I have to do this exercise:
($Z(t)=I(t)$, it's printed wrong).
I have a doubt about the first item. To find all resonance when $R=1$, I found the particular solution $I_{p}(t)=A\sin(\omega t)+B\cos(\omega t)$.
I got the first and second derivates $$I_{p}'=A\omega\cos(\omega t)-B\omega\sin(\omega t)$$ $$I_{p}''=-A\omega^{2}\sin(\omega t)-B\omega^{2}\cos(\omega t)$$ then, when I substitute in the equation $$I''(t)+I'(t)+I(t)=\sin(\omega t)$$ I got the system on $(A,B)$: $$\left\{\begin{array}{l} -A\omega^{2}-B\omega+A=1 \\ -B\omega^{2}+A\omega+B=0 \end{array}\right.$$ so $$A=\frac{(1-\omega^{2})}{(1-\omega^{2})^{2}+\omega^{2}}$$ I calculated $A$ because it follows $\sin(\omega t)$ and with this I can find all resonance frequencies.
Well, to find them, I gotta find the values of $\omega$ which makes $A$ as large as possible. To $A\to+\infty$, so, its denominator must goes to 0: $$(1-\omega^{2})^{2}+\omega^{2}\to0$$ But that's not possible.
The answer is $\displaystyle\frac{1}{2\pi\sqrt{2}}$. Comparing with the exercise, I know that $\omega=1/\sqrt{2}$.
If the denominator was $(1-\omega^{2})^{2}-\omega^{2}$, I got the result, but I don't know what's wrong here.
EDIT: I found after $$I_{p}(t)=\frac{1}{\sqrt{(1-\omega^{2})^{2}+\omega^{2}}}\sin(\omega t-\alpha)$$ where $\alpha=\arctan(\omega/(1-\omega^{2})$. So, $$A=\frac{1}{\sqrt{(1-\omega^{2})^{2}+\omega^{2}}}$$ and it's large just if its denominator is small... but how?!