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Section2.4Complex Expressions

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

Example2.4.1Line 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.

Example2.4.3Quadratic 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 above 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 \(\sqrt{-1} = i\text{.}\)

Example2.4.4Find the solution(s) of \(z^2 - (3+3i)z = 2-3i\)

The given equation is quadratic, and can be written as \begin{equation*} z^2-(3+3i)z-(2-3i)=0. \end{equation*} So to use the quadratic formula we set \(\alpha = 1\text{,}\) \(\beta = -(3+3i)\text{,}\) and \(\gamma = -(2-3i)\text{,}\) which gives us solutions \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.

<<SVG image is unavailable, or your browser cannot render it>>

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 \(z = i\text{.}\)

Example2.4.5Circle 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 is the case, 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 \((3,2)\text{.}\)

Example2.4.6Complex expressions and regions in the plane

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.

<<SVG image is unavailable, or your browser cannot render it>>

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*}

SubsectionExercises

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 \(z^2 + iz - (2+6i) = 0.\)