Electrical technology › Alternating current 3

Alternating current 3: admittance, parallel RLC circuits, current resonance

Sinusoidal Alternating Current III

Alternating Current 3

Admittance. Parallel RLC Circuits. Current Resonance

Summary. We study the circuit obtained by connecting the basic components, resistor, inductor and capacitor (R, L and C), in parallel. We then develop the concept of admittance, and analyze the behavior of the circuit as a function of frequency and the phenomenon of current resonance.

Before you start.
To fully understand the concepts presented here, you need some background knowledge, which you can build from the technical literature and also from the previous module in this series, Alternating Current 2: Reactance and Impedance, which presents and discusses the fundamental concepts of reactance and impedance, as well as other important topics such as series RLC circuits and voltage resonance (frequency response).

1 - The parallel RLC connection

Let us study the circuit in which the basic components, resistor R, inductor L and capacitor C, are connected in parallel. Each of them is then subjected to the same voltage \(\small v(t)\), with RMS value \(\small V = \dfrac{V_M}{\sqrt{2}}\), and carries a current \(\small i(t) \) whose intensity is determined by the admittance \(\small \dot{Y}\) ⁽¹⁾ of each branch.

1.1 - Admittance, conductance, susceptance

Admittance is a complex operator that performs the transformation \(\small \dot{Y} : \dot{V} \rightarrow \dot{I}\). It is essentially the inverse of the transformation performed by impedance, so admittance is the reciprocal of impedance:

(1) \( \qquad \qquad \qquad \qquad \dot{Y} = \dfrac{1}{\dot{Z}} \).

Its unit is the \(\small \Omega^{-1}\) or the siemens, S: 1 S = \(\small \dfrac{1}{\Omega} = \Omega^{-1}\).
As is well known, products and quotients of complex numbers are easier to work out in exponential or polar form⁽²⁾:

(2) \( \qquad \qquad \qquad \qquad \dot{Y} = Y\,e^{j\,\theta} \).

in Steinmetz notation:

(3) \( \qquad \qquad \qquad \qquad \dot{Y} = Y\,\angle \theta \).

Its relationship with impedance is:

(4) \( \qquad \qquad \qquad \qquad \dot{Y} = \dfrac{1}{Z\,e^{j\,\varphi}} = \dfrac{1}{Z}\,e^{-j\,\varphi} \).

or, in polar form:

(5) \( \qquad \qquad \qquad \qquad \dot{Y} = \dfrac{1}{Z}\,\angle -\varphi \).

with \(\small Y = \dfrac{1}{Z}\) for the magnitude and \(\small \theta = -\varphi\) for the argument.
Admittance can also be written in algebraic form:

(6) \( \qquad \qquad \qquad \qquad \dot{Y} = G + j\,B \).

Its real part \(\small G\) is called conductance and its imaginary part \(\small B\) susceptance.
If an impedance \(\small \dot{Z} = R + j\,X\) has a real part \(\small R\) (resistance) and an imaginary part \(\small X\) (inductive or capacitive reactance), a few simple steps⁽³⁾ give the expressions for the conductance \(\small G\) and the susceptance \(\small B\) in terms of the components of the impedance:

(7) \( \qquad \qquad \qquad \qquad G = \dfrac{R}{R^2 + X^2} \qquad \qquad B = \dfrac{- X}{R^2 + X^2} \).

Pay attention to the minus sign in the expression for \(\small B\). Equation (7) says that, in every case, the sign of the imaginary part of the admittance is the opposite of that of the corresponding impedance. In fact, as (4) and (5) also show, the angle of the admittance has the same absolute value as that of the impedance, but the opposite sign, and this determines the sign of the imaginary part of the admittance itself: negative for an inductive susceptance, positive for a capacitive susceptance. Denoting by \(\small B_L\) and \(\small B_C\) the absolute values of the inductive and capacitive susceptance, we have:

\( \qquad \qquad \qquad \qquad \dot{Y} = G - j\,B_L \qquad \qquad \dot{Y} = G + j\,B_C \).

