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.
Theorem 2.4.2
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}) \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 \(z^2  (3+3i)z = 23i\text{.}\) To solve this equation for \(z\) we first rewrite it as
\begin{equation*}
z^2(3+3i)z(23i)=0.
\end{equation*}
We use the quadratic formula with \(\alpha = 1\text{,}\) \(\beta = (3+3i)\text{,}\) and \(\gamma = (23i)\text{,}\) to obtain the solution(s)
\begin{align*}
z \amp = \frac{3+3i \pm \sqrt{(3+3i)^2+4(23i)}}{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^2y^2+2xy i \amp = 8 + 6i.
\end{align*}
Thus, we have two equations and two unknowns:
\begin{align*}
x^2y^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 halfangle 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 = (xh)+(yk)i\\
\amp =\sqrt{(xh)^2 + (yk)^2}.
\end{align*}
So \(z  z_0 = r\) is equivalent to the equation \((xh)^2 + (yk)^2 = r^2\) of the circle centered at \(z_0\) with radius \(r\text{.}\)
For instance, \(z  32i = 3\) 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.

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

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.

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*}