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Section 11.1 Complex numbers

Note: half a lecture

Subsection 11.1.1 The complex plane

In this chapter we consider approximation of functions, or in other words functions as limits of sequences and series. We will extend some results we already saw to a somewhat more general setting, and we will look at some completely new results. In particular, we consider complex-valued functions. We gave complex numbers as examples before, but let us start from scratch and properly define the complex number field.

A complex number is just a pair \((x,y) \in \R^2\) on which we define multiplication (see below). We call the set the complex numbers and denote it by \(\C\text{.}\) We identify \(x \in \R\) with \((x,0) \in \C\text{.}\) The \(x\)-axis is then called the real axis and the \(y\)-axis is called the imaginary axis. The set \(\C\) is sometimes called the complex plane.


\begin{equation*} \begin{aligned} & (x,y) + (s,t) := (x+s,y+t) , \\ & (x,y) (s,t) := (xs-yt,xt+ys) . \end{aligned} \end{equation*}

Under the identification above, we have \(0 = (0,0)\) and \(1 = (1,0)\text{.}\) These two operations make the plane into a field (exercise).

We write a complex number \((x,y)\) as \(x+iy\text{,}\) where we define 1 

\begin{equation*} i := (0,1) . \end{equation*}

Notice that \(i^2 = (0,1)(0,1) = (0-1,0+0) = -1\text{.}\) That is, \(i\) is a solution to the polynomial equation

\begin{equation*} z^2+1=0 . \end{equation*}

From now on, we will not use the notation \((x,y)\) and use only \(x+iy\text{.}\) See Figure 11.1.

Figure 11.1. The points \(1\text{,}\) \(i\text{,}\) \(x\text{,}\) \(iy\text{,}\) and \(x+iy\) in the complex plane.

We generally use \(x,y,r,s,t\) for real values and \(z,w,\xi,\zeta\) for complex values, although that is not a hard and fast rule. In particular, \(z\) is often used as a third real variable in \(\R^3\text{.}\)

Definition 11.1.1.

Suppose \(z= x+iy\text{.}\) We call \(x\) the real part of \(z\text{,}\) and we call \(y\) the imaginary part of \(z\text{.}\) We write

\begin{equation*} \Re\, z := x , \qquad \Im\, z := y . \end{equation*}

Define complex conjugate as

\begin{equation*} \bar{z} := x-iy , \end{equation*}

and define modulus as

\begin{equation*} \sabs{z} := \sqrt{x^2+y^2} . \end{equation*}

Modulus is the complex analogue of the absolute value and has similar properties. For example, \(\sabs{zw} = \sabs{z} \, \sabs{w}\) (exercise). The complex conjugate is a reflection of the plane across the real axis. The real numbers are precisely those numbers for which the imaginary part \(y=0\text{.}\) In particular, they are precisely those numbers which satisfy the equation

\begin{equation*} z = \bar{z} . \end{equation*}

As \(\C\) is really \(\R^2\text{,}\) we let the metric on \(\C\) be the standard euclidean metric on \(\R^2\text{.}\) In particular,

\begin{equation*} \sabs{z} = d(z,0) , \qquad \text{and also} \qquad \sabs{z-w} = d(z,w) . \end{equation*}

So the topology on \(\C\) is the same exact topology as the standard topology on \(\R^2\) with the euclidean metric, and \(\sabs{z}\) is equal to the euclidean norm on \(\R^2\text{.}\) Importantly, since \(\R^2\) is a complete metric space, then so is \(\C\text{.}\) As \(\sabs{z}\) is the euclidean norm on \(\R^2\text{,}\) we have the triangle inequality of both flavors:

\begin{equation*} \sabs{z+w} \leq \sabs{z}+\sabs{w} \qquad \text{and} \qquad \big\lvert \sabs{z}-\sabs{w} \big\rvert \leq \sabs{z-w} . \end{equation*}

The complex conjugate and the modulus are even more intimately related:

\begin{equation*} \sabs{z}^2 = x^2+y^2 = (x+iy)(x-iy) = z \bar{z} . \end{equation*}

Remark 11.1.2.

There is no natural ordering on the complex numbers. In particular, no ordering that makes the complex numbers into an ordered field. Ordering is one of the things we lose when we go from real to complex numbers.

Subsection 11.1.2 Complex numbers and limits

It is not hard to show that the algebraic operations are continuous. This is because convergence in \(\R^2\) is the same as convergence for each component and we already know that the real algebraic operations are continuous. For example, write \(z_n = x_n + i\,y_n\) and \(w_n = s_n + i\,t_n\text{,}\) and suppose that \(\lim \, z_n = z = x+i\,y\) and \(\lim \, w_n = w = s+i\,t\text{.}\) Let us show

\begin{equation*} \lim_{n\to\infty} z_n w_n = zw . \end{equation*}


\begin{equation*} z_n w_n = (x_ns_n-y_nt_n) + i(x_nt_n+y_ns_n) . \end{equation*}

The topology on \(\C\) is the same as on \(\R^2\text{,}\) and so \(x_n \to x\text{,}\) \(y_n \to y\text{,}\) \(s_n \to s\text{,}\) and \(t_n \to t\text{.}\) Hence,

\begin{equation*} \lim_{n\to\infty} (x_ns_n-y_nt_n) = xs-yt \qquad \text{and} \qquad \lim_{n\to\infty} (x_nt_n+y_ns_n) = xt+ys . \end{equation*}

As \((xs-yt)+i(xt+ys) = zw\text{,}\) then

\begin{equation*} \lim_{n\to\infty} z_n w_n = zw . \end{equation*}

Similarly the modulus and the complex conjugate are continuous functions. We leave the proof of the following proposition as an exercise.