In any case, in the circuit we are studying there is only one component in each branch, so the admittance is much simpler to write. Each branch has only a conductance \(\small G\) (R branch) or a susceptance \(\small B\) (L and C branches), which are obtained from (7) by setting \(\small X\) or \(\small R\) to zero:

(8) \( \qquad \qquad \qquad \qquad \dot{Y}_R = \dfrac{1}{R} = G \)

(9) \( \qquad \qquad \qquad \qquad \dot{Y}_L = \dfrac{1}{j\,X_L} = -j\,\dfrac{1}{X_L} = -j\,B_L \)

(10) \( \qquad \qquad \qquad \quad \; \dot{Y}_C = \dfrac{1}{-j\,X_C} = + j\,\dfrac{1}{X_C} = +j\,B_C \)

1.2 - Calculating the currents

The currents in the branches of the parallel circuit are given by:

(11) \( \qquad \qquad \qquad \quad \; \dot{I}_R = \dot{V}\,\dot{Y}_R \)

(12) \( \qquad \qquad \qquad \quad \; \dot{I}_L= \dot{V}\,\dot{Y}_L \)

(13) \( \qquad \qquad \qquad \quad \; \dot{I}_C = \dot{V}\,\dot{Y}_C \)

The total current delivered by the source is the sum of the branch currents.

(14) \( \qquad \dot{I} = \dot{I}_R + \dot{I}_L + \dot{I}_C \)

which, using (11-13), we can also write as:

(14') \( \qquad \dot{I} = \dot{V}\,\dot{Y}_R + \dot{V}\,\dot{Y}_L + \dot{V}\,\dot{Y}_C = \dot{V}\,(\dot{Y}_R + \dot{Y}_L + \dot{Y}_C) \)

This shows that the total current can also be found if we know the total admittance, that is, if we consider the equivalent two-terminal circuit of the parallel connection:

(14'') \( \qquad \dot{Y} = \dot{Y}_R + \dot{Y}_L + \dot{Y}_C = \dfrac{1}{R} - j\,\dfrac{1}{X_L} + j\,\dfrac{1}{X_C} = \dfrac{1}{R} + j\, \left(\dfrac{1}{X_C} - \dfrac{1}{X_L} \right) \)

(15) \( \qquad \qquad \qquad \quad \; \dot{I} = \dot{V}\,\dot{Y} \)

For the argument of \(\small \dot{Y}\) we have:

(16) \( \qquad tan(\theta) = \dfrac{\dfrac{1}{X_C} - \dfrac{1}{X_L}}{\dfrac{1}{R}} = \dfrac{R}{X_C} - \dfrac{R}{X_L} \)

(17) \( \qquad \qquad \qquad \theta = atan \left(\dfrac{R}{X_C} - \dfrac{R}{X_L}\right) \)

Let us now look at the particular cases of R-L, R-C and R-L-C in parallel, also from the point of view of the energy and power balance.

2 - The parallel R-L circuit

In the parallel R-L circuit the equivalent admittance is given by the following formulas:

(18) \( \qquad \qquad \qquad \qquad \dot{Y} = \dfrac{1}{R} - j\,\dfrac{1}{X_L} \)

(19) \( \qquad \qquad \qquad \theta = atan \left( \dfrac{ -\dfrac{1}{X_L}}{\dfrac{1}{R}}\right) = atan \left(- \dfrac{R}{X_L}\right) \)

The current \(\small \dot{I}_R\) in the resistor branch is in phase with the voltage and is given by:

(20) \( \qquad \qquad \qquad \quad \; \dot{I}_R = \dot{V}\,\dot{Y}_R = \dfrac{\dot{V}}{R} \)

In the inductor branch, instead, the current \(\small \dot{I}_L\) lags the voltage by \(\small \dfrac{\pi}{2}\), as clearly shown on the board on the left. Its value is:

(21) \( \qquad \qquad \qquad \quad \; \dot{I}_L= \dot{V}\,\dot{Y}_L = \dfrac{\dot{V}}{j\,X_L} = -j\,\dfrac{\dot{V}}{X_L}\)

The total current is the sum (Kirchhoff's current law):

(22) \( \qquad \qquad \qquad \quad \; \dot{I} = \dot{I}_R +\dot{I}_L \)

Increasing \(\small L\) increases \(\small X_L\) and decreases \(\small Y_L\), so the current \(\small I_L\) decreases. The total current tends to come into phase with the voltage \(\small V\).

