Complex numbers: Introduction

-

We review how complex numbers are introduced, starting from a simple linear equation. We also review the derivation of the quadratic formula.



Consider the following equation:

\[3x = 12.\]

The coefficient of \(x\) is 3, and the right-hand side is 12. 3 and 12 are natural numbers: \(3, 12 \in \mathbb{N}\). Solving this for \(x\), we have

\[x = 4 \in \mathbb{N}.\]

So, the solution is obtained within \(\mathbb{N}\).

Next, consider

\[2x + 4 = 0.\]

Again we have \(2, 4 \in \mathbb{N}\). But the solution

\[x = -2 \in \mathbb{Z}\]

is not found in \(\mathbb{N}\), but in \(\mathbb{Z}\).

Next, consider

\[2x - 3 = 0\]

where \(2, -3 \in \mathbb{Z}\). The solution is

\[x = \frac{3}{2} \in \mathbb{Q}\]

which is not in \(\mathbb{Z}\).

In general, consider

\[ax + b = 0, ~ a, b \in \mathbb{Q}, a \neq 0.\]

Then, the solution is always found within \(\mathbb{Q}\):

\[x = -\frac{b}{a} \in \mathbb{Q}.\]

As long as all the coefficients are in \(\mathbb{Q}\) and the equations are linear, we can always find the solution, if it exists, within \(\mathbb{Q}\). The algebra that has grown out of linear equations is called linear algebra. Of course, we can use \(\mathbb{R}\) in place of \(\mathbb{Q}\). We will study linear algebra to some extent in later posts.

Consider the following non-linear equation:

\[x^2 = 2.\]

Note \(1, 2\in \mathbb{Q}\). Its solutions are

\[x = \pm\sqrt{2} \in \mathbb{R}.\]

When we deal with non-linear (in this case, quadratic) equations, we cannot stay in \(\mathbb{Q}\), but need (at least) \(\mathbb{R}\). In fact, we need more.

Let us define a function \(f: \mathbb{R} \to \mathbb{R}\) by

\[f(x) = 2x^2 - 5x + 1.\]

The solutions of the equation \(f(x) = 0\) are the elements of

\[\{ \alpha \mid \alpha \in \mathbb{R}, f(\alpha) = 0\}.\]

The solutions of an equation are sometimes called roots.

Theorem (Solutions of the quadratic equation)

Consider the equation

\[ax^2 + bx + c = 0\]

where \(a,b,c\in\mathbb{R}\) and \(a \neq 0\). This equation has no solutions in \(\mathbb{R}\) if \(b^2 - 4ac < 0\). The real solutions are given by the formula

\[\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

if \(b^2 - 4ac \geq 0\).

Proof. Suppose that \(\alpha \in \mathbb{R}\) is a solution, then \(a\alpha^2 + b\alpha + c =0\). Since \(a \neq 0\), we can divide both sides by \(a\) to have

\[\alpha^2 + \frac{b}{a}\alpha = -\frac{c}{a}.\]

Complete the square on the left-hand side to get

\[\left(\alpha + \frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2 = -\frac{c}{a} \]

which can be rearranged into

\[\left(\alpha + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}.\]

If the right-hand side is negative, we have a contradiction since the square of a real number on the left-hand side is non-negative. In this case, our initial assumption that there is a solution in \(\mathbb{R}\) is false. If the right-hand side is non-negative, we have

\[\left(\alpha + \frac{b}{2a}\right)^2 = \left(\frac{\sqrt{b^2 - 4ac}}{2a}\right)^2.\]

Note that in \(\mathbb{R}\), \(u^2 = v^2\) if and only if \(u = \pm v\). Thus we have

\[\alpha + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a}\]

so

\[\alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\tag{eq:qsol}\]

and \(b^2 - 4ac \geq 0\).

