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Section 2.4 Complex Expressions

In this section we look at some equations and inequalities that will come up throughout the text.

Example 2.4.1 Line equations

The standard form for the equation of a line in the \(xy\)-plane is \(ax + by + d = 0\text{.}\) This line may be expressed via the complex variable \(z = x + yi\text{.}\) For an arbitrary complex number \(\beta = s + ti\text{,}\) note that

\begin{align*} \beta z + \overline{\beta z} \amp = \big[(sx - ty)+(sy+tx)i\big] + \big[(sx - ty) - (sy+tx)i\big]\\ \amp = 2sx - 2ty. \end{align*}

It follows that the line \(ax + by + d = 0\) can be represented by the equation

\begin{gather*} \alpha z + \overline{\alpha z} + d = 0 \tag{equation of a line} \end{gather*}

where \(\alpha = \frac{1}{2}(a - bi)\) is a complex constant and \(d\) is a real number.

Conversely, for any complex number \(\alpha\) and real number \(d\text{,}\) the equation

\begin{equation*} \alpha z + \overline{\alpha z} + d = 0 \end{equation*}

determines a line in \(\mathbb{C}\text{.}\)

We may also view any line in \(\mathbb{C}\) as the collection of points equidistant from two given points.

Given two points \(\gamma\) and \(\beta\) in \(\mathbb{C}\text{,}\) \(z\) is equidistant from both if and only if \(|z - \gamma|^2 = |z - \beta |^2\text{.}\) Expanding this equation, we obtain

\begin{align*} (z - \gamma)(\overline{z - \gamma}) \amp = (z - \beta)(\overline{z - \beta})\\ |z|^2 - \overline{\gamma}z - \gamma\overline{z} + |\gamma|^2 \amp = |z|^2 - \overline{\beta}z - \beta\overline{z} + |\beta|^2\\ \overline{(\beta-\gamma)}z + (\beta-\gamma)\overline{z} + (|\gamma|^2 - |\beta|^2) \amp = 0. \end{align*}

This last equation has the form of a line, letting \(\alpha = \overline{(\beta - \gamma)}\) and \(d = |\gamma|^2 - |\beta|^2\text{.}\)

Conversely, starting with a line we can find complex numbers \(\gamma\) and \(\beta\) that do the trick. In particular, if the given line is the perpendicular bisector of the segment \(\gamma\beta\text{,}\) then \(|z - \gamma| = |z - \beta|\) describes the line. We leave the details to the reader.

Example 2.4.3 Quadratic equations

Suppose \(z_0\) is a complex constant and consider the equation \(z^2 = z_0.\) A complex number \(z\) that satisfies this equation will be called a square root of \(z_0\), and will be written as \(\sqrt{z_0}\text{.}\)

If we view \(z_0 = r_0e^{i\theta_0}\) in polar form with \(r_0 \geq 0\text{,}\) then a complex number \(z = re^{i\theta}\) satisfies the equation \(z^2 = z_0\) if and only if

\begin{equation*} re^{i\theta}\cdot re^{i\theta} =r_0e^{i\theta_0}. \end{equation*}

In other words, \(z\) satisfies the equation if and only if \(r^2 = r_0\) and \(2\theta = \theta_0\) (modulo \(2\pi\)).

As long as \(r_0\) is greater than zero, we have two solutions to the equation, so that \(z_0\) has two square roots:

\begin{equation*} \pm \sqrt{r_0}e^{i\theta_0/2}. \end{equation*}

For instance, \(z^2 = i\) has two solutions. Since \(i =1 e^{i\pi/2}\text{,}\) \(\sqrt{i} = \pm e^{i\pi/4}\text{.}\) In Cartesian form, \(\sqrt{i} = \pm (\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i)\text{.}\)

More generally, the complex quadratic equation \(\alpha z^2 + \beta z + \gamma = 0\) where \(\alpha, \beta, \gamma\) are complex constants, will have one or two solutions. This marks an important difference from the real case, where a quadratic equation might not have any real solutions. In both cases we may use the quadratic formula to hunt for roots, and in the complex case we have solutions

\begin{equation*} z = \frac{-\beta \pm \sqrt{\beta^2 - 4\alpha\gamma}}{2\alpha}. \end{equation*}