2.1 - How the R-L circuit behaves as frequency changes

Since \(\small Y_L = \dfrac{1}{2\,\pi\,L\,f}\), increasing the frequency decreases \(\small Y_L\) and the current in the inductor branch.
Low-frequency currents flow preferably through the inductor branch (low reactance, high admittance), so an inductor in parallel with a load, represented by R, acts as a high-pass filter: high-frequency currents flow through the load more easily.
Changing the frequency or the inductance affects only the current \(\small \dot{I_L}\). The component of the total current along the voltage, \(\small \dot{I_R}\), does not change. This also means that these changes take place without affecting the active power.

2.2 - The current triangle and the admittance triangle

The three vectors \(\small \dot{I}\), \(\small \dot{I_R}\) and \(\small \dot{I_L}\) form the current triangle. The angle between \(\small \dot{I_R}\) and \(\small \dot{I}\) is the phase angle \(\small \varphi\) between voltage and current. The elements of the triangle are related by the familiar trigonometric relationships:

(23) \( \qquad \qquad \qquad \quad \; I_R = I\,cos\,\varphi \qquad \qquad I_L = I\,sin\,\varphi \)

Just as we did for impedance, we can show the admittance triangle, whose sides are related by the obvious trigonometric relationships.

Activity 1

2.3 - Power in the parallel R-L circuit

The resistive branch is characterized by the active power:

(24) \( \qquad \qquad \qquad \quad \;P = V\,I_R = V\,I\,cos\,\varphi \)

and the inductive branch by the reactive power:

(25) \( \qquad \qquad \qquad \quad \;Q = V\,I_L = V\,I\,sin\,\varphi \)

where we have used (23).
Alternatively, using the magnitudes in (20) and (21), the active power can also be calculated as:

(26) \( \qquad \qquad \qquad \quad \;P = V\,\dfrac{V}{R} = \dfrac{V^2}{R} \)

and the reactive power as:

(27) \( \qquad \qquad \qquad \quad \;Q = V\,\dfrac{V}{X_L} = \dfrac{V^2}{X_L} \)

To learn more:

Power in sinusoidal alternating current

2.4 - Equivalent impedance of the parallel R-L circuit

The reciprocal relationship between admittance and impedance leads to the following considerations. We start from the familiar relationship:

\( \qquad \qquad \qquad \qquad \dot{Y} = \dfrac{1}{\dot{Z}} \)

For each branch we can write a relationship between the admittance and the impedance of the branch:

\( \qquad \qquad \qquad \qquad \dot{Y}_R = \dfrac{1}{R} = \dfrac{1}{\dot{Z}_R} \)

\( \qquad \qquad \qquad \qquad \dot{Y}_L = \dfrac{1}{j\,X_L} = \dfrac{1}{\dot{Z}_L} \)

As we saw in section 1.2, the equivalent admittance is the sum of the admittances:

\( \qquad \qquad \dot{Y} = \dfrac{1}{\dot{Z}} = \dot{Y}_R + \dot{Y}_L = \dfrac{1}{R} + \dfrac{1}{j\,X_L} = \dfrac{R + j\,X_L}{R\,j\,X_L} \)

and finally:

(28) \( \qquad \qquad \dot{Z} = \dfrac{R\,\,j\,X_L}{R + j\,X_L} = \dfrac{\dot{Z}_R\, \dot{Z}_L}{\dot{Z}_R+\dot{Z}_L} \)

So the equivalent impedance of the parallel R-L circuit is calculated with a formula formally identical to the one for resistors in parallel in direct current, with the obvious difference that here we use impedances and complex numbers instead of resistances and real numbers.

3 - The parallel R-C circuit

In the parallel R-C circuit the equivalent admittance is given by the following formula:

(29) \( \qquad \qquad \qquad \qquad \dot{Y} = \dfrac{1}{R} + j\,\dfrac{1}{X_C} \)

The current \(\small \dot{I}_R\) through the resistor is in phase with the voltage and its value is:

(30) \( \qquad \qquad \qquad \quad \; \dot{I}_R = \dot{V}\,\dot{Y}_R = \dfrac{\dot{V}}{R} \)

In the capacitor branch, instead, the current \(\small \dot{I}_C\) leads the voltage by \(\small \dfrac{\pi}{2}\), as you can see on the board on the left. Its value is:

(31) \( \qquad \qquad \qquad \quad \; \dot{I}_C= \dot{V}\,\dot{Y}_C = \dfrac{\dot{V}}{-j\,X_C} = j\,\dfrac{\dot{V}}{X_C}\)

The total current is the sum of the branch currents (Kirchhoff's current law):

(32) \( \qquad \qquad \qquad \quad \; \dot{I} = \dot{I}_R +\dot{I}_C \)

Decreasing the capacitance \(\small C\) increases \(\small X_C\) and decreases \(\small Y_C\), so the current \(\small \dot{I}_C\) decreases. The total current tends to come into phase with the voltage \(\small \dot{V}\).

