# 2.4: Complex Expressions

$$\newcommand{\vecs}{\overset { \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$

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

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

This page titled 2.4: Complex Expressions is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Michael P. Hitchman via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.