• $\omega C=1/(\omega L)$ and L is in parallel with 10 Ohms. – Tony Stewart Sunnyskyguy EE75 May 21 at 22:20
$$\Z_L=10+\frac{1}{j\omega C}\$$ turns out 10+j20, well it can't. $$\\frac{1}{j\omega C} = \frac{-j}{\omega C}\$$ , so it has to be negative. On the source side the conjugate complex would give a positive imaginary number, so an inductor.
$$Z_L=R_L+\frac{1}{j\omega C_L}=R_L-\frac{j}{\omega C_L}$$ $$Z_L={Z_S}^*$$ $$Z_S=R_L-\frac{1}{j\omega C_L}=R_L+\frac{j}{\omega C_L}$$ $$Z_S=R_S+j\omega L_S$$ $$R_S=R_L , \ \omega L_S=\frac{1}{\omega C_L}$$