2.4: Complex Expressions
In this section we look at some equations and inequalities that will come up throughout the text.
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.
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
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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.
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 \(\sqrt{-1} = i\text{.}\)
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 \(z = i\text{.}\)
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 \((3,2)\text{.}\)
Describe each complex expression below as a region in the plane.
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\(\left|\dfrac{1}{z}\right| \gt 2\text{.}\)
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 \( \dfrac{1}{2} \text{.}\)
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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.
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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 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
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
\[ (m+i)z + (m-i)\overline{z} + 2b = 0\text{.} \]
In each case, sketch the set of complex numbers \(z\) satisfying the given condition.
- \(|z + i| = 3\text{.}\)
- \(|z+i|=|z-i|\text{.}\)
- Re\((z) = 1\text{.}\)
- \(|z/10 + 1 - i| \lt 5\text{.}\)
- Im\((z) >\) Re\((z)\text{.}\)
- Re\((z) = | z - 2 |\text{.}\)
- Hint
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It may be helpful to set \(z = x + yi\) and rewrite the expression in terms of \(x\) and \(y\text{.}\)
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
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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.
Find all solutions to the quadratic equation \(z^2 + iz - (2+6i) = 0.\)