Do complex numbers behave like ordinary numbers?

Jun 4, 2016

See explanation...

Explanation:

Yes and no.

Complex numbers form a field - that is a set equiped with addition and multiplication that behave in the ways with which you are familiar. In technical language, a field is an abelian group under addition, its non-zero elements form an abelian group under multiplication and multiplication is distributive over addition.

Like the rational numbers, which it contains, it is a field of characteristic $0$. That is, there is no positive integer $n$ such that ${\overbrace{1 + 1 + \ldots + 1}}^{\text{n times}} = 0$.

This leads to the Complex numbers behaving much like the rationals or Reals when simply treated as numbers.

For example, you can use the quadratic formula directly to provide the roots of $a {x}^{2} + b x + c = 0$, even if $a , b , c$ are Complex numbers.

There are some interesting things that happen with radicals and Complex numbers.

For example, we are used to thinking that Complex numbers are naturally expressible in the form $a + b i$ where $a$ and $b$ are Real numbers. We find that there are natural ways to express $\sqrt{a + b i}$ in the form $c + \mathrm{di}$ with $c$ and $d$ expressed as formulas in $a$ and $b$ involving square roots. The same is not true of cube roots however.

For example, $\sqrt{\frac{1}{2} + \frac{\sqrt{3}}{2} i} = \cos \left(\frac{\pi}{9}\right) + i \sin \left(\frac{\pi}{9}\right)$

but $\cos \left(\frac{\pi}{9}\right)$ and $\sin \left(\frac{\pi}{9}\right)$ are not expressible in terms of $n$th roots of rational numbers.

Note however, that if you are happy to just think of Complex numbers as numbers, leaving expressions like $\sqrt{\frac{1}{2} + \frac{\sqrt{3}}{2} i}$ as is, not trying to get into the form $a + b i$ with $a$, $b$ as algebraic expressions, then things generally work well.