For instance, \(z^2 + 2z + 4 = 0\) has two solutions:

\begin{equation*} z = \frac{-2 \pm \sqrt{-12}}{2} = -1 \pm \sqrt{3}i \end{equation*}

since \(\sqrt{-1} = i\text{.}\)

Example 2.4.4 Solving a quadratic equation

Consider the equation \(z^2 - (3+3i)z = 2-3i\text{.}\) To solve this equation for \(z\) we first rewrite it as

\begin{equation*} z^2-(3+3i)z-(2-3i)=0. \end{equation*}

We use the quadratic formula with \(\alpha = 1\text{,}\) \(\beta = -(3+3i)\text{,}\) and \(\gamma = -(2-3i)\text{,}\) to obtain the solution(s)

\begin{align*} z \amp = \frac{3+3i \pm \sqrt{(3+3i)^2+4(2-3i)}}{2}\\ z \amp = \frac{3+3i\pm \sqrt{8+6i}}{2}. \end{align*}

To determine the solutions in Cartesian form, we need to evaluate \(\sqrt{8+6i}\text{.}\) We offer two approaches. The first approach considers the following task: Set \(x + yi = \sqrt{8+6i}\) and solve for \(x\) and \(y\) directly by squaring both sides to obtain a system of equations.

\begin{align*} x+yi \amp = \sqrt{8+6i}\\ (x+yi)^2 \amp = 8+6i\\ x^2-y^2+2xy i \amp = 8 + 6i. \end{align*}

Thus, we have two equations and two unknowns:

\begin{align*} x^2-y^2 \amp = 8 \tag{1}\\ 2xy \amp = 6. \tag{2} \end{align*}

In fact, we also know that \(x^2+y^2 = |x+yi|^2 = |(x+yi)^2| =|8+6i| = 10\text{,}\) giving us a third equation

\begin{align*} x^2+y^2 \amp = 10. \tag{3} \end{align*}

Adding equations (1) and (3) yields \(x^2 = 9\) so \(x = \pm 3\text{.}\) Substituting \(x = 3\) into equation (2) yields \(y = 1\text{;}\) substituting \(x = -3\) into (2) yields \(y = -1\text{.}\) Thus we have two solutions:

\begin{equation*} \sqrt{8+6i} = \pm (3+i). \end{equation*}

We may also use the polar form to determine \(\sqrt{8+6i}\text{.}\) Consider the right triangle determined by the point \(8+6i = 10e^{i\theta}\) pictured in the following diagram.

We know \(\sqrt{8+6i} = \pm \sqrt{10}e^{i\theta/2}\text{,}\) so we want to find \(\theta/2\text{.}\) Well, we can determine \(\tan(\theta/2)\) easily enough using the half-angle formula

\begin{equation*} \tan(\theta/2) = \frac{\sin(\theta)}{1+\cos(\theta)}. \end{equation*}

The right triangle in the diagram shows us that \(\sin(\theta) = 3/5\) and \(\cos(\theta)=4/5\text{,}\) so \(\tan(\theta/2) = 1/3\text{.}\) This means that any point \(re^{i\theta/2}\) lives on the line through the origin having slope 1/3, and can be described by \(k(3+i)\) for some scalar \(k\text{.}\) Since \(\sqrt{8+6i}\) has this form, it follows that \(\sqrt{8+6i} = k(3+i)\) for some \(k\text{.}\) Since \(|\sqrt{8+6i}| = \sqrt{10}\text{,}\) it follows that \(|k(3+i)| = \sqrt{10}\text{,}\) so \(k = \pm 1\text{.}\) In other words, \(\sqrt{8+6i} = \pm (3+i)\text{.}\)

Now let's return to the solution of the original quadratic equation in this example:

\begin{align*} z \amp = \frac{3+3i\pm \sqrt{8+6i}}{2}\\ z \amp = \frac{3+3i\pm (3+i)}{2}. \end{align*}

