Eventually, you are going to encounter a situation where you will have to deal with square
roots of negative numbers. How can this be done ? After all, a positive number squared or a
negative number squared will always equal a positive number.

Mathematicians have designated a special
number 'i' which is equal to the square root of minus 1. Then, it follows that i^{2} =
-1. To determine the square root of a negative number (-16 for example), take the
square root of the absolute value of the number (square root of 16 = 4) and then multiply it by
'i'. So, the square root of -16 is 4i. As a double check, we can square 4i (4*4 = 16 and
i*i =-1), producing -16. All negative square roots are called
"imaginary numbers" (now you know where that letter 'i' comes from).

Complex Numbers

When a number has the form a + bi (a real number plus an imaginary number) it is called a
"complex number". How do complex numbers "crop up" in mathematics? A good example would be the
roots of the quadratic equation x^{2} -6x + 25 = 0 where the 2 roots are 3 + 4i and
3 - 4i. Can we be sure these are the roots of the equation?
As a double-check, using those roots, we can "rebuild" the original equation by

Since i^{2}= -1 then -(16)*i^{2} becomes -(-16) = 16 and so:

X^{2} -6X +25 =0

Complex Number Multiplication

Addition and subtraction of complex numbers pretty much follow the rules of basic arithmetic and
so we won't discuss these. Multiplication starts getting a little tricky. Consider:

(5 + 6i) * (7 + 8i)

This equals 35 + 40i + 42i + 48i^{2}

As we saw above, i^{2} = -1 so 48i^{2} = -48

So answer= -13 + 82i

Complex Number Division

Were you wondering - is division more difficult than multiplication? Sure is. First we must define
a new term - conjugate, whereby the conjugate of a + bi = a-bi.
Example - the conjugate of (3 + 4i) is (3 - 4i).
The main principle to remember in complex number division is that
we multiply the "top" and "bottom" of the fraction by the conjugate of the denominator.
Time for an example: let's divide (9 +3i) by (7 +5i)

(9 + 3i)
—————
(7 + 5i)

The denominator is (7 + 5i) and its conjugate is (7- 5i)

Answer = 1.054054054054054 & -0.32432432432432434 i

Square Root of a Complex Number

Now we move on to even greater difficulty.
Time to define another term - modulus, whereby the modulus of a complex number a + bi equals the square root of
(a^{2} + b^{2}).
The modulus of a complex number is generally represented by the letter 'r' and so:

r = Square Root (a^{2} + b^{2})

Next we'll define these 2 quantities:

y = Square Root ((r-a)/2)

x = b/2y

Finally, the 2 square roots of a complex number are:

root _{1} = x + yi

root _{2} = -x - yi

An example should make this procedure much clearer.
Find the square root of 12 + 16i

r = Square Root (12^{2} + 16^{2})

r = Square Root (144 + 256) = 20

y = Square Root ((20-12)/2) = 2

x = 16/(2*2) = 4

root _{1} = 4 + 2i

root _{2} = -4 - 2i

Even though you have a calculator that can do these calculations for
you, now you know the procedures for complex number arithmetic.