3.1 - How the R-C circuit behaves as frequency changes

Since \(\small \dot{Y}_C = 2\,\pi\,C\,f\), increasing the frequency increases \(\small Y_C\) and the current in the capacitor branch.
High-frequency currents flow preferably through the capacitor branch (low reactance, high admittance), so a capacitor in parallel with a load acts as a low-pass filter: low-frequency currents flow through the load more easily.
Changing the frequency or the capacitance affects only the current \(\small \dot{I}_C\). The component of the total current along the voltage (\(\small \dot{I}_R\)) does not change. This also means that these changes take place without affecting the active power.

3.2 - The current triangle and the admittance triangle

The three vectors \(\small \dot{I}\), \(\small \dot{I_R}\) and \(\small \dot{I_C}\) form the current triangle. The angle between \(\small \dot{I_R}\) and \(\small \dot{I}\) is the phase angle \(\small \varphi\) between voltage and current. The elements of the triangle are related by the familiar trigonometric relationships:

(33) \( \qquad \qquad \qquad \quad \; I_R = I\,cos\,\varphi \qquad \qquad I_C = I\,sin\,\varphi \)

Just as we did for impedance, we can show the admittance triangle, whose sides are related by the obvious trigonometric relationships.

Activity 2

3.3 - Power in the parallel R-C circuit

The resistive branch is characterized by the active power:

(34) \( \qquad \qquad \qquad \quad \;P = V\,I_R = V\,I\,cos\,\varphi \)

and the capacitive branch by the reactive power:

(35) \( \qquad \qquad \qquad \quad \;Q = -V\,I_C = -V\,I\,sin\,\varphi \)

where we used formulas (33).
Alternatively, using the magnitudes in (30) and (31), the active power can also be calculated as:

(36) \( \qquad \qquad \qquad \quad \;P = V\,\dfrac{V}{R} = \dfrac{V^2}{R} \)

and the reactive power as:

(37) \( \qquad \qquad \qquad \quad \;Q = -V\,\dfrac{V}{X_C} = -\dfrac{V^2}{X_C} \)

The minus sign in formulas (35) and (37) reminds us that capacitive reactive power behaves in the opposite way to the corresponding inductive power. This means that when one is absorbed, the other is delivered, and vice versa (we will see this in more detail later on, when discussing the parallel RLC circuit). We have kept the same sign convention established in the previous module (Alternating Current 2: series RLC): inductive Q positive and capacitive Q negative.

3.4 - Equivalent impedance of the parallel R-C circuit

Given that:

\( \qquad \qquad \qquad \qquad \dot{Z}_R = R \qquad \qquad \dot{Z}_C = -j\,X_C \)

reasoning exactly as in section 2.4 leads to a similar conclusion:

(38) \( \qquad \qquad \qquad \dot{Z} = \dfrac{R\,\,(-j\,X_C)}{R - j\,X_C} = \dfrac{\dot{Z}_R\, \dot{Z}_C}{\dot{Z}_R+\dot{Z}_C} \)

4 - The parallel R-L-C circuit

Let us now go back to the analysis of the circuit with R, L and C connected in parallel. We briefly summarize what we saw in sections 1.1 and 1.2, keeping the same formula numbers.
For the admittances of the branches:

(8,9,10) \( \qquad \dot{Y}_R = \dfrac{1}{R} = G \qquad \dot{Y}_L = -j\,\dfrac{1}{X_L} = -j\,B_L \qquad \dot{Y}_C = + j\,\dfrac{1}{X_C} = +j\,B_C \)

The branch currents:

(11, 12, 13) \( \qquad \qquad \dot{I}_R = \dot{V}\,\dot{Y}_R \qquad \dot{I}_L= \dot{V}\,\dot{Y}_L \qquad \dot{I}_C = \dot{V}\,\dot{Y}_C \)

By Kirchhoff's current law, the total current is:

