# The Relationship Between Real and Complex Numbers

In mathematics involving waves, it is common to express cosine (or sine) with a different representation using the inverse Euler formula:

$$\cos(\theta)=\frac{e^{i\theta}+e^{-i\theta}}{2} \tag{1}$$

For example, the following expression can be rewritten using the inverse Euler formula:

$$\cos(5\pi t)\cos(6\pi t) = \left(\frac{e^{i5\pi t} + e^{-i5\pi t}}{2}\right)\left(\frac{e^{i6\pi t} + e^{-i6\pi t}}{2}\right)$$

And this makes the simplification/decomposition of this expression much simpler.

I was just wondering why Euler's formula can be used to seemingly convert a real number to a complex number. When I think of real numbers, I think of just a single number/scalar. I think of imaginary numbers, however, as a tuple of sorts, $$(a,b),$$ where $$a$$ is the real part and $$b$$ is imaginary part (but both $$a$$ and $$b$$ are real numbers of course).

In other words, I see the left side of $$(1)$$ as just a number and the right side as an ordered pair. How can a single number in any expression be replaced by this ordered pair and not "change the result"?

Of course, the imaginary part of the ordered pair in $$(1)$$ is $$0,$$ and I have heard people say that real numbers can be treated as complex numbers with an imaginary part equal to $$0.$$ However, how does one go about proving this? How does one know that any operation performed on the real and complex number will have the same result? I feel comfortable proving this on a case-by-case basis, but is there a general way to show that this can always be done?

If you want to get into petty pedantry, the real number and the complex number are different objects. One is an ordered pair of real numbers, and one is just a real number. To make this "equivalence" between them more formal, you can define an injective homomorphism from $$\mathbb{R}$$ to $$\mathbb{C}$$ by $$\phi : \mathbb{R} \to \mathbb{C} : x \mapsto (x, 0).$$ Note that this is a homomorpism, as \begin{align*} \phi(x +_\mathbb{R} y) &= (x +_\mathbb{R} y, 0) = (x +_\mathbb{R} y, 0 +_\mathbb{R} 0) = (x, 0) +_\mathbb{C} (y,0) = \phi(x) +_\mathbb{C} \phi(y) \\ \phi(x \times_\mathbb{R} y) &= (x \times_\mathbb{R} y, 0) = (x \times_\mathbb{R} y - 0 \times_\mathbb{R} 0, x \times_\mathbb{R} 0 + y \times_\mathbb{R} 0) = (x, 0) \times_\mathbb{C} (y, 0) = \phi(x) \times_\mathbb{C} \phi(y). \end{align*} We also have a perfectly well-defined function $$\exp$$ from $$\mathbb{C}$$ to $$\mathbb{C}$$. It so happens that $$\exp((0, x)) = (\cos(x), \sin(x))$$, and hence \begin{align*} &(1/2, 0) \times_\mathbb{C} (\exp((0, x)) +_\mathbb{C} \exp(0, -x))) \\ = \, &(1/2, 0) \times_\mathbb{C}((\cos(x), \sin(x)) +_\mathbb{C} (\cos(x), -\sin(x))) \\ = \, &(1/2, 0) \times_\mathbb{C} (2 \times_\mathbb{R} \cos(x), 0) \\ = \, &\phi(1/2) \times_\mathbb{C} \phi(2\cos(x)) \\ = \, &\phi(\cos(x)). \end{align*} If you wish to compute $$\cos(5\pi t)\cos(6 \pi t)$$, then you may elect to compute \begin{align*} &\phi(\cos(5\pi t) \times_\mathbb{R} \cos(6 \pi t)) \\ = \, &\phi(\cos(5\pi t)) \times_\mathbb{C} \phi(\cos(6\pi t)) \\ = \, &(1/4, 0) \times_\mathbb{C} (\exp((0, 5\pi t)) +_\mathbb{C} \exp(0, -5 \pi t))) \times_\mathbb{C} (\exp((0, 6 \pi t)) +_\mathbb{C} \exp(0, -6 \pi t))). \end{align*} The complex exponential has its own exponential rules, all of which have their own proofs, and after applying them, you'll get something along the lines of, $$((\cos(11 \pi t) + \cos(\pi t)) / 2, 0),$$ and the only $$x \in \mathbb{R}$$ such that $$\phi(x)$$ is the above expression is $$x = \frac{\cos(11 \pi t) + \cos(\pi t)}2.$$ So, from this, we can conclude that $$\cos(5\pi t)\cos(6 \pi t) = \frac{\cos(11 \pi t) + \cos(\pi t)}2.$$ Other excursions into complex numbers can be explained similarly. We simply use this embedding of $$\mathbb{R}$$ into $$\mathbb{C}$$, and use the properties proven for all complex numbers to help us with real numbers.
There are two operations in $$\mathbb R$$ and $$\mathbb C$$ that you need to worry about, the addition and the multiplication. Everything else is derived from those. What you would need to prove is that the complex numbers with imaginary part $$0$$ (or real numbers) form a subgroup. That is by adding or multiplying these numbers the result is also a member of the subgroup. So for $$z_1=(x_1,0)$$ and $$z_2=(x_2,0)$$, $$z_1+z_2=(x_1,0)+(x_2,0)=(x_1+x_2,0+0)$$ and $$(x_1,0)\cdot(x_2,0)=(x_1\cdot x_2-0\cdot 0,x_1\cdot 0+0\cdot x_2)=(x_1x_2,0)$$