Interesting results easily achieved using complex numbers

Solution 1:

There are too many examples to count. Let me just mention one that is particularly concrete: how many subsets of an $n$-element set have a cardinality divisible by $3$ (or any positive integer $k$)? In other words, how do we evaluate

$$\sum_{k=0}^{\lfloor \frac{n}{3} \rfloor} {n \choose 3k}$$

in closed form? Although the statement of this problem does not involve complex numbers, the answer does: the key is what is known in high school competition circles as the roots of unity filter and what is known among real mathematicians as the discrete Fourier transform. Starting with the generating function

$$(1 + x)^n = \sum_{k=0}^n {n \choose k} x^k$$

we observe that the identity

$$1 + \omega^k + \omega^{2k} = \begin{cases} 3 \text{ if } 3 \mid k \\\ 0 \text{ otherwise} \end{cases}$$

where $\omega = e^{ \frac{2 \pi i}{3} }$ is a primitive third root of unity implies that

$$\sum_{k=0}^{ \lfloor \frac{n}{3} \rfloor} {n \choose 3k} = \frac{(1 + 1)^n + (1 + \omega)^n + (1 + \omega^2)^n}{3}.$$

Since $1 + \omega = -\omega^2$ and $1 + \omega^2 = - \omega$, this gives

$$\sum_{k=0}^{ \lfloor \frac{n}{3} \rfloor} {n \choose 3k} = \frac{2^n + (-\omega)^n + (-\omega^2)^n}{3}.$$

This formula can be stated without complex numbers (either by using cosines or listing out cases) but both the statement and the proof are much cleaner with it. More generally, complex numbers make their presence known in combinatorics in countless ways; for example, they are pivotal to the theory of asymptotics of combinatorial sequences. See, for example, Flajolet and Sedgewick's Analytic Combinatorics.

Solution 2:

Without complex numbers, it's a mystery why the power series for $1/(1+ x^2)$ centered at the origin has radius of convergence 1. The function is infinitely differentiable, no strange behavior at 1 or -1, etc.

But in the complex domain, there are singularities at $\pm i$ which explains why the radius of convergence is 1.

Solution 3:

I came across this slick proof of Heron's formula on artofproblemsolving.com the other day. Heron's formula yields the area of a triangle given the lengths of its three sides: $$A = \sqrt{s(s-a)(s-b)(s-c)}$$ where $ s = \frac{1}{2}(a+b+c)$. The entire proof, by high schooler Miles Dillon Edwards, is reproduced here.

Let $I$ be the center of the incircle of $\triangle ABC$. Let $a = y + z$, $b = x + z$, and $c = x + y$ be the lengths of the sides opposite $A$, $B$, and $C$, respectively, and let $s = x + y + z$ be the semiperimeter of the triangle. Clearly $2 \alpha + 2 \beta + 2 \gamma = 2\pi$, so $\alpha + \beta + \gamma = \pi$. Now notice that $$(r + ix)(r + iy)(r + iz) = (u e^{i \alpha})(v e^{i \beta})(w e^{i \gamma}) = u v w e^{i(\alpha+\beta+\gamma)} = u v w e^{\pi i} = −uvw.$$ Therefore $$0 = \text{Im}[(r + ix)(r + iy)(r + iz)] = r^2(x + y + z) − xyz,$$ so $$r = \sqrt{\frac{x y z}{x + y + z}} = \sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$$. Thus the area of $\triangle ABC$ is $$\frac{r a}{2} + \frac{r b}{2} + \frac{r c}{2} = r s = \sqrt{s(s-a)(s-b)(s-c)}$$

Solution 4:

As Dylan Wilson comments in Byron's link, you can derive most of trigonometric identities thanks to Euler's formula:

$$ e^{i\theta} = \cos \theta + i \sin \theta \ . $$

For instance, I never could learn by heart the formula for the cosine or the sine of the sum of two angles, but on one hand

$$ e^{i(\alpha + \beta)} = e^{i\alpha} \cdot e^{i\beta} = (\cos\alpha \cos\beta -\sin\alpha\sin\beta) + i (\sin\alpha\cos\beta + \cos\alpha \sin\beta) \ . $$

And on the other hand:

$$ e^{i(\alpha + \beta)} = \cos(\alpha + \beta) + i\sin(\alpha + \beta) \ . $$

So

$$ \cos(\alpha + \beta) = \cos\alpha \cos\beta -\sin\alpha\sin\beta $$

and

$$ \sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha \sin\beta \ . $$