The results of Section AppQuadEqus tell us that the equation \(x^{2} + 1 = 0\) has no real number solutions. However, it would have solutions if we could make sense of \(\sqrt {-1}\). The Complex Numbers do just that - they give us a mechanism for working with \(\sqrt {-1}\). As such, the set of complex numbers fill in an algebraic gap left by the set of real numbers.

Here’s the basic plan. There is no real number \(x\) with \(x^2 = -1\), since for any real number \(x^2 \geq 0\). However, we could formally extract square roots and write \(x = \pm \sqrt {-1}\). We build the complex numbers by relabeling the quantity \(\sqrt {-1}\) as \(i\), the unfortunately misnamed imaginary unit. The number \(i\), while not a real number, is defined so that it plays along well with real numbers and acts very much like any other radical expression. For instance, \(3(2i) = 6i\), \(7i-3i = 4i\), \((2-7i) + (3 + 4i) = 5-3i\), and so forth. The key properties which distinguish \(i\) from the real numbers are listed below.

Property 1 in Definition idefn establishes that \(i\) does act as a square root of \(-1\), and property 2 establishes what we mean by the ‘principal square root’ of a negative real number. In property 2, it is important to remember the restriction on \(c\). For example, it is perfectly acceptable to say \(\sqrt {-4} = i \sqrt {4} = i(2) = 2i\). However, \(\sqrt {-(-4)} \neq i \sqrt {-4}\), otherwise, we’d get

\[ 2 = \sqrt {4} = \sqrt {-(-4)} = i \sqrt {-4} = i (2i) = 2i^2 = 2(-1) = -2,\]

which is unacceptable. The moral of this story is that the general properties of radicals do not apply for even roots of negative quantities. With Definition idefn in place, we can define the set of complex numbers.

Complex numbers include things you’d normally expect, like \(3+2i\) and \(\frac {2}{5} - i\sqrt {3}\). However, don’t forget that \(a\) or \(b\) could be zero, which means numbers like \(3i\) and \(6\) are also complex numbers. In other words, don’t forget that the complex numbers include the real numbers, so \(0\) and \(\pi - \sqrt {21}\) are both considered complex numbers. The arithmetic of complex numbers is as you would expect. The only things you need to remember are the two properties in Definition idefn. The next example should help recall how these animals behave.

In the previous example, we used the ‘conjugate’ idea from Section AppRadEqus to divide two complex numbers. More generally, the complex conjugate of a complex number \(a+bi\) is the number \(a-bi\). The notation commonly used for complex conjugation is a ‘bar’: \(\overline {a+bi} = a-bi\). For example, \(\overline {3+2i} = 3-2i\) and \(\overline {3-2i} = 3+2i\). To find \(\overline {6}\), we note that \(\overline {6} = \overline {6+0i}= 6 - 0i = 6\), so \(\overline {6} = 6\). Similarly, \(\overline {4i} = -4i\), since \(\overline {4i} = \overline {0 + 4i} = 0 - 4i = -4i\). Note that \(\overline {3+\sqrt {5}} = 3 + \sqrt {5}\), not \(3 - \sqrt {5}\), since \(\overline {3+\sqrt {5}} = \overline {3+\sqrt {5} + 0i} = 3+\sqrt {5} - 0i = 3+\sqrt {5}\). Here, the conjugation specified by the ‘bar’ notation involves reversing the sign before \(i = \sqrt {-1}\), not before \(\sqrt {5}\). The properties of the conjugate are summarized in the following theorem.

Theorem conjugateprops says in part that complex conjugation works well with addition, multiplication and powers. The proofs of these properties can best be achieved by writing out \(z = a+bi\) and \(w = c+di\) for real numbers \(a\), \(b\), \(c\) and \(d\). Next, we compute the left and right sides of each equation and verify that they are the same.

The proof of the first property is a very quick exercise. To prove the second property, we compare \(\overline {z+w}\) with \(\overline {z} + \overline {w}\). We have \(\overline {z} + \overline {w} = \overline {a+bi} + \overline {c+di} = a-bi + c-di\). To find \(\overline {z+w}\), we first compute

\[z+w = (a+bi) + (c+di) = (a+c)+(b+d)i\]

so

\[\overline {z+w} = \overline {(a+c)+(b+d)i} = (a+c) - (b+d)i = a+c - bi - di = a - bi + c - di = \overline {z} + \overline {w}\]

As such, we have established \(\overline {z+w} = \overline {z}+\overline {w}\). The proof for multiplication works similarly. The proof that the conjugate works well with powers can be viewed as a repeated application of the product rule, and is best proved using a technique called Mathematical Induction. The last property is a characterization of real numbers. If \(z\) is real, then \(z = a + 0i\), so \(\overline {z} = a - 0i = a = z\). On the other hand, if \(z=\overline {z}\), then \(a+bi = a - bi\) which means \(b=-b\) so \(b=0\). Hence, \(z = a +0i = a\) and is real.

We now return to the business of solving quadratic equations. Consider \(x^2-2x+5 = 0\). The discriminant \(b^2 - 4ac = -16\) is negative, so we know by Theorem discriminanttheoremrealversion there are no real solutions, since the Quadratic Formula would involve the term \(\sqrt {-16}\). Complex numbers, however, are built just for such situations, so we can go ahead and apply the Quadratic Formula to get:

\[ x = \frac {-(-2) \pm \sqrt {(-2)^2-4(1)(5)}}{2(1)} = \frac {2 \pm \sqrt {-16}}{2} = \frac {2 \pm 4i}{2} = 1 \pm 2i.\]

It is no coincidence that the non-real solutions to the equations in Example complexsolnsreviewex appear in complex conjugate pairs. Any time we use the Quadratic Formula to solve an equation with real coefficients, the answers will form a complex conjugate pair owing to the \(\pm \) in the Quadratic Formula. This leads us to a generalization of Theorem discriminanttheoremrealversion which we state below.

We will have much more to say about complex solutions to equations in Section ComplexZeros and we will revisit Theorem discriminanttheoremcomplexversion then.