Use these identities :-
\$z = R.e^{j\theta}\$
\$Re(z) = R\cos(\theta) = a\$
\$Im(z) = R\sin(\theta) = b\$
\$z = a + jb\$
\$R = |z| = \sqrt{a^2 + b^2}\$
\$\theta = Arg(z) = \arctan(\frac{b}{a})\$
For example:
\$56.e^{j40} = 56\cos(40) + 56j\sin(40) = 42.9 + 36.0j\$
\$75 - j22 = \sqrt{75^2 + 22^2}.e^\arctan(\frac{-22}{75}) = 78.16.e^{-16.3j}\$
Reactive power would put no extra load on a generator shaft if everything were perfect. However, real generators have real losses, with some of those proportional to the square of the current. The reactive load causes more current in the wires than there would be with a purely resistive load of the same real power. The extra current causes additional real power to be lost.
So the answer is that the engine will see a somewhat higher load and therefore use a bit more fuel. This is because of more inefficiencies and losses in the system, not be reactive power itself makes the generator harder to turn.
Added:
I should have mentioned this before, but somehow it slipped thru my mind at the time.
A reactive load on a perfect generator does not require more shaft power averaged over a cycle, but it does add "bumps" to the torque. One attribute of a 3 phase AC generator is that the torque is constant over a cycle with a resistive load. However, with a reactive load parts of the cycle will require more power and other parts less. The average power is still the same, but constant pushing forwards and backwards relative to the average torque can cause undesirable mechanical stresses and vibrations.
You can think of this a bit like moving two magnets past each other. Let's say they are oriented to repell. At a distance there is little force. You have to apply force to move them close together, meaning you put energy into the system. The magnets push in the direction of motion as they move away, thereby giving you back the energy you put in earlier. The net energy spent is 0, but there was definitely energy flow back and forth. There is always some loss as energy is moved around or converted back and forth in real systems.
Again, the reactive power itself doesn't cause the problem, but real power is lost because energy can't be moved around and converted with perfect efficiency. This real power loss has to be made up with more real power input. In addition, the extra mechanical forces can decrease the life of the generator and engine driving it.
Best Answer
I think you mean induction motor.
Electric motors are energy conversion devices and an ideal electric motor converts 100% of the electrical power it receives from the circuit to mechanical power - the ideal motor itself does not consume power but, rather, changes its form.
Of course, for real motors, there are loses due to e.g., winding resistance, friction, etc.
Now an inductor and a capacitor are energy storage circuit elements - they alternately store and then release energy from and to the electric circuit to which they are attached.
So, for ideal inductors and capacitors, there is no loss - on average, their associated electric power is zero. The instantaneous power associated with an ideal reactive element is alternately positive and negative.
However, for a resistance, the electric power from the attached circuit is dissipated - converted to heat. The instantaneous power is always positive.