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Mathematics LibreTexts

2.4: Complex Expressions

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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.

im-square-root.svg

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.

im-plane-region.svg

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.

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