We now look at an example that shows complex numbers are indeed useful and not just a mathematical curiosity for allowing equations such as \(x^2+1=0\) to have a solution.

You might have heard of the Fibonacci sequence: \(0,1,1,2,3,5,8,13,21,\ldots\). The Fibonacci sequence is well studied and appears, perhaps surprisingly, in nature and in many facets of life.

To obtain the next number in the sequence, simply take the sum of the current number and the previous number. Obviously, for this rule to make sense, you need to specify the two numbers of the sequence. Using this rule, the number after \(21\) is \(13 + 21 = 34\) and the one after that is \(21 + 34 = 55\).

Mathematically, the Fibonacci numbers can be defined inductively as follows: \begin{eqnarray*} F_0 & = & 0 \\ F_1 & = & 1 \\ F_n & = & F_{n-1} + F_{n-2}~~n = 2, 3,\ldots \\ \end{eqnarray*}

There is a closed-form formula for \(F_n\): \[ F_n = \frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^n - \frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^n ~~~~\text{ for all } n = 0,1,2\ldots.\]

We now describe one way to obtain the above the formula. The method is also applicable to many sequences of numbers defined in a way similar to the Fibonacci sequence.

Suppose that the sequence of numbers \(a_0, a_1, a_2\ldots\) is defined by \begin{eqnarray*} a_0 & = & a \\ a_1 & = & b \\ a_n & = & p a_{n-1} + q a_{n-2}~~n = 2, 3,\ldots \\ \end{eqnarray*} where \(a,b,p,q\) are real numbers.

Form the **characteristic equation**
\[x^2 - px - q = 0.\]
If this quadratic equation has two *distinct*
solutions \(\gamma_1\) and \(\gamma_2\), then
\[a_n = \alpha \gamma_1^n + \beta \gamma_2^n\]
for some \(\alpha\) and \(\beta\).
Since \(a_0\) and \(a_1\) are given, one can solve for \(\alpha\) and
\(\beta\) by solving the system
\begin{eqnarray*}
\alpha + \beta & = & a \\
\alpha\gamma_1 + \beta \gamma_2 & = & b \\
\end{eqnarray*}
Since \(\gamma_1\neq \gamma_2\), this system has a unique solution.

For the Fibonacci sequence, the characteristic equation is \(x^2 - x -1 =0\) with solutions \(\frac{1+\sqrt{5}}{2}\) and \(\frac{1-\sqrt{5}}{2}\). Continuing from here gives the formula displayed above.

Let us now consider the sequence of numbers \(a_0, a_1, a_2\ldots\) defined by \begin{eqnarray*} a_0 & = & 0 \\ a_1 & = & 1 \\ a_n & = & a_{n-1} - 2a_{n-2}~~n = 2, 3,\ldots \\ \end{eqnarray*} The first ten numbers in this sequence are \(0,1,1,-1,-3,-1,5,7,-3,-17\).

The characteristic equation is \(x^2 -x + 2 =0\) and has solutions \(\frac{1 \pm \sqrt{7} i}{2}\). Solving \(\alpha + \beta = 0\) and \(\alpha \frac{1+\sqrt{7}i}{2} + \beta \frac{1-\sqrt{7}i}{2} = 1 \) for \(\alpha\) and \(\beta\) gives \(\alpha = -\frac{i}{\sqrt{7}}\) and \(\beta = \frac{i}{\sqrt{7}}\).

Thus, \[a_n = -\frac{i}{\sqrt{7}}\left(\frac{1+\sqrt{7}i}{2}\right)^n + \frac{i}{\sqrt{7}}\left(\frac{1-\sqrt{7}i}{2}\right)^n~~~~\text{ for all } n = 0,1,2,\ldots.\] This formula involves complex numbers even though \(a_n\) is an integer for each \(n\). Can we obtain a formula without mentioning complex numbers? The answer is “yes”.

Let \(\theta = \tan^{-1}\sqrt{7}\) and \(r = \sqrt{2}\). Then the polar form of \(\frac{1+\sqrt{7}i}{2}\) is \(r\operatorname{cis} \theta\) and the polar form of \(\frac{1-\sqrt{7}i}{2}\) is \(r\operatorname{cis} (2\pi - \theta)= r\operatorname{cis} (- \theta).\)

Thus, \begin{eqnarray*} & & -\frac{i}{\sqrt{7}}\left(\frac{1+\sqrt{7}i}{2}\right)^n + \frac{i}{\sqrt{7}}\left(\frac{1-\sqrt{7}i}{2}\right)^n \\ & = & -\frac{i}{\sqrt{7}} \left (\left(r\operatorname{cis} \theta\right)^n - \left(r\operatorname{cis} (-\theta)\right)^n\right) \\ & = & -\frac{i}{\sqrt{7}} r^n \left (\operatorname{cis} n\theta - \operatorname{cis} (-n\theta)\right) \\ & = & -\frac{i}{\sqrt{7}} r^n \left ( (\cos n\theta + i\sin n\theta) - (\cos (-n\theta) + i\sin (-n\theta))\right) \\ & = & -\frac{i}{\sqrt{7}} r^n \left ( (\cos n\theta - \cos(-n\theta) + i(\sin n\theta + \sin (-n\theta))\right) \\ & = & -\frac{i}{\sqrt{7}} r^n \left ( 2i\sin n\theta\right) \\ & = & \frac{2}{\sqrt{7}} r^n \left ( \sin n\theta\right) \\ \end{eqnarray*} since \(\cos (-n\theta) = \cos n\theta\) and \(\sin (-n\theta) = -\sin n\theta\) for all \(\theta\). Thus, \[a_n = \frac{2\sqrt{2}^n \sin \left ( n \tan^{-1} \sqrt{7} \right)}{\sqrt{7}} ~~~\text{ for all } n = 0,1,2,\ldots.\] This formula for \(a_n\) does not involve complex numbers. However, the complexity of this formula is quite overwhelming. In fact, the presence of the sine and the inverse tangent functions makes one wonder if \(a_n\) could even be a rational number, let alone an integer! Also, it is not clear if there is a simple method without referencing complex numbers that one could use to obtain such a formula. Is this formula better than the one involving complex numbers? You be the judge.