A complex number is an expression of the form $z=a+bi$, where a and b are real numbers called the real part and the imaginary part of z, respectively.

The sum and product of two complex numbers $z=a+bi$ and $w=c+di$ (where a, b, c, and d are real numbers) are defined, respectively, as follows:

and

Exercise.

The (complex) conjugate of a complex number $a+bi$ is the complex number $a-bi$. We denote the conjugate of the complex number z by $\overline{z}$.

We leave the proofs of (a), (d), and (e) to the reader.

(b) Let $z=a+bi$ and $w=c+di$, where a, b, c, $d\in R$. Then

(c) For z and w, we have

For any complex number $z=a+bi,z\overline{z}$ is real and nonnegative, for

This fact can be used to define the absolute value of a complex number.

Let $z=a+bi$, where a, $a,b\in R$. The absolute value (or modulus) of z is the real number $\sqrt{a^2+b^2}$. We denote the absolute value of z by $\left|z\right|$.

Observe that $z\overline{z}=|z{|}^2$. The fact that the product of a complex number and its conjugate is real provides an easy method for determining the quotient of two complex numbers; for if $c+di\ne 0$, then

(a) By Theorem D.2, we have

proving (a).

(b) For the proof of (b), apply (a) to the product $\left({\displaystyle \frac{z}{w}}\right)w$.

(c) For any complex number $x=a+bi$, where $a,b\in R$, observe that

Thus $x+\overline{x}$ is real and satisfies the inequality $x+\overline{x}\le 2\left|x\right|$. Taking $x=w\overline{z}$, we have, by Theorem D.2 and (a),

Using Theorem D.2 again gives

By taking square roots, we obtain (c).

(d) From (a) and (c), it follows that

So

proving (d).

It is interesting as well as useful that complex numbers have both a geometric and an algebraic representation. Suppose that $z=a+bi$, where a and b are real numbers. We may represent z as a vector in the complex plane (see Figure D.1(a)). Notice that, as in ${\text{R}}^2$, there are two axes, the real axis and the imaginary axis. The real and imaginary parts of z are the first and second coordinates, and the absolute value of z gives the length of the vector z. It is clear that addition of complex numbers may be represented as in ${\text{R}}^2$ using the parallelogram law.

In Section 2.7 (p.133), we introduce Euler’s formula. The special case $e^{i\theta }=\cos \theta +i\sin \theta$ is of particular interest. Because of the geometry we have introduced, we may represent the vector $e^{i\theta }$ as in Figure D.1(b); that is, $e^{i\theta }$ is the unit vector that makes an angle $\theta$ with the positive real axis. From this figure, we see that any nonzero complex number z may be depicted as a multiple of a unit vector, namely, $z=\left|z\right|e^{i\varphi }$, where $\varphi$ is the angle that the vector z makes with the positive real axis. Thus multiplication, as well as addition, has a simple geometric interpretation: If $z=\left|z\right|e^{i\theta }$ and $w=\left|w\right|e^{iw}$ are two nonzero complex numbers, then from the properties established in Section 2.7 and Theorem D.3, we have

So zw is the vector whose length is the product of the lengths of z and w, and makes the angle $\theta +w$ with the positive real axis.

Our motivation for enlarging the set of real numbers to the set of complex numbers is to obtain a field such that every polynomial with nonzero degree having coefficients in that field has a zero. Our next result guarantees that the field of complex numbers has this property.

We want to find $z_0$ in C such that $p(z_0)=0$. Let m be the greatest lower bound of $\{|p(z)|\text{: }z\in C\}$. For $|z|=s>0$, we have

Because the last expression approaches infinity as s approaches infinity, we may choose a closed disk D about the origin such that $|p(z)|>m+1$ if z is not in D. It follows that m is the greatest lower bound of $\{|p(z)|\text{: }z\in D\}$. Because D is closed and bounded and p(z) is continuous, there exists $z_0$ in D such that $|p(z_0)|=m$. We want to show that $m=0$. We argue by contradiction.

Assume that $m\ne 0$. Let $q(z)={\displaystyle \frac{p(z+z_0)}{p(z_0)}}$. Then q(z) is a polynomial of degree n, $q(0)=1$, and $|q(z)|\ge 1$ for all z in C. So we may write

where $b_k\ne 0$. Because $-{\displaystyle \frac{|b_k|}{b_k}}$ has modulus one, we may pick a real number $\theta$ such that $e^{ik\theta }=-{\displaystyle \frac{|b_k|}{b_k}}$, or $e^{ik\theta }{b}_k=-|b_k|$. For any $r>0$, we have

Choose r small enough so that $1-|b_k|r^k>0$. Then

Now choose r even smaller, if necessary, so that the expression within the brackets is positive. We obtain that $|q(r{e}^{i\theta })|<1$. But this is a contradiction.

If $p(z)=a_n{z}^n+a_{n-1}{z}^{n-1}+\cdots +a_{1}z+a_0$ is a polynomial of degree $n\ge 1$ with complex coefficients, then there exist complex numbers $c_1,c_2,\dots ,c_n$ (not necessarily distinct) such that

Exercise.

A field is called algebraically closed if it has the property that every polynomial of positive degree with coefficients from that field factors as a product of polynomials of degree 1. Thus the preceding corollary asserts that the field of complex numbers is algebraically closed.