Is there any tip

If you know the power on Z1 then you know that that is dissipated by the "2" part of 2 - j5 and this leads on to be able to state the current through that 2 ohms: -

$$20 = I^2 \cdot 2$$

Hence current is \$\sqrt{10}\$ = 3.162 amps.

From that you can calculate the line voltage using Z1's impedance. That impedance is \$\sqrt{2^2+5^2}\$ = 5.285 ohms. So you have the line voltage of 17.03 volts.

Can you take it from here?

In engineering, power is real power, apparent power is volts x amps and reactive power is \$\sqrt{(V\cdot A)^2 - (watts)^2}\$.

Is there any tip

If you know the power on Z1 then you know that that is dissipated by the "2" part of 2 - j5 and this leads on to be able to state the current through that 2 ohms: -

$$20 = I^2 \cdot 2$$

Hence current is \$\sqrt{10}\$ = 3.162 amps.

From that you can calculate the line voltage using Z1's impedance. That impedance is \$\sqrt{2^2+5^2}\$ = 5.285 ohms. So you have the line voltage of 17.03 volts.

Can you take it from here?

In engineering, power is real (or active) power, apparent power is volts x amps and, reactive power is \$\sqrt{(V\cdot I)^2 - (watts)^2}\$: -

Is there any tip

If you know the power on Z1 then you know that that is dissipated by the "2" part of 2 - j5 and this leads on to be able to state the current through that 2 ohms: -

$$20 = I^2 \cdot 2$$

Hence current is \$\sqrt{10}\$ = 3.162 amps.

From that you can calculate the line voltage using Z1's impedance. That impedance is \$\sqrt{2^2+5^2}\$ = 5.285 ohms. So you have the line voltage of 17.03 volts.

Can you take it from here?

In engineering, power is real power, apparent power is volts x amps and reactive power is what it is.

Is there any tip

If you know the power on Z1 then you know that that is dissipated by the "2" part of 2 - j5 and this leads on to be able to state the current through that 2 ohms: -

$$20 = I^2 \cdot 2$$

Hence current is \$\sqrt{10}\$ = 3.162 amps.

From that you can calculate the line voltage using Z1's impedance. That impedance is \$\sqrt{2^2+5^2}\$ = 5.285 ohms. So you have the line voltage of 17.03 volts.

Can you take it from here?

In engineering, power is real power, apparent power is volts x amps and reactive power is \$\sqrt{(V\cdot A)^2 - (watts)^2}\$.