Finally, retracing the argument in reverse, we can show that these \(\alpha\)'s are indeed the solutions. ■

Remark. The last paragraph is necessary because we assumed that \(\alpha\) was a solution, but this assumption may well be false. From a false assumption, anything can be ``proved'' to be true. Therefore, after we have obtained Eq. (eq:qsol), we need to verify that this is indeed the solution (i.e., the assumption is true). □

What happens if \(b^2 -4ac < 0\)? Since there are no real numbers such that their square is negative, we do not have real solutions in this case. Nevertheless, we introduce a new ``number'' \(i = \sqrt{-1}\) called the imaginary unit. Formally,  the square of the imaginary unit is -1: \(i^2 = (\sqrt{-1})^2 = -1\) (hence, it cannot be real!) Using the imaginary unit, we define complex numbers such as \(a + bi\) with \(a, b \in \mathbb{R}\). In the complex number \(a + bi\), \(a\) is called the real part, and \(b\) is called the imaginary part. A real multiple of \(i=\sqrt{-1}\), such as the \(b\) in \(bi\), is called an imaginary number. However, there are a bunch of questions arising. Is this new number system consistent? What does \(a + bi\) (addition between real and imaginary numbers) mean at all? We will make the notion of complex numbers more precise in the next post by constructing them from scratch.


Comments

Post a Comment

Popular posts from this blog

Open sets and closed sets in \(\mathbb{R}^n\)

-
Image
Open sets In \(\mathbb{R}\), we have the notion of an open interval such as \((a, b) = \{x \in \mathbb{R} | a < x < b\}\). We want to extend this idea to apply to \(\mathbb{R}^n\). We also introduce the notions of bounded sets and closed sets in \(\mathbb{R}^n\). Recall that the \(\varepsilon\)-neighbor of a point \(x\in\mathbb{R}^n\) is defined as \(N_{\varepsilon}(x) = \{y \in \mathbb{R}^n | d(x, y) < \varepsilon \}\) where \(d(x,y)\) is the distance between \(x\) and \(y\). Definition (Open set) A subset \(U\) of \(\mathbb{R}^n\) is said to be an  open set  if the following holds: \[\forall x \in U ~ \exists \delta > 0 ~ (N_{\delta}(x) \subset U).\tag{Eq:OpenSet}\] That is, for every point in an open set \(U\), we can always find an open ball centered at that point, that is included in \(U\). See the following figure. Perhaps, it is instructive to see what is  not  an open set. Negating (Eq:OpenSet), we have \[\exists x \in U ~ \forall \delta > 0 ~ (N_{\delta}(x) \not
Read more

Euclidean spaces

-
Image
We would like to study multivariate functions (i.e., functions of many variables), continuous multivariate functions in particular. To define continuity, we need a measure of "closeness" between points. One measure of closeness is the Euclidean distance. The set \(\mathbb{R}^n\) (with \(n \in \mathbb{N}\)) with the Euclidean distance function is called a Euclidean space. This is the space where our functions of interest live. The real line is a geometric representation of \(\mathbb{R}\), the set of all real numbers. That is, each \(a \in \mathbb{R}\) is represented as the point \(a\) on the real line. The coordinate plane , or the \(x\)-\(y\) plane , is a geometric representation of \(\mathbb{R}^2\), the set of all pairs of real numbers. Each pair of real numbers \((a, b)\) is visualized as the point \((a, b)\) in the plane. Remark . Recall that \(\mathbb{R}^2 = \mathbb{R}\times\mathbb{R} = \{(x, y) | x, y \in \mathbb{R}\}\) is the Cartesian product of \(\mathbb{R}\) with i
Read more

Applications of multiple integrals

-
Image
We can use multiple integrals to compute areas and volumes of various shapes. Area of a planar region Definition (Area) Let \(D\) be a bounded closed region in \(\mathbb{R}^2\). \(D\) is said to have an area if the multiple integral of the constant function 1 over \(D\), \(\iint_Ddxdy\), exists. Its value is denoted by \(\mu(D)\): \[\mu(D) = \iint_Ddxdy.\] Example . Let us calculate the area of the disk \(D = \{(x,y)\mid x^2 + y^2 \leq a^2\}\). Using the polar coordinates, \(x = r\cos\theta, y = r\sin\theta\), \(dxdy = rdrd\theta\), and the ranges of \(r\) and \(\theta\) are \([0, a]\) and \([0, 2\pi]\), respectively. Thus, \[\begin{eqnarray*} \mu(D) &=& \iint_Ddxdy\\ &=&\int_0^a\left(\int_0^{2\pi}rd\theta\right)dr\\ &=&2\pi\int_0^a rdr\\ &=&2\pi\left[\frac{r^2}{2}\right]_0^a = \pi a^2. \end{eqnarray*}\] □ Volume of a solid figure Definition (Volume) Let \(V\) be a solid figure in the \((x,y,z)\) space \(\mathbb{R}^3\). \(V\) is sai
Read more