# How To Simplify Imaginary Numbers

By Kathleen Cantor, 03 Jul 2020

An imaginary number is essentially a complex number - or two numbers added together. The difference is that an imaginary number is the product of a real number, say *b,* and an imaginary number, *j*. The imaginary unit is defined as the square root of -1. Here's an example: sqrt(-1).

So the square of the imaginary unit would be -1. Here's an example: *j2* = -1.

The square of an imaginary number, say *bj*, is (*bj*)2 = -*b*2. An imaginary number can be added to a real number to form another complex number. For example, *a *+ *bj* is a complex number with *a* as the *real part* of the complex number and *b* as the *imaginary part* of the complex number.

Complex numbers are sometimes represented using the Cartesian plane. The x-axis represents the real part, with the imaginary part on the y-axis. From this representation, the magnitude of a complex number is defined as the point on the Cartesian plane where the real and the imaginary parts intersect.

Care must be taken when handling imaginary numbers expressed in the form of square roots of negative numbers. For example:

Sqrt(-6) = sqrt(-1) * sqrt(6)

= sqrt(6)*j*.

However, this does not apply to the square root of the following,

Sqrt(-4 * -3) = sqrt(12)

And not sqrt(-4) * sqrt(-3) = 2*j * sqrt(3)j*

So when the negative signs can be neutralized before taking the square root, it becomes wrong to simplify to an imaginary number.

**Simplifying Imaginary Numbers**

The nature of problems solved these days has increased the chances of encountering complex numbers in solutions. And since imaginary numbers are not physically real numbers, simplifying them is important if you want to work with them. We'll consider the various ways you can simplify imaginary numbers.

**Powers of the Imaginary Unit**

The imaginary unit, *j,* is the square root of -1. Hence the square of the imaginary unit is -1. This follows that:

*j0 =*1*j1*=*j**j2 =*-1*j3 = j2*x*j =*-1 x*j*= -*j**j4 = j2*x*j2 =*-1 x -1 = 1*j5 = j4*x*j =*1 x*j = j**j6 = j4*x*j2 =*1 x -1 = -1

Understanding the powers of the imaginary unit is essential in understanding imaginary numbers. Following the examples above, it can be seen that there is a pattern for the powers of the imaginary unit. It always simplifies to -1, -*j*, 1, or *j*. A simple shortcut to simplify an imaginary unit raised to a power is to divide the power by 4 and then raise the imaginary unit to the power of the reminder.

For example: to simplify *j*23, first divide 23 by 4.

23/4 = 5 remainder 3. So *j23 =* *j3 = -j *…… as already shown above.

Consider another example

(*2j)6* = 26 x *j6* = 64 x -1 = -64

**Conjugates**

Simply put, a conjugate is when you switch the sign between the two units in an equation. The conjugate of a complex number would be another complex number that also had a real part, imaginary part, the same magnitude. However, it has the opposite sign from the imaginary unit.

For example, if *x* and *y* are real numbers, then given a complex number, *z = x + yj*, the complex conjugate of *z* is *x – yj.*

Complex conjugates are very important in complex numbers because the product of complex conjugates is a real number of the form *x2* + *y2*. They are important in finding the roots of polynomials.

To illustrate the concept further, let us evaluate the product of two complex conjugates.

(*x + yj*)(*x – yj*)

= x2 – xy*j* + xy*j* – y2*j2*

= x2 – (-y2)

= x2 + y2

**Example**

*Simplify the expression* 2 / (1 + 3*j*)

The above expression is a complex fraction where the denominator is a complex number. As it is, we can't simplify it any further except if we rationalized the denominator. The concept of conjugates would come in handy in this situation.

When dealing with fractions, if the numerator and denominator are the same, the fraction is equal to 1.

Hence (1 – 3*j*) / (1 – 3*j*) = 1

Also, when a fraction is multiplied by 1, the fraction is unchanged. So we will multiply the complex fraction 2 / (1 + 3*j*) by (1 – 3*j*) / (1 – 3*j*) where (1 – 3*j*) is the complex conjugate of (1 + 3*j*).

(2 / (1 + 3*j*)) * ((1 – 3*j*) / (1 – 3*j*))

= 2(1 – 3*j*) / (1 + 3*j*)(1 – 3*j*)

The denominator of the fraction is now the product of two conjugates. As stated earlier, the product of the two conjugates will simplify to the sum of two squares.

Hence

2(1 - 3*j*) / (1 + 3*j*)(1 – 3*j*) = 2(1 - 3*j*) / (12 + 32)

= 2(1 - 3*j*) / (1 + 9)

= (2 - 6*j*) / 10

= 0.2 - 0.6*j*

We've been able to simplify the fraction by applying the complex conjugate of the denominator.

**De Moivre’s Theorem**

Complex numbers can also be written in polar form. The earlier form of *x* + y*j* is the rectangular form of complex numbers. Given a complex number z = x + y*j*, then the complex number can be written as z = r(cos(*n*) + *j*sin(*n*))

Where r = sqrt(x2 + y2)

*n* = arctan (y/x).

De Moivre’s theorem states that r(cos(*n*) + *j*sin(*n*))p = rp(cos(p*n*) + *j*sin(p*n*))

Here's an example that can help explain this theory.

**Example**

*Given z = 1 + j, find z2*

Let us convert the complex number to polar form.

r = sqrt(12 + 12) = sqrt (2)

*n* = arctan (1/1) = 45o

So *z* in polar form is z = sqrt(2)(cos(45) + *j*sin(45)).

Z2 = sqrt(2)(cos(45) + *j*sin(45))2.

You can see what happens when we apply De Moivre’s theorem:

sqrt(2)(cos(45) + *j*sin(45))2 = (sqrt(2))2(cos(2 x 45) + *j*sin(2 x 45))

= 2(cos(90) + *j*sin(90))

= 2(0 + *j*) or = 2*j*

So z2 = (1 + *j*)2 = 2*j*

You can verify the answer by expanding the complex number in rectangular form.

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