Thus, \(z = 3+2i\) or \(z = i\text{.}\)

Example 2.4.5 Circle equations

If we let \(z = x + yi\) and \(z_0 = h + ki\text{,}\) then the complex equation

\begin{gather*} |z - z_0| = r \tag{equation of a circle} \end{gather*}

describes the circle in the plane centered at \(z_0\) with radius \(r> 0\text{.}\)

To see this, note that

\begin{align*} |z - z_0| \amp = |(x-h)+(y-k)i|\\ \amp =\sqrt{(x-h)^2 + (y-k)^2}. \end{align*}

So \(|z - z_0| = r\) is equivalent to the equation \((x-h)^2 + (y-k)^2 = r^2\) of the circle centered at \(z_0\) with radius \(r\text{.}\)

For instance, \(|z - 3-2i| = 3\) describes the set of all points that are 3 units away from \(3+2i\text{.}\) All such \(z\) form a circle of radius 3 in the plane, centered at the point \((3,2)\text{.}\)

Example 2.4.6 Complex expressions as regions

Describe each complex expression below as a region in the plane.

  1. \(|1/z| \gt 2\text{.}\)

    Taking the reciprocal of both sides, we have \(|z| \lt 1/2\text{,}\) which is the interior of the circle centered at 0 with radius \(1/2\text{.}\)

  2. Im\((z)\lt\) Re\((z)\text{.}\)

    Set \(z = x + yi\) in which case the inequality becomes \(y \lt x\text{.}\) This inequality describes all points in the plane under the line \(y = x\text{,}\) as pictured below.

  3. Im\((z) = |z|\text{.}\)

    Setting \(z = x + yi\text{,}\) this equation is equivalent to \(y = \sqrt{y^2 + x^2}\text{.}\) Squaring both sides we obtain \(0 = x^2\text{,}\) so that \(x = 0\text{.}\) It follows that \(y = \sqrt{y^2} = |y|\) so the equation describes the points \((0,y)\) with \(y \geq 0\text{.}\) These points determine a ray on the positive imaginary axis.

Moving forward, lines and circles will be especially important objects for us, so we end the section with a summary of their descriptions in the complex plane.

Lines and circles in \(\mathbb{C}\)

Lines and circles in the plane can be expressed with a complex variable \(z = x + yi\text{.}\)

  • The line \(ax + by + d = 0\) in the plane can be represented by the equation

    \begin{equation*} \alpha z + \overline{\alpha z} + d = 0 \end{equation*}

    where \(\alpha = \frac{1}{2}(a - bi)\) is a complex constant and \(d\) is a real number.

  • The circle in the plane centered at \(z_0\) with radius \(r \gt 0\) can be represented by the equation

    \begin{equation*} |z - z_0| = r. \end{equation*}

Subsection Exercises

1

Use a complex variable to describe the equation of the line \(y = mx + b\text{.}\) Assume \(m \neq 0\text{.}\) In particular, show that this line is described by the equation

\begin{equation*} (m+i)z + (m-i)\overline{z} + 2b = 0. \end{equation*}
2

In each case, sketch the set of complex numbers \(z\) satisfying the given condition.

a. \(|z + i| = 3\text{.}\)

b. \(|z+i|=|z-i|\text{.}\)

c. Re\((z) = 1\text{.}\)

d. \(|z/10 + 1 - i| \lt 5\text{.}\)

e. Im\((z) >\) Re\((z)\text{.}\)

f. Re\((z) = | z - 2 |\text{.}\)

3

Suppose \(u, v, w\) are three complex numbers not all on the same line. Prove that any point \(z\) in \(\mathbb{C}\) is uniquely determined by its distances from these three points. Hint: Suppose \(\beta\) and \(\gamma\) are complex numbers such that \(|u - \beta| = |u - \gamma|\text{,}\) \(|v - \beta| = |v - \gamma|\) and \(|w - \beta| = |w - \gamma|\text{.}\) Argue that \(\beta\) and \(\gamma\) must in fact be equal complex numbers.

4

Find all solutions to the quadratic equation \(z^2 + iz - (2+6i) = 0.\)