9.4: Complex Numbers

In AC circuits, parameters such as voltage and current are vectors, that is, they have both a magnitude and a phase shift or angle. For example, a voltage might be “12 volts at an angle of 30 degrees” (or more compactly, $12\angle 30^{\circ}$). This is known as polar form or magnitude-angle form. Alternately, a vector can be broken into rectangular form, that is, its right angle components.

This can be visualized as a right triangle where the magnitude is the hypotenuse, the angle is the rotation above or below the horizontal, the horizontal component is the side adjacent to the angle and the vertical component is the side opposite of the angle. This is shown in Figure $\PageIndex{1}$.

Figure $\PageIndex{1}$: Polar and rectangular forms.

Properly, voltage and current vectors are complex numbers that lay on a complex plane consisting of a real part and an imaginary part. The horizontal axis is the real number axis and the vertical axis is the imaginary number axis. The imaginary axis denotes values times the imaginary operator $j$ (and often referred to as $i$ outside of electrical analysis). The $j$ operator is the square root of −1. An example of such a value is $3 + j4$, in other words, 3 units along the horizontal real axis and 4 units up the vertical imaginary axis. This is depicted in Figure $\PageIndex{2}$.

Figure $\PageIndex{2}$: Polar and rectangular forms on the complex plane.

Converting from one form to another relies on basic trigonometric relations. For convenience, the relationships between the magnitude, angle, real component and imaginary component are reproduced below:

$$\text{Magnitude } = \sqrt{\text{Real}^2 + \text{Imaginary}^2} \label{1.6a}$$

$$\theta = \tan^{−1} \frac{\text{Imaginary}}{\text{Real}} \label{1.6b}$$

$$\text{Real } = \text{ Magnitude } \cos \theta \label{1.6c}$$

$$j\text{Imaginary } = \text{ Magnitude } \sin \theta \label{1.6d}$$

To add or subtract complex quantities, first put them into rectangular form and then combine the reals with the reals and the $j$ terms with the $j$ terms as in $(3 + j5) + (13 − j1) = 16 + j4$. These real and imaginary terms must be kept separate. $3 + j5$ does not equal 8 (or even $j8$). That would be like saying that moving 3 feet to your right and 5 feet forward puts you in the same location as moving 8 feet to your right (or 8 feet forward).

The direct way to multiply or divide complex values is to first put them in polar form and then multiply or divide the magnitudes. The angles are added together for multiplication and subtracted for division. For example, $12\angle 30^{\circ}$ times $2\angle 45^{\circ}$ is $24\angle 75^{\circ}$ while dividing them yields $6\angle −15^{\circ}$. The need for complex numbers will become more obvious as we move through the upcoming material. It is imperative that you have mastered the manipulation of complex values before moving on to subsequent chapters.

Example $\PageIndex{1}$

Convert $15 + j20$ and $1 k − j2 k$ into polar form, and $10\angle 45^{\circ}$ and $0.2\angle −30^{\circ}$ into rectangular form.

For the first two conversions, use Equations $\ref{1.6a}$ and $\ref{1.6b}$.

$$\text{Magnitude } = \sqrt{15^2 + 20^2} = 25 \nonumber$$

$$\theta = \tan^{−1} \left( \frac{20}{15} \right) = 53.1^{\circ} \nonumber$$

$$\text{Magnitude } = \sqrt{1 k^2 + (−2 k)^2} = 2.236 k \nonumber$$

$$\theta = \tan^{−1} \left( \frac{−2k}{1 k} \right) = −63.4^{\circ} \nonumber$$

The answers for the first part are $25\angle 53.1^{\circ}$ and $2.236 k\angle −63.4^{\circ}$.

For the second pair of conversions use Equations $\ref{1.6c}$ and $\ref{1.6d}$.

$$\text{Real } = 10 \cos 45^{\circ} = 7.07 \nonumber$$

$$j\text{Imaginary } = 10 \sin 45^{\circ} = j 7.07 \nonumber$$

$$\text{Real } = 0.2 \cos −30^{\circ} = 0.173 \nonumber$$

$$j\text{Imaginary } = 0.2 \sin −30^{\circ} = − j0.1 \nonumber$$

The answers for the second part are $7.07 + j7.07$ and $0.173 − j0.1$.

Example $\PageIndex{2}$

a) Add $7 + j8 to + j6$, b) Subtract $5\angle 53.1^{\circ}$ from $10\angle −45^{\circ}$, c) Divide $20\angle 90^{\circ}$ by $4\angle −50^{\circ}$, d) Multiply $90 + j75$ by $6 + j10$.

a) Add real to real and imaginary to imaginary: $7 + j14$.

b) First convert the values into rectangular: $3 + j4$ and $7.07 − j7.07$. Now subtract the first pair from the second pair: $4.07 − j11.07$ (or $11.8\angle −69.8^{\circ}$).

c) Divide the magnitudes and subtract the angles: $5\angle 140^{\circ}$.

d) First convert the values into polar: $117.15\angle 39.8^{\circ}$ and $11.66\angle 59^{\circ}$. Now multiply the magnitudes and add the angles: $1.366 k\angle 98.8^{\circ}$.