Before imaginary numbers were introduced to our system of numbers, there was no solution to the equation *x*² = −1, because the square of 0 is 0 and the square of any positive or negative real number is a positive number, so any number we put in for *x* makes the equation false. This situation became inconvenient, so mathematicians invented the number *i*, called the imaginary unit, as the solution to this equation. If *i *² = −1, then *i *= , and to be consistent with the rest of our number system, we need to express the sqare root of any negative number in terms of *i* before applying any operation to it. Examples: , , . A complex number is a number that can be expressed in the form *a* + *bi*, where *a* and *b* are real numbers. Examples: 2 + 3*i*, 5 − 4*i*, . All real numbers, *a*, can be expressed as *a *+ 0*i* and all imaginary numbers, *bi*, can be expressed as 0 + *bi*, so all real numbers and all imaginary numbers are complex numbers. *a *+ *bi* cannot be combined into one term, as *a* and *bi* are unlike terms, but we can do a lot of other operations with imaginary and complex numbers, for example, addition, subtraction, multiplication, division, exponentiation. For addition and subtraction, combine like terms: (*a *+ *bi *) + (*c* + *di *) = (*a *+ *c*) + (*b *+ *d*)*i*, (*a *+ *bi *) − (*c *+ *di *) = (*a *− *c*) + (*b **− d*)*i*. For multiplication, use FOIL and combine like terms: (*a *+ *bi *)(*c *+ *di *) = *ac *+ *adi *+ *bci *+ *bdi**² = (ac − bd*) + (*ad *+ *bc*)*i*. For division, : Multiply numerator and denominator by *c **− di*, which will result in something in the form of , which can be written as . I won’t go into exponentiation here, as nothing in this essay requires exponentiation beyond squaring. Lastly, two complex numbers are equal if, and only if, their real parts are equal and their imaginary parts are equal: *a *+ *bi *= *c *+ *di* ↔ *a *= *c* and *b =* *d*.

While every textbook that covers complex numbers deals with basic arithmetic operations, in 23 years of teaching high school math, I’ve never seen a textbook with instructions on how to take the square root of an imaginary or complex number. These roots must exist; the result of squaring a complex number, *a *+ *bi*, is (*a *+ *bi *)² = (*a*² − *b*²) *+ *2*abi*, so . Example: (5 + 2*i *)² = 21 + 20*i*, so * * must be 5 + 2*i*. But this does not tell us how to find the square root of a random complex number.

Let’s start with something that seems as basic as we can get it: how can we find ? Each expansion of our number system came about only when we needed a new kind of number to solve some problem. So let’s assume for a moment that is not a new kind of number, but is a member of the most inclusive set of numbers that we have so far, the set of complex numbers. Therefore, = *p *+ *qi* for some real numbers *p* and *q*. Squaring both sides, we get *i *= (*p*² − *q*²) *+* 2pqi. Using the rule for equality of complex numbers, we get 0 = *p*²* − **q*², so *q*² = *p**² *and *q* = ±*p*. Using the same rule, we get 2*pq* = 1. Substituting ±*p* for *q* and solving, we get * * , so , or , so . Checking the four resultant possible solutions for , we determine that , with the other three being equal to , , and .

Imaginary numbers other than *i* can be expressed as *bi*, and . Example: . You can check it if you like.

*TO BE CONTINUED*