# 1.17 Complex numbers

$$\def\Real{\mathbb{R}}\def\Comp{\mathbb{C}}\def\Rat{\mathbb{Q}}\def\Field{\mathbb{F}}\def\Fun{\mathbf{Fun}}\def\e{\mathbf{e}}\def\f{\mathbf{f}}\def\bv{\mathbf{v}}\def\i{\mathbf{i}} \def\eye{\left(\begin{array}{cc}1&0\\0&1\end{array}\right)}$$

Complex numbers are expressions $$z=x+\i y$$, where $$\i$$ is the imaginary unit, defined by $$\i^2=-1$$. Note the ambiguity ($$-\i$$ would work just as well).

Complex numbers can be added subtracted, multiplied the usual way, with the usual properties of associativity, commutativity and distributivity. There is a zero, and a norm, $$\|x+y\i\|=x^2+y^2$$.

Flipping $$\i$$ gives a conjugate complex number:
$\overline{x+\i y}=x-\i y.$

The norm satisfies $$\|z\|^2=z\bar{z}$$.

Complex conjugacy is multiplicative, $$\overline{z_1z_2}=\bar{z}_1\bar{z}_2$$, and so is the norm.

(This makes the complex numbers $$\Comp$$ a normed algebra.)

It is useful also to see a complex number as a matrix,

$z=x+\i y\mapsto\left(\begin{array}{cc}x&y\\-y&x\end{array}\right)=:M(z).$

It follows that the inverse of a complex number $$z\neq 0$$ is
$\frac{1}{z}=\frac{1}{x+\i y}=\frac{x-\i y}{x^2+y^2}.$

Existence of an inverse for any non-zero element makes $$\Comp$$ a field.

If $$r=\sqrt{x^2+y^2}$$ and $$x=r\cos\phi, y=r\sin\phi$$, then the matrix $$M(z)$$ represents a dilation by $$r$$ and rotation (counterclockwise in the standard frame) by $$\phi$$. One often refers to the notation $$z=r(\cos\phi+\sin\phi$$) as polar coordinate representation.

The exponential of a complex number $$z$$ can be understood as an exponential of $$M(z)$$.

A digression: how to define an exponential of anything you can add or multiply? Two ways – via power series, or using the limit
$\exp(a):=\lim_{n\to\infty}(1+a/n)^n.$

One can also define the exponential as a function that is a character, that is has the property
$$f(a+b)=f(a)f(b)$$. Assuming differentiability of the function quickly leads to the definition above.

In this way,
$\exp\left(\begin{array}{cc}0&b\\-b&0\end{array}\right)=\left(\begin{array}{cc} \cos{b}&\sin{b}\\-\sin{b}&\cos{b}\end{array}\right)=M(\cos(b)+\sin(b)\i).$

Most importantly, the complex numbers are algebraically closed: any polynomial
$z^n+a_1z^{n-1}+\ldots+a_{n-1}z+a_n$
with coefficients in $$\Comp$$ has at least one root in $$\Comp$$.

The same is true neither over rationals ($$z^2-2$$), nor over reals ($$z^2+2$$).
Exercises:
Describe the curve $$1/z$$, where $$z=1+t\i, t\in(-\infty,\infty)$$.

Represent $$z^4+4$$ as a product of two quadratic polynomials with real coefficients.