}}_{\space\text{R}_3}+\underline{\text{Z}}_{\space\text{L}}\right)\right]\\ \\ &=\text{R}_1+\frac{\displaystyle1}{\displaystyle\frac{\displaystyle1}{\displaystyle\frac{\displaystyle1}{\displaystyle\text{j}\omega\text{C}}}+\frac{\displaystyle1}{\displaystyle\text{R}_2}+\frac{\displaystyle1}{\displaystyle\text{R}_3+\text{j}\omega\text{L}}}\\ \\ &=\text{R}_1+\frac{\displaystyle1}{\displaystyle\text{j}\omega\text{C}+\frac{\displaystyle1}{\displaystyle\text{R}_2}+\frac{\displaystyle1}{\displaystyle\text{R}_3+\text{j}\omega\text{L}}}\\ \\ &=\text{R}_1+\frac{\displaystyle1}{\displaystyle\text{j}\omega\text{C}+\frac{\displaystyle1}{\displaystyle\text{R}_2}+\frac{\displaystyle1}{\displaystyle\text{R}_3+\text{j}\omega\text{L}}}\cdot\frac{\displaystyle\text{R}_3+\text{j}\omega\text{L}}{\displaystyle\text{R}_3+\text{j}\omega\text{L}}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_3+\text{j}\omega\text{L}}{\displaystyle\text{j}\omega\text{C}\left(\text{R}_3+\text{j}\omega\text{L}\right)+\frac{\displaystyle\text{R}_3+\text{j}\omega\text{L}}{\displaystyle\text{R}_2}+\frac{\displaystyle\text{R}_3+\text{j}\omega\text{L}}{\displaystyle\text{R}_3+\text{j}\omega\text{L}}}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_3+\text{j}\omega\text{L}}{\displaystyle\text{j}\omega\text{C}\left(\text{R}_3+\text{j}\omega\text{L}\right)+\frac{\displaystyle\text{R}_3+\text{j}\omega\text{L}}{\displaystyle\text{R}_2}+1}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_3+\text{j}\omega\text{L}}{\displaystyle\text{j}\omega\text{C}\left(\text{R}_3+\text{j}\omega\text{L}\right)+\frac{\displaystyle\text{R}_3+\text{j}\omega\text{L}}{\displaystyle\text{R}_2}+1}\cdot\frac{\displaystyle\text{R}_2}{\displaystyle\text{R}_2}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\left(\text{R}_3+\text{j}\omega\text{L}\right)}{\displaystyle\text{j}\omega\text{CR}_2\left(\text{R}_3+\text{j}\omega\text{L}\right)+\frac{\displaystyle\text{R}_2\left(\text{R}_3+\text{j}\omega\text{L}\right)}{\displaystyle\text{R}_2}+\text{R}_2}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\left(\text{R}_3+\text{j}\omega\text{L}\right)}{\displaystyle\text{j}\omega\text{CR}_2\left(\text{R}_3+\text{j}\omega\text{L}\right)+\text{R}_3+\text{j}\omega\text{L}+\text{R}_2}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\left(\text{R}_3+\text{j}\omega\text{L}\right)}{\displaystyle\text{R}_2+\text{R}_3+\text{j}\omega\text{CR}_2\text{R}_3+\text{j}\omega\text{CR}_2\text{j}\omega\text{L}+\text{j}\omega\text{L}}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\left(\text{R}_3+\text{j}\omega\text{L}\right)}{\displaystyle\text{R}_2+\text{R}_3-\text{CLR}_2\omega^2+\omega\left(\text{CR}_2\text{R}_3+\text{L}\right)\text{j}}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\left(\text{R}_3+\text{j}\omega\text{L}\right)}{\displaystyle\text{R}_2+\text{R}_3-\text{CLR}_2\omega^2+\omega\left(\text{CR}_2\text{R}_3+\text{L}\right)\text{j}}\cdot\frac{\displaystyle\text{R}_2+\text{R}_3-\text{CLR}_2\omega^2-\omega\left(\text{CR}_2\text{R}_3+\text{L}\right)\text{j}}{\displaystyle\text{R}_2+\text{R}_3-\text{CLR}_2\omega^2-\omega\left(\text{CR}_2\text{R}_3+\text{L}\right)\text{j}}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\left(\text{R}_3+\text{j}\omega\text{L}\right)\left(\text{R}_2+\text{R}_3-\text{CLR}_2\omega^2-\omega\left(\text{CR}_2\text{R}_3+\text{L}\right)\text{j}\right)}{\displaystyle\left(\text{R}_2+\text{R}_3-\text{CLR}_2\omega^2\right)^2+\left(\omega\left(\text{CR}_2\text{R}_3+\text{L}\right)\right)^2} \end{split}\tag1 \end{equation}
Where \$\alpha\space\text{||}\space\beta:=\frac{\displaystyle\alpha\beta}{\displaystyle\alpha+\beta}\$.
Using your values, I found:
$$\underline{\text{Z}}_{\space\text{i}}\left(1\right)=3\space\Omega\tag2$$
So, we see that: \$\displaystyle\Im\left(\underline{\text{Z}}_{\space\text{i}}\left(1\right)\right)=0\space\Omega\$.
EDIT, the voltage across the inductor is given by:
$$\text{V}_\text{L}\left(t\right)=\hat{\text{u}}\cos\left(t+\varphi\right)\tag3$$
Where:
- $$\hat{\text{u}}=\left|\text{j}\omega\text{L}\cdot\underbrace{\frac{\displaystyle\hat{\text{u}}_\text{i}\exp\left(\varphi_\text{i}\right)}{\displaystyle\underline{\text{Z}}_{\space\text{i}}\left(\omega\right)}\cdot\underbrace{\frac{\displaystyle\text{R}_2\space\text{||}\space\frac{1}{\text{j}\omega\text{C}}}{\displaystyle\text{R}_3+\text{j}\omega\text{L}+\left(\text{R}_2\space\text{||}\space\frac{1}{\text{j}\omega\text{C}}\right)}}_{=\space\text{current divider}}}_{=\space\text{current through inductor}}\right|\tag4$$
- $$\varphi=\arg\left(\text{j}\omega\text{L}\cdot\frac{\displaystyle\hat{\text{u}}_\text{i}\exp\left(\varphi_\text{i}\right)}{\displaystyle\underline{\text{Z}}_{\space\text{i}}\left(\omega\right)}\cdot\frac{\displaystyle\text{R}_2\space\text{||}\space\frac{1}{\text{j}\omega\text{C}}}{\displaystyle\text{R}_3+\text{j}\omega\text{L}+\left(\text{R}_2\space\text{||}\space\frac{1}{\text{j}\omega\text{C}}\right)}\right)\tag5$$
Using your values, we find: trough
- $$\hat{\text{u}}=\frac{4 \sqrt{2}}{3}\approx1.88562\space\text{V}\tag6$$
- $$\varphi=\frac{\pi}{2}\tag7$$