1. **Problem statement:** We are given the impedance $Z$ of a parallel circuit as $$\frac{1}{Z} = \frac{1}{R} + j\omega L + j\omega C$$ where $R=1$, $L=2$, $C=0.5$, and $\omega > 0$. We need to find the value of $\omega$ for resonance, i.e., when $Z$ is purely real. 2. **Recall:** Resonance in a parallel circuit occurs when the imaginary part of the admittance $\frac{1}{Z}$ is zero, making $Z$ real. 3. **Substitute values:** $$\frac{1}{Z} = \frac{1}{1} + j\omega \times 2 + j\omega \times 0.5 = 1 + j\omega(2 + 0.5) = 1 + j2.5\omega$$ 4. **Separate real and imaginary parts:** $$\text{Real part} = 1$$ $$\text{Imaginary part} = 2.5\omega$$ 5. **Set imaginary part to zero for resonance:** $$2.5\omega = 0 \implies \omega = 0$$ 6. **But given $\omega > 0$, so no resonance occurs with this direct sum.** 7. **Re-examine the formula:** The problem states $$\frac{1}{Z} = \frac{1}{R} + j\omega L + j\omega C$$ which is unusual because the admittance of an inductor is $\frac{1}{j\omega L} = -j\frac{1}{\omega L}$ and of a capacitor is $j\omega C$. 8. **Correct admittance formula for parallel circuit:** $$\frac{1}{Z} = \frac{1}{R} + \frac{1}{j\omega L} + j\omega C = \frac{1}{R} - j\frac{1}{\omega L} + j\omega C$$ 9. **Substitute values:** $$\frac{1}{Z} = 1 - j\frac{1}{\omega \times 2} + j\omega \times 0.5 = 1 + j\left(0.5\omega - \frac{1}{2\omega}\right)$$ 10. **Set imaginary part to zero for resonance:** $$0.5\omega - \frac{1}{2\omega} = 0$$ 11. **Solve for $\omega$:** $$0.5\omega = \frac{1}{2\omega}$$ Multiply both sides by $2\omega$: $$2\omega \times 0.5\omega = 2\omega \times \frac{1}{2\omega}$$ $$\omega^2 = 1$$ 12. **Since $\omega > 0$,** $$\boxed{\omega = 1}$$ **Final answer:** The resonance angular frequency is $\omega = 1$ rad/s.