What is The complex field? — Mathematics

Definition 3.1 University Mathematics — Year 1 · Chapter 3 — Complex Numbers

$\C = \{a + \iu b : a, b \in \R\}$ with the usual addition and the multiplication determined by $\iu^2 = -1$ . Every nonzero $z = a + \iu b$ has an inverse: $z^{-1} = \frac{a - \iu b}{a^2 + b^2}$ — in the language of Chapter 7, $\C$ is a field. One writes $a = \Re(z)$ , $b = \Im(z)$ , $\conj{z} = a - \iu b$ (the conjugate) and $\abs{z} = \sqrt{a^2 + b^2}$ (the modulus).

Examples

Example 3.3 (A complete modulus-and-argument workout)

Put $w = \dfrac{3 + 4\iu}{1 - 2\iu}$ in algebraic form and compute its modulus twice. Multiplying by the conjugate of the denominator:

w = \frac{(3 + 4\iu)(1 + 2\iu)}{(1 - 2\iu)(1 + 2\iu)} = \frac{3 + 6\iu + 4\iu - 8}{1 + 4} = \frac{-5 + 10\iu}{5} = -1 + 2\iu .

Directly: $\abs w = \sqrt{1 + 4} = \sqrt5$ . Via the quotient rule (Proposition 3.2 (3)): $\abs w = \frac{\abs{3 + 4\iu}}{\abs{1 - 2\iu}} = \frac5{\sqrt5} = \sqrt5$ — same answer, no algebraic form needed. The lesson generalizes: moduli and arguments travel well through products and quotients, real and imaginary parts travel well through sums. Choose the representation that matches the operations at hand, and convert only when forced.

Example 3.4 (Equations involving the conjugate)

Solve in $\C$ : $\;z + 2\conj z = 6 + 2\iu$ . An equation mixing $z$ and $\conj z$ is not polynomial in $z$ ; the reliable move is to split into real coordinates. With $z = x + \iu y$ :

z + 2\conj z = 3x - \iu y ,

so the equation reads $3x = 6$ and $-y = 2$ : the unique solution is $z = 2 - 2\iu$ . (Check: $(2 - 2\iu) + 2(2 + 2\iu) = 6 + 2\iu$ .) Alternatively, conjugate the whole equation to get $\conj z + 2z = 6 - 2\iu$ and solve the linear system in the unknowns $z, \conj z$ — same answer, and a useful trick when the coefficients are complex. Equations in $z$ and $\conj z$ are really systems of two real equations; expecting “degree $1$ , one solution” is safe here, but $z\conj z = -1$ (no solution) shows the polynomial intuition failing as soon as products appear.

Example 3.15 (Cube roots of $-27$ )

Solve $z^3 = -27$ . Exponential form of the right side: $-27 = 27\,\eu^{\iu\pi}$ , so the three roots are

z_k = 3\,\eu^{\iu(\frac\pi3 + \frac{2k\pi}3)}, \quad k = 0, 1, 2 : \qquad z_0 = 3\eu^{\iu\pi/3} = \frac32 + \frac{3\sqrt3}2\,\iu, \quad z_1 = -3, \quad z_2 = \conj{z_0} .

Two checks. First, the real root $-3$ is the obvious one, and the other two are its rotations by $\pm\frac{2\pi}3$ — equivalently $-3j$ and $-3j^2$ . Second, algebra confirms: $z^3 + 27 = (z + 3)(z^2 - 3z + 9)$ , and the quadratic has discriminant $9 - 36 = -27 < 0$ with roots $\frac{3 \pm 3\iu\sqrt3}2 = z_0, \conj{z_0}$ . The insight: for real right-hand sides, the non-real roots always come in conjugate pairs, so a picture of the solution set is symmetric about the real axis — a preview of the real factorization theorem of Chapter 8.

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