(14) \( \qquad \qquad \qquad \dot{I} = \dot{I}_R + \dot{I}_L + \dot{I}_C \)

and, going one step further:

(14') \( \qquad \qquad \dot{I} = \dot{V}\,(\dot{Y}_R + \dot{Y}_L + \dot{Y}_C) =\dot{V}\,\dot{Y} \)

where we recognize the equivalent (or total) admittance \(\small \dot{Y}\):

(14") \( \qquad \qquad \dot{Y} = \dot{Y}_R + \dot{Y}_L + \dot{Y}_C = \dfrac{1}{R} + j\, \left(\dfrac{1}{X_C} - \dfrac{1}{X_L} \right) \)

The argument of the equivalent admittance \(\small \dot{Y}\) is given by:

(17) \( \qquad \qquad \qquad \theta = atan \left(\dfrac{R}{X_C} - \dfrac{R}{X_L}\right) \)

4.1 - Energy and power exchanges in the parallel RLC circuit

The parallel RLC circuit (like the series one, for that matter) is particularly interesting because of the energy exchanges that take place between its components.
Of course the resistor absorbs active energy, and therefore active power, which is instantly converted into heat.

(39) \( \qquad \qquad \qquad \ P = R\,I^2_R = \dfrac {V^2}{R}= V\,I\, cos \varphi \)

As we know, this is the only form of power that can be 'extracted' from the electric system, for example to do work (instead of the resistor we could have, say, an electric motor).
Things are different for the reactive components, which alternately store energy while building up the magnetic field (the inductor) or the electric field (the capacitor), and release it when the fields collapse.
As a consequence, there is a phase shift of \(\small \dfrac{\pi}{2}\) between the voltages and the corresponding currents. But the interesting thing about this circuit is that part of the energy exchange takes place locally between the inductor and the capacitor, without involving the source. Indeed, under a particular operating condition, resonance, we will see that the exchange between the two components is complete: the energy absorbed by the inductor in a given time interval is supplied by the capacitor, which in the same interval releases it, and vice versa.
This is also a consequence of the fact that the currents \(\small \dot{I}_L\) and \(\small \dot{I}_C\) are opposite, as the phasor diagram clearly shows.

5 - Current resonance

Let us now examine how the three basic components in parallel behave as a function of frequency, paying particular attention to whether a resonance condition exists.
As we have seen, resonance occurs when, from an energy point of view, the two reactive components exactly cancel each other out, which is the same as saying that the two reactive powers are equal and opposite in sign. The reactive power (or energy) is exchanged without involving the source.

Activity 3: resonance

Figure 1 - Inductive and capacitive susceptance as a function of frequency

Figure 1 shows how the inductive and capacitive susceptances and the total admittance vary; the total admittance has a minimum at the resonant frequency.
The total current, like the admittance, has a minimum at resonance. When the system is in resonance, the currents \(\small \dot{I}_L\) and \(\small \dot{I}_C\) are equal in magnitude and flow in opposite directions in their branches. With respect, for example, to node N, at resonance we always have \(\small \dot{I}_L = - \dot{I}_C\), that is, \(\small \dot{I}_L + \dot{I}_C = 0\). So there can be no current in the other branch connected to N. The currents \(\small \dot{I}_L\) and \(\small \dot{I}_C\) can be regarded as representing a single mesh current that circulates in the \(\small L\) and \(\small C\) branches. The current in the source branch, or total current, is then determined by R alone.
As the resistance \(\small R\) increases, the total current decreases, and we can infer that, without the resistive branch (\(\small R = \infty\)), at the resonant frequency the source would deliver no current at all, while a current would circulate only in the closed mesh formed by the inductor and the capacitor. Paradoxical as it may seem, at resonance current flows in the two branches of the parallel circuit, but no current flows in the source branch. This could only happen, however, with an ideal inductor and capacitor, completely free of ohmic resistance and therefore of power losses. Real components always dissipate some energy (they "heat up"), so the source must still deliver a current that carries the necessary active power.
Current resonance is potentially dangerous: while the total current is modest, and almost zero if \(\small R\) is very large, extremely high currents can flow in the reactive branches.

