2.4: Complex Expressions

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

Example $\PageIndex{1}$: 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

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

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

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

$$\alpha z + \overline{\alpha z} + d = 0$$

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.

Theorem $\PageIndex{1}$

Any line in $\mathbb{C}$ can be expressed by the equation $\displaystyle |z - \gamma| = |z - \beta|$ for suitably chosen points $\gamma$ and $\beta$ in $\mathbb{C}\text{,}$ and the set of all points (Euclidean) equidistant from distinct points $\gamma$ and $\beta$ forms a line.

Proof

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

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

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 $\PageIndex{2}$: Quadratic Equations

Suppose $z_0$ is a complex constant and consider the equation $z^2 = z_0\text{.}$ 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

$$re^{i\theta}\cdot re^{i\theta} =r_0e^{i\theta_0}\text{.}$$

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:

$$\pm \sqrt{r_0}e^{i\theta_0/2}\text{.}$$

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 $\alpha\text{,}$ $\beta\text{,}$ and $\gamma$ 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

$$z = \frac{-\beta \pm \sqrt{\beta^2 - 4\alpha\gamma}}{2\alpha}\text{.}$$

For instance, $z^2 + 2z + 4 = 0$ has two solutions:

$$z = \frac{-2 \pm \sqrt{-12}}{2} = -1 \pm \sqrt{3}i$$

since $\sqrt{-1} = i\text{.}$

Example $\PageIndex{3}$: 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

$$z^2-(3+3i)z-(2-3i)=0\text{.}$$

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

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

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.

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

Thus, we have two equations and two unknowns:

$$x^2-y^2 = 8 \label{2.4.1}$$

$$2xy = 6 \text{.} \label{2.4.2}$$

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

$$x^2+y^2 = 10 \text{.} \label{2.4.3}$$

Adding equations ($\ref{2.4.1}$) and ($\ref{2.4.3}$) yields $x^2 = 9$ so $x = \pm 3\text{.}$ Substituting $x = 3$ into equation ($\ref{2.4.2}$) yields $y = 1\text{;}$ substituting $x = -3$ into ($\ref{2.4.2}$) yields $y = -1\text{.}$ Thus we have two solutions:

$$\sqrt{8+6i} = \pm (3+i)\text{.}$$

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 $\dfrac{\theta}{2}\text{.}$ Well, we can determine $\tan \left(\dfrac{\theta}{2} \right)$ easily enough using the half-angle formula

$$\tan \left(\dfrac{\theta}{2} \right) = \dfrac{\sin(\theta)}{1+\cos(\theta)}\text{.}$$

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

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

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

Example $\PageIndex{4}$: 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

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

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 $\PageIndex{5}$: Complex Expressions as Regions

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

1. $\left|\dfrac{1}{z}\right| \gt 2\text{.}$

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

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.

Note: 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

$$\alpha z + \overline{\alpha z} + d = 0$$

where $\alpha = \dfrac{1}{2}(a - bi)$ 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

$$|z - z_0| = r\text{.}$$

Exercises

Exercise $\PageIndex{1}$

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

$$(m+i)z + (m-i)\overline{z} + 2b = 0\text{.}$$

Exercise $\PageIndex{2}$

In each case, sketch the set of complex numbers $z$ satisfying the given condition.

1. $|z + i| = 3\text{.}$
2. $|z+i|=|z-i|\text{.}$
3. Re$(z) = 1\text{.}$
4. $|z/10 + 1 - i| \lt 5\text{.}$
5. Im$(z) >$ Re$(z)\text{.}$
6. Re$(z) = | z - 2 |\text{.}$

Hint

It may be helpful to set $z = x + yi$ and rewrite the expression in terms of $x$ and $y\text{.}$

Exercise $\PageIndex{3}$

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.

Exercise $\PageIndex{4}$

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