Complex Numbers

The eigenvalue unit reported that a quarter-turn rotation has eigenvalues $\pm i$, and that a real matrix's non-real eigenvalues arrive in conjugate pairs. Both claims were used before the numbers involved had been defined.

This unit supplies them. Adjoining a single solution of $x^2=-1$ produces a field in which every polynomial factors completely, so a characteristic polynomial always splits, and an $n\times n$ matrix always has $n$ eigenvalues counted with multiplicity. Nothing further ever needs adding.

The arithmetic is ordinary once $i^2=-1$ is applied. What repays attention is the second description: a complex number as a point in the plane, with multiplication rotating and scaling. In that reading $\pm i$ as the eigenvalues of a rotation stops being a curiosity and becomes a statement of what the matrix does.

The imaginary part is real. For $z=3+4i$, $\mathrm{Im}z=4$, not $4i$. Both $\mathrm{Re}z$ and $\mathrm{Im}z$ are real numbers. They are the coordinates of the point $z$ in the plane, and a coordinate is a number rather than a vector. Reporting $4i$ makes $\mathrm{Im}$ a map from $\mathbb{C}$ to $\mathbb{C}$ instead of to $\mathbb{R}$, and the type error propagates into every identity built on it.

Why $z\overline{z}$ is the key to division. $\mathbb{C}$ has no order, so there is no "dividing by the larger part". What makes division possible is that

is always a nonnegative real. Multiplying a quotient above and below by $\overline{w}$ therefore converts the denominator into a real number, after which dividing is ordinary:

This is the same manoeuvre as rationalising $\frac{1}{1+\sqrt{2}}$, and the conjugate plays the same role.

Conjugation is a field homomorphism. $\overline{z+w}=\overline{z}+\overline{w}$ and $\overline{zw}=\overline{z}\overline{w}$. It respects both operations, and it fixes exactly the real numbers. That combination is what forces conjugate pairing later: a polynomial with real coefficients is unchanged by conjugating its coefficients, so conjugating a root produces another root.

Geometrically it is reflection across the real axis, which is why $|\overline{z}|=|z|$ and $\arg \overline{z}=-\arg z$.

No order. is undefined for non-real numbers. Any ordering compatible with the arithmetic would have to place $i$ on one side of 0, and squaring gives a contradiction either way. Only moduli, which are real, can be compared, so "larger complex number" is never a well-formed phrase.

The argument is not unique. $\arg z$ is determined only up to multiples of $2\pi$, so $1+i$ has argument $\pi /4$, $9\pi /4$, $-7\pi /4$ and so on. Conventions fix a principal value in $(-\pi ,\pi ]$, and computing $n$th roots requires deliberately using several representatives, which is why the roots come out distinct.

Multiply by expanding and replacing $i^2$ with $-1$:

Collect the real terms and the $i$ terms separately. The sign flip on $bd$ is the only place $i^2$ enters, and missing it is the commonest slip.

Divide by multiplying above and below by the conjugate of the denominator, then dividing each part by the resulting real number:

Convert to polar. $r=\sqrt{a^2+b^2}$, and $\theta$ from $\tan \theta =b/a$ adjusted for the quadrant. A calculator's $\arctan$ returns values in $(-\pi /2,\pi /2)$, so for add or subtract $\pi$. Sketching the point first is the reliable guard.

Convert back. $a=r\cos \theta$, $b=r\sin \theta$.

Powers, by De Moivre. $z^n=r^n{e}^{in\theta }$, raise the modulus, multiply the argument. For a large exponent this is far cheaper than repeated multiplication, and it is exact where repeated multiplication accumulates error.

Roots. The $n$ distinct $n$th roots of $r{e}^{i\theta }$ are

They sit at equal modulus and equal angular spacing $2\pi /n$. Taking only $k=0$ finds one root and misses the rest, which is the standard omission.

Checks worth running.

Let $z=3+4i$ and $w=1-2i$.

Product.

Expanding term by term: $3-6i+4i-8i^2=3-2i+8=11-2i$, the $+8$ coming from $-8i^2$.

Check by modulus. $|z|=\sqrt{9+16}=5$ and $|w|=\sqrt{1+4}=\sqrt{5}$, so $|z||w|=5\sqrt{5}\approx 11.180$. And $|zw|=\sqrt{121+4}=\sqrt{125}=5\sqrt{5}$.

Quotient. Multiply above and below by $\overline{w}=1+2i$, noting $|w{|}^2=5$:

Check. $(-1+2i)(1-2i)=-1+2i+2i-4i^2=-1+4i+4=3+4i=z$.

Conjugate and modulus. $\overline{z}=3-4i$, and

real and nonnegative, as it must be.

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Polar form. Take $u=1+i$. Then $r=\sqrt{1+1}=\sqrt{2}$ and, since $u$ lies in the first quadrant, $\theta =\pi /4$. So

Powers by De Moivre. $u^n=(\sqrt{2}{)}^n{e}^{in\pi /4}$:

Check $u^2$ directly. $(1+i{)}^2=1+2i+i^2=2i$. And $u^8=16$, a positive real, which the polar form predicts without any multiplication, since eight quarter-turns return to the starting direction.

Cube roots of unity. Solving $z^3=1$ means $r^3=1$ and $3\theta =2\pi k$, so $r=1$ and $\theta =0,2\pi /3,4\pi /3$:

Each cubes to 1, they are equally spaced around the unit circle, and they sum to 0. The two non-real ones are conjugates, and their real parts $-\frac{1}{2}$ each cancel half of the first root's 1.

