9.3: Complex Numbers

Imaginary Numbers

Although square roots of negative numbers such as \(\sqrt{-4}\) are not real numbers, they occur often in Mathematics and its applications. Mathematicians began working with square roots of negative numbers in the sixteenth century in their attempts to solve quadratic and cubic equations. René Descartes gave them the name imaginary numbers, which reflected the mistrust with which mathematicians regarded them at the time. Today, however, such numbers are well understood and used routinely by scientists and engineers.

We begin by defining a new number, called \(i\), whose square is \(-1\).

The letter \(i\) used in this way is not a variable; it is the name of a specific number and hence is a constant. We have seen this sort of designation before—\(\pi\) represents the approximation \(3.14159\ldots\) and \(e\) represents the approximation \(2.71828\ldots\).

The square root of any negative number can be written as the product of a real number and \(i\). For example,\[\begin{array}{rcl} \sqrt{-4} & = & \sqrt{-1 \cdot 4} \\[6pt] & = & \sqrt{-1} \sqrt{4} \\[6pt] & = & i \cdot 2 \\[6pt] & = & 2i \\[6pt] \end{array}\nonumber\]Any number that is the product of \(i\) and a real number is called an imaginary number.

Examples of imaginary numbers are \(3 i\), \(\frac{7}{8} i\), \(-38 i\), and \(i \sqrt{5}\).

Just as each positive number has two real-valued square roots, every negative number has two imaginary square roots. For example, the two square roots of \(-9\) are \(3 i\) and \(-3 i\).

Complex Numbers

Consider the quadratic equation\[x^2-2 x+5=0.\nonumber\]Using the Quadratic Formula to solve the equation, we find\[x=\dfrac{-(-2) \pm \sqrt{(-2)^2-4(1)(5)}}{2}=\dfrac{2 \pm \sqrt{-16}}{2}.\nonumber\]If we now replace \(\sqrt{-16}\) by \(4 i\), we have\[x=\dfrac{2 \pm 4 i}{2}=1 \pm 2 i.\nonumber\]The two solutions are \(1+2 i\) and \(1-2 i\). These are examples of complex numbers.

Examples of complex numbers are\[3-5 i, \quad 2+\sqrt{7} i, \quad \dfrac{4-i}{3}, \quad 6 i, \text{ and} \quad -9.\nonumber\]The real and imaginary parts of a complex number cannot be combined. Thus, two complex numbers \(z_1\) and \(z_2\) are equal if and only if their real parts are equal and their imaginary parts are equal.

Arithmetic of Complex Numbers

We add and subtract complex numbers by combining their real and imaginary parts separately. For example,\[\begin{array}{rcl} (4+5 i)+(2-3 i) & = & (4+2)+(5-3) i \\[6pt] & = & 6+2 i \\[6pt] \end{array}\nonumber\]The algebraic form of this rule can be stated as follows.

Products of Complex Numbers

We use the following theorem to help us find the product of two imaginary numbers.

For example,\[\begin{array}{rcl} (3 i)(4 i) & = & 3 \cdot 4 i^2 \\[6pt] & = & 12(-1) \\[6pt] & = & -12 \\[6pt] \end{array}\nonumber\]To find the product of two complex numbers, we use the Distributive Law as if the numbers were binomials.

Quotients of Complex Numbers

To find the quotient of two complex numbers, we use the technique of rationalizing the denominator. First, consider division by a pure imaginary number.

Perhaps you recall that to rationalize a binomial denominator, we multiply by its conjugate. For example, to rationalize the denominator of \(\frac{5}{2+\sqrt{3}}\), we multiply numerator and denominator by \(2-\sqrt{3}\). A similar technique works for dividing complex numbers. We first define the conjugate of a complex number.

The conjugate of a complex number has several useful properties. In particular, the product of a nonzero complex number and its conjugate is always a positive real number.

Now that we have an idea of what a complex conjugate is, and what the product of a complex number and its conjugate become, we are ready to define division of complex numbers.

To illustrate, we calculate the quotient \(\frac{5}{2+3 i}\). We multiply the numerator and denominator by the complex conjugate of the denominator, \(2-3 i\), to obtain\[\dfrac{5}{2+3 i} \cdot \dfrac{2-3 i}{2-3 i}.\nonumber\]The denominator is then a real number, because\[(2+3 i)(2-3 i)=4-6 i+6 i-9 i^2=4+9=13\nonumber\]and the quotient is\[\dfrac{5}{2+3 i} \cdot \dfrac{2-3 i}{2-3 i}=\dfrac{10-15 i}{13}=\dfrac{10}{13}-\dfrac{15}{13} i.\nonumber\]

Graphing Complex Numbers

Real numbers can be plotted on a number line, but we use the complex plane to graph a complex number.

