Remark 2

Since $\mathbf{C}$ is of degree 2 over $\mathbf{R}$, it follows that $\mathbf{C}$ is, up to isomorphism, the only algebraic extension of $\mathbf{R}$ other than $\mathbf{R}$ itself, and that there is no field contained in $\mathbf{C}$ which contains $\mathbf{R}$, other than $\mathbf{R}$ and $\mathbf{C}$.

We know that $\mathbf{R}$ may be identified with a subfield of $\mathbf{C}$, and that every $z \in \mathbf{C}$ can be written uniquely in the form $x + iy$, where $x$ and $y$ are real; $x$ is called the real part of $z$ and is denoted by $\Re(z)$; $y$ the imaginary part of $z$, denoted by $\Im(z)$. Complex numbers of the form $iy$ ($y$ real) are called pure imaginary. The relation $x + iy = 0$ ($x, y$ real) is equivalent to $x = 0$ and $y = 0$.

Since $i^2 = -1$, the elements of $\mathbf{C}$ (when given by their real and imaginary parts) satisfy the following rules of calculation:

In particular, $(x + iy)(x - iy) = x^2 + y^2 \in \mathbf{R}$, so that, if $x + iy \neq 0$,

The second root of the polynomial $X^2 + 1$ in $\mathbf{C}$ is $-i$; consequently the only automorphism of $\mathbf{C}$, other than the identity mapping, which leaves all real numbers invariant, is that which maps $z = x + iy$ to $x - iy$; the latter is denoted by $\overline{z}$ and (in agreement with the general definitions) is called the complex number conjugate to $z$. We have

By reason of this automorphism, if $f(z)$ is a polynomial with real coefficients, we have $f(\overline{z}) = \overline{f(z)}$ for all $z \in \mathbf{C}$.

The real number $z \overline{z} = x^2 + y^2$ is called the algebraic norm of $z$, or simply the norm whenever there is no risk of confusion; it is a real number $\geq 0$, which vanishes only if $z = 0$. The real number $\sqrt{z \overline{z}} = \sqrt{x^2 + y^2} \geq 0$ reduces to the absolute value of $z$ when $z$ is real, and we still call it the absolute value of $z$, and denote it by $|z|$, when $z$ is any complex number. The relation $|z| = 0$ is equivalent to $z = 0$. If $z$ and $z'$ are two complex numbers, the conjugate of $zz'$ is $\overline{zz'}$, hence $|zz'|^2 = zz'\overline{zz'} = |z|^2|z'|^2$ and therefore $|zz'| = |z|\cdot|z'|$: the absolute value of a product is the product of the absolute values of the factors. In particular, if $z \neq 0$ and $z' = 1/z$, we have $|1/z| = |1/z|$.

Finally, for all complex numbers $z, z'$, we have the triangle inequality