Observe o circuito a seguir.A impedância nos terminais A-B é...

a parte em paralelo é:

=( 20 + j30 - j10 ) // ( 10 - j20 )

= (( 20 + j20 ) * ( 10 - j20)) / ( 20 + j20 ) + ( 10 - j20 )

=(Aplica distributiva e simplifica, lembrando que quando "jX * jY = -X.Y")

=(600 - j200)/30

=(20 - j20/3)

Somando com os ramos em série:

=(20 - j20/3) + 5 + j2/3

=25 - j6

ZAB = R3 + (R2-j/wC)//(R1+j(wL1-1/(wC1)) + jwL2

Substituindo:

ZAB = 5 + [(10-j20)(20+j20)/30] + j2/3 :.

ZAB = 5 + [(200+j200-j400+400)/30] + j2/3 :.

ZAB = 5 + [(600-j200)/30] + j2/3 :.

ZAB = 5 + 20 - j20/3 + j2/3 :.

ZAB = 25 - j20/3 + j2/3

ZAB = 25 - j18/3 

ZAB = 25 - j6

Zeq1(R1,L1,C1) = 20+j20

Zeq2(R2,C2) = 10-j20

Zeq3(Zeq1,Zeq2) = [(20+j20)(10-j20)*]/30

Zeq3(Zeq1,Zeq2) = [(60-j20)/3] = 20 - j20/3

*: 200 - j400 + j200 + 400

Zeq4(R3,L2) = 5+j2/3

Zeq5(Zeq3,Zeq4) = (20+5)+j(2/3-20/3):.

Zeq5(Zeq3,Zeq4) = 25 - j6

Opção [C]