We will first find the Thevenin’s equivalent circuit between terminals A and B. When the load is removed , the circuit become as shown in Fig..
Vth = drop across (2 + j10) = 50 ∠45º × (2 + j10)/(7 + j10)
= 41.8 × 68.7º = 15.2 + j 38.9 Rth = 5 | | (2 + j 10) = 4.1∠23.7º = 3.7 + j 1.6
The Thevenin’s equivalent source is shown in Fig.
Since for MPT, conjugate match is required hence, XC = 1.6 Ω and RL = 3.7 Ω