\(\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.

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