Now for maximum power transfer,

Applying Nodal analysis, we get

\begin{gathered} \frac{V_{ Th }-100 \angle 0}{6+8 j}+\frac{V_{ Th }-90 \angle 0}{8 j+6}=0 \\ V _{ Th }=100 \angle 0^{\circ} \end{gathered}

So,    Z_{ Th }=3+4 j  \Omega

The equivalent circuit is

But here, only R_{ L } is given

\begin{aligned} R_{ L } & =|3+4 j |=5  \Omega \\ I & =\frac{100 \angle 0}{3+4 j+5}=\frac{100 \angle 0}{3+4 j} \\ P_{\max } & =I^2 R_{ L }=\left(|I|^2 \times 5\right)=625  W \end{aligned}