To plot a complex number \(a+b i\), we move \(a\) units from the origin in the horizontal direction and \(b\) units in the vertical direction.

The modulus, or length, of a complex number is its distance from the origin in the complex plane. The modulus of a complex number is analogous to the absolute value of a real number. It is similarly denoted by \(|z|\). If \(z=a+b i\), we can use the Pythagorean Theorem to compute its modulus.

For example, the modulus of \(z=2+3 i\) is\[|z|=\sqrt{2^2+3^2}=\sqrt{13}.\nonumber\]

We can think of the graph of a complex number \(z=a+b i\) as the vector that starts at the origin and ends at the point \((a, b)\) in the complex plane, as shown in Figure \(\PageIndex{5a}\).

Then, the sum of two complex numbers corresponds to the sum of the two vectors representing them. Figure \(\PageIndex{5b}\) illustrates the sum of two complex numbers, \(z_1\) and \(z_2\), by vector addition using the parallelogram law: we form a parallelogram with the vectors \(z_1\) and \(z_2\) as adjacent sides. Their sum is the vector that forms the diagonal of the parallelogram, starting at the origin.

To visualize subtraction, we add the opposite of the second vector because\[z_1-z_2=z_1+\left(-z_2\right).\nonumber\]Recall that the opposite of a vector has the same length, but it points in the opposite direction.

Zeros of Polynomials

A polynomial with real-number coefficients may or may not have real-valued zeros. For example, the polynomial \(x^4+4\) has no real-valued zeros. But a polynomial always has a zero if we allow complex numbers as inputs. Because we can add, subtract, and multiply any two complex numbers, we can evaluate a polynomial function at a complex number. Thus, we can extend the domain of any polynomial to include all complex numbers.

The zeros of a quadratic polynomial \(a x^2+b x+c\), of course, are the solutions of the quadratic equation \(a x^2+b x+c=0\), and those solutions are given by the Quadratic Formula,\[x=\dfrac{-b}{2 a}+\dfrac{\sqrt{b^2-4 a c}}{2 a} \text{ and } x=\dfrac{-b}{2 a}-\dfrac{\sqrt{b^2-4 a c}}{2 a}.\nonumber\]If the discriminant \(D=b^2-4 a c\) is negative, the two solutions are complex conjugates. For example, the solutions of the equation \(x^2-4 x+5=0\) are\[\dfrac{-(-4) \pm \sqrt{(-4)^2-4(1)(5)}}{2(1)}=\dfrac{4 \pm \sqrt{-4}}{2}\nonumber\]or \(z=2+i\) and \(\bar{z}=2-i\). Thus, if we know that \(z=a+b i\) is one complex solution of a quadratic equation, we know that \(\bar{z}=a-b i\) is the other solution.

We can now write a quadratic polynomial, with real coefficients, having any complex number as one of its zeros. The factored form of the quadratic polynomial with zeros \(z\) and \(\bar{z}\) is\[p(x)=(x-z)(x-\bar{z}).\nonumber\]Expanding the right side, we find\[p(x)=x^2-(z+\bar{z}) x+z \bar{z}.\nonumber\]Because \((z+\bar{z})\) and \(z \bar{z}\) are both real numbers, this polynomial has real-valued coefficients.

One of the most important results in mathematics is the Fundamental Theorem of Algebra.

Despite the name, however, the Fundamental Theorem of Algebra needs one linked theorem (called a corollary) to be useful for our current conversation.

Before we continue, let's define the multiplicity of a zero of a polynomial.

For example, the polynomial \(p(x)=(x-5)^2\) has a zero of multiplicity 2 at \(x=5\).

As a result of the Fundamental Theorem of Algebra and the Complete Factorization Theorem, every polynomial can be factored as the product of linear terms.

For example, although the graph of \(y=x^4+4\) shown below has no \(x\)-intercepts, the Fundamental Theorem of Algebra tells us that there are four complex solutions to \(x^4+4=0\), and that \(x^4+4\) can be factored. You can check that the four solutions to \(x^4+4=0\) are\[1+i, \, -1+i, \,-1-i,\, \text{and} \, 1-i.\nonumber\]

For example, if \(x=1+i\), then\[x^2=(1+i)^2=1+2 i+i^2=2 i\nonumber\]and\[x^4=\left(x^2\right)^2=(2 i)^2=-4.\nonumber\]Because each zero corresponds to a factor of the polynomial, the factored form of \(x^4+4\) is\[x^4+4=[x-(1+i)][x-(-1+i)][x-(-1-i)][x-(1-i)].\nonumber\]The four solutions to \(x^4+4=0\) form two complex conjugate pairs, namely \(1 \pm i\) and \(-1 \pm i\). We see again that, for every polynomial with real coefficients, the nonreal zeros always occur in complex conjugate pairs.