# Complex number in A/C circuit analysis

I am going to do a series of posts and try explaining what impedance is and how to analyse a passive Low and High Pass filters. But for that we need to know what a complex number is and how to handle it. A complex number is in the format of A + Bj, where B is an imaginary number and A was real. That was the entire story I knew. Until recently, I didn’t know why it is necessary or what it represented. I will be going to great depth on this subject and begin with the absolute basics.

Graphical representation of a Complex Number

Let’s say we have a graph as shown in figure 1.

We have three points, A = (0, 15), B = (10, 0) and another point Z. We defined Z as the sum of A and B. So what is the value of Z.? What point does it represent? That is simple:

Z = A + B = (0 + 10, 15 + 0) = (10, 15) .We just add the X values and Y values together and that’s the answer.

A complex number can also be thought of just a point in a graph. The Y axis is represents the imaginary numbers and X real.

For example the graph in figure 1, the point A is represents the imaginary part and point B the real part of a complex number. Therefore, that complex number must be B + Aj, or, 10 + 15j. And this complex number is what Z represents.

Another way of thinking is that all points in a graph represent some complex number. A point that lies on the Y axis is purely imaginary, and, that which lies on the X axis is purely real.

Thus A is a complex number, with no real part and B with no imaginary part. To find A + B we simply do a complex number addition.

Z = B + A = ( 10 + 0j ) + ( 0 + 15j ) = 10 + 15j.

Polar and Cartesian form of a complex number

We can represent a point on a graph either in polar or Cartesian (or rectangular) format. And because a complex number is thought as nothing more than just a point on a graph, we can represent a complex number in either of these forms.

In Cartesian format, we represent a complex number by the point on the graph it represents. For example in the figure 1, the point Z represents the point (10, 15). So its Cartesian form is Z = 10 + 15j.

In the polar format, we represent a complex by its distance from the centre and an angle. The distance from the origin, is, the magnitude of the complex number. The complex number can also be thought as a vector; with a magnitude and a direction.

The magnitude $r = \sqrt[2]{{y^2} + {x^2}}$

And the angle $\phi = tan^{-1} (\frac{y}{x})$

In the polar format we can demote Z with its magnitude and it’s angle, so $Z = \sqrt[2]{{y^2} + {x^2}}$ $\angle tan^{-1} (\frac{y}{x})$

1. Say we want to represent Z = 10 + 15j in polar format. This is in Cartesian format, so we know, that it denotes a point of (10,15) on the graph.

$Z = \sqrt[2]{{15^2} + {10^2}}$ $\angle tan^{-1} (\frac{15}{10})$

$\Rightarrow Z = 18$ $\angle 56.3^{ \circ}$

1. Take another example. Convert P = 15.5 – 10.5j in polar format.

Again, we know that this complex number denotes the point (15.5 , 10.5) in a graph. So

$Z = \sqrt[2]{{15.5^2} + {10.5^2}}$ $\angle tan^{-1} (\frac{-10.5}{15.5})$

$\Rightarrow Z = 18.72$ $\angle -34.1^{ \circ}$

We can convert a complex number in polar format to Cartesian format in the following way.

We know that the Cartesian format of a complex number is in the form

Z = x + yj.

From the basic trigonometry, $cos \phi = \frac{x}{r}$ and $sin \phi = \frac{y}{r}$

$\Rightarrow x=r.cos\phi$ and $y = r.sin\phi$

thus the Cartesian form of the complex number is

$Z = r.cos\phi + r.sin\phi j$

For example: A the Cartesian form of the complex number

$Z=5.385 \angle68.2^{\circ}$ is $Z = 5.385 * cos(68.2) + 5.385 * sin(68.2)j = 2 + 5j$.

Multiplication, Division and Reciprocal of a complex number

Multiplication and division is easier when the complex number is represented in the polar format. For example:

Multiplication

We just multiply the scalar and add up the angles.

For example:

1. $(18\angle90^{ \circ})(10\angle-45^{ \circ}) = 180\angle45^{ \circ}$

2. $(176\angle-320^{ \circ})(5\angle-10^{ \circ}) = 880\angle-330^{ \circ}$ or $880\angle30^{\circ}$

Division

We divide the magnitude as usual, then to find the angle, we subtract the denominator from the numerator.

For example:

1. $\frac{18\angle90^\circ}{10\angle-45^\circ} =1.8\angle45^\circ$

Reciplocal

Reciprocal of a complex number is also easy to do, when the complex number is in the polar format. Just follow the division procedure, except the angle in numerator is zero.

1. $\frac{1}{23\angle45^\circ}=\frac{1\angle0^\circ}{23\angle45^\circ}=0.043\angle-45^\circ$

2. $\frac{1}{15\angle-19^\circ}=\frac{1\angle0^\circ}{15\angle-19^\circ}=0.0667\angle19^\circ$

That’s that.

In the next post, I will design a passive high pass and a low pass filter, and then you will see how important complex numbers are. It makes our lives a lot easier.

PS: I am not an expert in this, so please inform me, if I have done anything wrong, or something that you didn’t understand.

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