Section 1.5 Complex Numbers
I do not expect complex numbers to be familiar to you. I’m going to give a very brief description of them in this preliminary section. I will not use complex numbers often in this course, but they are key for a major result in
Chapter 3.
Definition 1.5.1.
The number \(\imath\) is defined by the property \(\imath^2 = -1\text{.}\) A complex number is any expression \(a + b \imath\) where \(a\) and \(b\) are real numbers. \(a\) is called the real part and \(b\) is called the imaginary part. The set of all complex number is written \(\CC\text{.}\)
Addition and multiplication of complex numbers are extended from \(\RR\) by linearity and the distribution of multiplication. Note in the multiplication I use the defining property that \(\imath^2 = -1\text{.}\)
\begin{align*}
(a + b \imath) + (c + d \imath) \amp = (a+b) + (c+d) \imath\\
(a + b \imath)(c + d \imath) \amp = ac + bc \imath + ad
\imath + bd \imath^2\\
\amp ac + (bc + ad) \imath + bd(-1) = (ac-bd) + (bc+ad)
\imath
\end{align*}
All non-zero complex numbers are invertible (have a reciprocal). The reciprocal of a non-zero complex number is
\begin{equation*}
\frac{1}{a+ b \imath} = \frac{a-b\imath}{a^2+b^2}\text{.}
\end{equation*}
Definition 1.5.2.
The conjugation of a complex number \(a+b\imath\) is \(a-b\imath\text{.}\) It is indicated with a vertical bar \(\overline{a+b\imath} = a-b\imath\text{.}\)
Complex numbers have many intriguing properties. The most important for this course is the guaranteed existence of roots of any polynomial.
Theorem 1.5.3. Fundamental Theorem of Algebra.
Let \(p(x)\) be a polynomial of degree \(d\) with real or complex coefficients. Then \(p(x)\) has exactly \(d\) roots in the complex numbers (though some may be repeated). Equivalently, \(p(x)\) factors completely over the complex numbers: there are complex numbers \(\alpha_1,
\ldots, \alpha_d\text{,}\) not necessariliy distinct from each other, such that
\begin{equation*}
p(x) = (x-\alpha_i)(x-\alpha_2)\ldots(x-\alpha_d)\text{.}
\end{equation*}
Complex exponentials are understood by Euler’s formula. This is particularly important for us, since the complex numbers in this course will naturally appear in exponents.
Proposition 1.5.4. Euler’s Formula.
\begin{equation*}
e^{\imath \theta} = \cos \theta + i \sin \theta\text{.}
\end{equation*}
Proof.
There are a variety of way to justify Euler’s formlua. One of the justifications only needs the known taylor series for \(e^x\) and the trig functions, which were developed in Calculus II. In the following argument via talyor series, the number \(\imath\) can just be treated like any other constant. First I’ll write the known taylor series for the three functions I need. Then I calculate \(e^{\imath t}\) via its taylor series. Various manipulations of the series, using linearity and splitting up the series into even and odd terms, leads to the final expression in the taylor series for sine and cosine, proving Euler’s formula.
\begin{align*}
e^t \amp = \sum_{n=0}^\infty \frac{t^n}{n!}\\
\sin t \amp = \sum_{n=0}^\infty \frac{(-1)^n
t^{2n+1}}{(2n+1)!}\\
\cos t \amp = \sum_{n=0}^\infty \frac{(-1)^n t^{2n}}{(2n)!}\\
e^{\imath t} \amp = \sum_{n=0}^\infty \frac{(\imath
t)^n}{n!} = \sum_{n=0}^\infty \frac{\imath^n t^n}{n!}\\
\amp = \sum_{n=0}^\infty \frac{\imath^{2n} t^{2n}}{{2n}!} +
\sum_{n=0}^\infty \frac{\imath^{2n+1} t^{2n+1}}{(2n+1)!}\\
\amp = \sum_{n=0}^\infty \frac{(-1)^n t^{2n}}{{2n}!} +
\imath \sum_{n=0}^\infty \frac{(-1)^n t^{2n+1}}{(2n+1)!}\\
\amp = \cos t + \imath \sin t
\end{align*}
