We are all taught that finding the square root of a number is impossible, but in fact, it is. That is the significance of imaginary numbers. Imaginary numbers are numbers that have a negative result when they are squared, and hence would be able to help in finding the square root of a negative number by serving as a constant. Often, when an imaginary number becomes linked with a real number, the result is a complex number, which has the basic form of a+bi, where a is the “real number” part and b is the bi is the “imaginary” part. Complex numbers have many applications in the real world, and that is why they are so important and prevalent in our lives.
I’d like to begin my explanation on complex numbers by talking about how to perform operations with them. The best way to add complex numbers is arithmetically. You add the numbers just as you would add real numbers or when you combine terms. You take the real numbers and add them, and you add the coefficients of the imaginary numbers and put i as the constant at the end. Look at the following example: Add: (7 + 5i) + (8 - 3i) Now let’s look at multiplication. You would multiple these numbers using foil (first, outer, inner last) just as you would when you have two expressions with two terms.
Example: (3 + 2i)(1 + 7i)
(3 + 2i)(1 + 7i) = 3×1 + 3×7i + 2i×1+ 2i×7i

= 3 + 21i + 2i + 14i2

= 3 + 21i + 2i - 14
(because i2 = -1)

= -11 + 23i

Now that you have seen a bit about how complex numbers would be used in equations, I’d like to talk about when complex numbers first started appearing.
e). Girolamo Cardano, from The Great Art On the Cube Equal to the First Power and Number Demonstration Let the cube be equal to the first power and constant and let DC and DF be two cubes the product of the sides of which, AB and BC, is equal to one-third the coefficient of x, and let the sum of these cubes be equal to the constant. I say that AC is the value of x. Now since AB × BC equals one-third the coefficient of x, 3(AB × BC ) will equal the coefficient of x, and the product of AC and 3(AB ×BC ) is the whole first power, AC having been assumed to be x. But AC × 3(AB ×BC ) makes six bodies, three of which are AB ×BC and the other three, BC × AB2 . Therefore these six bodies are equal to the whole first power, and these [six bodies] plus the cubes DC and DF constitute FIGURE 5.8. Cardano’s demonstration. 238 5. Algebra: The Search for an Elusive Formula FIGURE 5.9. Cardano’s cube. the cube AE, according to the first proposition of Chapter VI. The cubes DC and DF are also equal to the given number. Therefore the cube AE is equal to the given first power and number, which was to be proved. It remains to be shown that 3AC (AB ×BC ) is equal to the six bodies. This is clear enough if I prove that AB(BC ×AC ) equals the two bodies AB ×BC 2 and BC ×AB2 , for the product of AC and (AB ×BC ) is equal to the product of AB and the surface BE — since all sides are equal to all sides — but this [i.e., AB × BE] is equal to the product of AB and (CD + DE); the product AB × DE is equal to the product CB × AB2 , since all sides are equal to all sides; and therefore AC (AB × BC ) is equal to AB × BC 2 plus BC × AB2 , as was proposed. Rule The rule, therefore, is: When the cube of one-third the coefficient of x is not greater than the square of one-half the constant of the equation, subtract the former from the latter and add the square root of the remainder to one-half the constant of the equation and, again, subtract it from the same half, and you will have, as was said, a binomium and its apotome, the sum of the cube roots of which constitutes the value of x. For example, x 3 = 6x + 40. Raise 2, one-third the coefficient of x, to the cube, which makes 8; subtract this from 400, the square of 20, one-half the constant, making 392; the square root of this added to 20 makes 20+√ 392, and subtracted from 20 makes 20− √ 392; and the sum of the cube roots of these, p3 20 + √ 392 + p3 20 − √