Resonance therefore occurs when the admittances (and so also the reactances) of the two reactive branches are equal in magnitude:

(39) \( \qquad \qquad \qquad \qquad Y_C = Y_L \)

After a few steps (\(\small \omega_r\) = angular frequency at resonance):

\( \qquad \qquad \qquad \omega_r\,C = \dfrac{1}{\omega_r\,L} \qquad \omega_r^2\,L\,C = 1 \qquad 4\,\pi^2\, f^2_r\,L\,C = 1 \)

we obtain:

(40) \( \qquad \qquad \qquad \qquad f_r = \dfrac{1}{2\,\pi\,\sqrt{L\,C}}\)

the same formula as for series resonance, or voltage resonance.
Figure 2 shows how the admittance and the phase \(\small \varphi\) (of the impedance) vary. Below the resonant frequency the angle \(\small \varphi\) is positive (the angle \(\small \theta\) of the admittance is negative) and the circuit is resistive-inductive; above \(\small f_r\), \(\small \varphi\) is negative and the circuit is resistive-capacitive.

Figure 2 - Admittance and phase as a function of frequency

From an energy point of view, we can say that resonance is characterized by a balanced exchange of power between the reactive components. The two powers are equal and opposite in sign, which means that while one component absorbs reactive energy (power), the other releases exactly the amount needed. As far as reactive power is concerned, the system becomes independent of the source.

Activity 4: resonance, power factor and power factor correction

As we have seen, resonance occurs when the inductive and capacitive susceptances, or the inductive and capacitive reactances, are equal. In this situation the effects of the reactive components cancel each other out and the circuit behaves as if it were purely resistive, with voltage and current in phase. For given values of \(\small L\) and \(\small C\), resonance can be obtained by setting the frequency to the value given by (40).

In industrial installations, where the frequency is fixed by standards (50 Hz in Europe), resonance can only be obtained by suitably changing the values of the reactive components. In fact, since industrial devices (such as electric motors) are generally resistive-inductive, the only parameter we can act on is, if anything, the capacitance of a capacitor suitably connected to the electrical load (a motor or other device).

But why would it be so interesting to achieve resonance?
The answer comes from what we said earlier: at resonance the total current and the voltage are in phase and the circuit behaves as if it were purely resistive; in other words, at resonance only active power is drawn (see also: Power in Alternating Current). The reactive power needed by the installation is exchanged directly with the capacitor rather than with the source. This reduces the total current, but not the current drawn by the load (the resistor in our case).
A key parameter is the power factor, which is simply the cosine of the angle \(\small \varphi\) between the voltage and the total current. It determines how much of the electric power can be used to do work (physical work: force x displacement), according to the familiar relationship: \[ P = V\,I\,cos\,\varphi \]
Let us go back to our example.
We have a resistive-inductive load in which the values of \(\small R\) and \(\small L\) give an angle \(\small \varphi\) corresponding to a power factor of:

cos \(\varphi\) =

By connecting a capacitor in parallel and changing its capacitance, we can increase cos \(\small \varphi\) and so progressively reduce the angle by which the current lags, until voltage and current are in phase. At that point we are at resonance and the power factor is 1. This is called full (or complete) power factor correction.

In practice, however, full correction is not required; a slightly lower value, for example 0.9, is enough.

Now, starting from the initial situation, connect a capacitor in parallel and change only its capacitance until the power factor is 0.92 (you can check cos \(\varphi\) a few lines above, or calculate it from the angle \(\small \varphi\)).
Watch how the quantities change:

What happens to the total current?
What does not change?
If no other parameters are changed, the active power stays the same. Why?

Exercise
Set the following values for the components of the parallel RL circuit:

\(\qquad \qquad \qquad \qquad R = 4\; \Omega\) \(\qquad L = 11.42\; mH\)

Find the value of the capacitance a) for full power factor correction and b) for a power factor of 0.9
Then answer the following questions (use a decimal point, not a comma).

close this box

Previous modules:

Rev. January 2021; English version October 2026

Books by the author

The author’s books

⁽¹⁾ Of course we could have said "whose intensity is determined by the impedance Z of each branch", but when dealing with parallel connections it is more convenient to use the admittance operator Y. In series connections impedance is preferable, because voltage drops add up; in parallel connections admittance is preferable, because currents add up.

⁽²⁾For the exponential form of complex numbers see Euler's formula. Complex powers and logarithms (in Italian)

⁽³⁾ They involve dividing complex numbers in algebraic form: multiply the numerator and the denominator by the complex conjugate of the denominator.
See complex numbers (in Italian)