As we have seen above, convergence in \(\C\) is the same as convergence in \(\R^2\text{.}\) In particular, a sequence in \(\C\) converges if and only if the real and imaginary parts converge. Therefore, feel free to apply everything you have learned about convergence in \(\R^2\text{,}\) as well as applying results about real numbers to the real and imaginary parts.

We also need convergence of complex series. Let \(\{ z_n \}\) be a sequence of complex numbers. The series

\begin{equation*} \sum_{n=1}^\infty z_n \end{equation*}

converges if the limit of partial sums converges, that is, if

\begin{equation*} \lim_{k\to\infty} \sum_{n=1}^k z_n \qquad \text{exists.} \end{equation*}

As before, we sometimes write \(\sum z_n\) for the series. A series converges absolutely if \(\sum \sabs{z_n}\) converges.

We say a series is Cauchy if the sequence of partial sums is Cauchy. The following two propositions have essentially the same proofs as for real series and we leave them as exercises.

The series \(\sum \sabs{z_n}\) is a real series. All the convergence tests (ratio test, root test, etc.) that talk about absolute convergence work with the numbers \(\sabs{z_n}\text{,}\) that is, they are really talking about convergence of series of nonnegative real numbers. You can directly apply these tests them without needing to reprove anything for complex series.

Subsection 11.1.3 Complex-valued functions

When we deal with complex-valued functions \(f \colon X \to \C\text{,}\) what we often do is to write \(f = u+i\,v\) for real-valued functions \(u \colon X \to \R\) and \(v \colon X \to \R\text{.}\)

Suppose we wish to integrate \(f \colon [a,b] \to \C\text{.}\) We write \(f = u+i\,v\) for real-valued \(u\) and \(v\text{.}\) We say that \(f\) is Riemann integrable if \(u\) and \(v\) are Riemann integrable, and in this case we define

\begin{equation*} \int_a^b f := \int_a^b u + i \int_a^b v . \end{equation*}

We make the same definition for every other type of integral (improper, multivariable, etc.).

Similarly when we differentiate, write \(f \colon [a,b] \to \C\) as \(f = u+i\,v\text{.}\) Thinking of \(\C\) as \(\R^2\) we say that \(f\) is differentiable if \(u\) and \(v\) are differentiable. For a function valued in \(\R^2\text{,}\) the derivative was represented by a vector in \(\R^2\text{.}\) Now a vector in \(\R^2\) is a complex number. In other words, we write the derivative as

\begin{equation*} f'(t) := u'(t) + i \, v'(t) . \end{equation*}

The linear operator representing the derivative is the multiplication by the complex number \(f'(t)\text{,}\) so nothing is lost in this identification.

Subsection 11.1.4 Exercises

Exercise 11.1.1.

Check that \(\C\) is a field.

Exercise 11.1.2.

Prove that for \(z,w \in \C\text{,}\) we have \(\sabs{zw} = \sabs{z} \, \sabs{w}\text{.}\)

Exercise 11.1.6.

Considering the definition of complex multiplication, given \(x +iy\) define the matrix \(\left[ \begin{smallmatrix} x & -y \\ y & x \end{smallmatrix} \right]\text{.}\) Prove that

  1. The action of this matrix on a vector \((s,t)\) is the same as the action of multiplying \((x+iy)(s+it)\text{.}\)

  2. Multiplying two such matrices is the same multiplying the underlying complex numbers and then finding the corresponding matrix for the product. In other words, we can think of the field \(\C\) as also a subset of the 2-by-2 matrices.

  3. Show that \(\left[ \begin{smallmatrix} x & -y \\ y & x \end{smallmatrix} \right]\) has eigenvalues \(x+iy\) and \(x-iy\text{.}\) Recall that \(\lambda\) is an eigenvalue of a matrix \(A\) if \(A-\lambda I\) (a complex matrix in our case) is not invertible, or in other words if it has linearly dependent rows: That is, one row is a (complex) multiple of the other

Exercise 11.1.7.

Prove the Bolzano–Weierstrass theorem for complex sequences. Suppose \(\{ z_n \}\) is a bounded sequence of complex numbers, that is, there exists an \(M\) such that \(\sabs{z_n} \leq M\) for all \(n\text{.}\) Prove that there exists a subsequence \(\{ z_{n_k} \}\) that converges to some \(z \in \C\text{.}\)

Exercise 11.1.8.

  1. Prove that there is no simple mean value theorem for complex-valued functions: Find a differentiable function \(f \colon [0,1] \to \C\) such that \(f(0) = f(1) = 0\text{,}\) but \(f'(t) \not= 0\) for all \(t \in [0,1]\text{.}\)

  2. However, there is a weaker form of the mean value theorem as there is for vector-valued functions. Prove: If \(f \colon [a,b] \to \C\) is continuous and differentiable in \((a,b)\text{,}\) and for some \(M\text{,}\) \(\sabs{f'(x)} \leq M\) for all \(x \in (a,b)\text{,}\) then \(\sabs{f(b)-f(a)} \leq M \sabs{b-a}\text{.}\)

Exercise 11.1.9.

Prove that there is no simple mean value theorem for integrals for complex-valued functions: Find a continuous function \(f \colon [0,1] \to \C\) such that \(\int_0^1 f = 0\) but \(f(t) \not= 0\) for all \(t \in [0,1]\text{.}\)

Note that engineers use \(j\) instead of \(i\text{.}\)
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