A real quadratic with complex roots. $x^2-2x+5=0$ has discriminant $4-20=-16$, so

The roots are conjugates, as real coefficients require. Checking against the coefficients: their sum is $2=-b/a$ and their product is $(1+2i)(1-2i)=1+4=5=c/a$. Both real, which is the pairing making itself visible.

Theorem. Let $p(x)=c_n{x}^n+\cdots +c_{1}x+c_0$ have real coefficients. If $p(\lambda )=0$ then $p(\overline{\lambda })=0$.

Proof. Conjugation respects sums and products, so

where the third equality uses $\overline{c_k}=c_k$. The hypothesis that the coefficients are real, and the only place it is used. Since $p(\lambda )=0$ and $\overline{0}=0$, we get $p(\overline{\lambda })=0$. $◼$

Corollary (eigenvalues of a real matrix). The characteristic polynomial of a real matrix has real coefficients, so its non-real eigenvalues occur in conjugate pairs $\lambda ,\overline{\lambda }$, with equal algebraic multiplicities.

Corollary (eigenvectors too). If $Av=\lambda v$ with $A$ real, conjugating every entry gives $A\overline{v}=\overline{\lambda }\overline{v}$, since $\overline{A}=A$. So the conjugate eigenvector belongs to the conjugate eigenvalue, and the pair arrives as a package.

Corollary (odd dimension). A real $n\times n$ matrix with $n$ odd has at least one real eigenvalue: the non-real ones pair off, leaving an odd count that cannot be exhausted. This is why every rotation of ${\mathbb{R}}^3$ has an axis. A real eigenvector with eigenvalue 1.

Fundamental theorem of algebra. Every non-constant polynomial with complex coefficients has a complex root, and hence factors completely into linear factors over $\mathbb{C}$. The field is algebraically closed, which is why adjoining $i$ to $\mathbb{R}$ is the last extension needed: solving $x^2=-1$ incidentally solves everything else.

For matrices this is the statement that a characteristic polynomial always splits over $\mathbb{C}$, so an $n\times n$ complex matrix always has $n$ eigenvalues with multiplicity. The diagonalization unit's first obstruction, a polynomial that does not split, therefore never arises over $\mathbb{C}$, and only the second obstruction, too few eigenvectors, can survive.

Worked instance. The rotation has $p(\lambda )={\lambda }^2+1$, with roots $i$ and $-i$. A conjugate pair, as the theorem requires. Their eigenvectors are $(1,-i)$ and $(1,i)$, conjugates of each other. Checking the first: $R(1,-i{)}^{\mathsf{T}}=(i,1{)}^{\mathsf{T}}$ and $i(1,-i{)}^{\mathsf{T}}=(i,-i^2{)}^{\mathsf{T}}=(i,1{)}^{\mathsf{T}}$.

And the dimension is even, so no real eigenvalue is forced, consistent with a plane rotation fixing no real direction.

The unit imaginary. $i=e^{i\pi /2}$ has modulus 1 and argument a quarter turn. Multiplying by it rotates without scaling: $1\mapsto i\mapsto -1\mapsto -i\mapsto 1$, returning after four steps. That cycle is $i^4=1$, and it is why powers of $i$ repeat with period four.

A real number. $-3=3e^{i\pi }$: modulus 3, argument $\pi$. Real numbers are the points with argument $0$ or $\pi$, and multiplying by a negative real scales and reflects through the origin. Conjugation fixes them, which is the defining property of $\mathbb{R}$ inside $\mathbb{C}$.

A number on the unit circle. $\frac{1}{\sqrt{2}}(1+i)=e^{i\pi /4}$ has modulus 1. Multiplying by it is pure rotation by $45°$, and its eighth power is $e^{2\pi i}=1$. Every modulus-1 number is a pure rotation, which is exactly the set whose powers stay bounded.

The cube roots of unity. $1$, $\omega =-\frac{1}{2}+\frac{\sqrt{3}}{2}i$ and $\overline{\omega }=-\frac{1}{2}-\frac{\sqrt{3}}{2}i$ sit at $0°$, $120°$ and $240°$ on the unit circle. Each cubes to 1, and they sum to zero. The two real parts of $-\frac{1}{2}$ cancel the 1 exactly.

They illustrate two things at once: that $n$th roots are equally spaced, and that the non-real ones come in a conjugate pair because $z^3-1$ has real coefficients.

A conjugate pair from a real quadratic. $x^2-2x+5$ has roots $1\pm 2i$, at modulus $\sqrt{5}$ and arguments $\pm \arctan 2$. Their sum is $2$ and their product is $5$. Both real, which is forced: a conjugate pair always sums and multiplies to real numbers, which is why a real quadratic can have complex roots without any complex coefficient appearing.

Eigenvalues of a rotation. has eigenvalues $\pm i$, on the unit circle at $\pm 90°$. The modulus 1 says the map preserves length; the argument $\pi /2$ says it turns by a quarter. For a real matrix a complex eigenvalue $r{e}^{i\theta }$ always reads this way, rotate by $\theta$, scale by $r$, in some plane.

What the plane picture supplies. Modulus is distance, argument is angle, conjugation is reflection in the real axis, multiplication is rotate-and-scale. Every fact in this unit is one of those four read off a diagram, which is why sketching the point is the first move when an arctan or a root count is in doubt.

Reporting the imaginary part with its $i$. Writing $\mathrm{Im}(3+4i)=4i$. The imaginary part is the real number 4. A coordinate of the point, like the 3. The error makes $\mathrm{Im}$ return complex values, and identities such as $z=\mathrm{Re}z+i\mathrm{Im}z$ then double-count the $i$.

Losing the sign flip in a product. Computing $(3+4i)(1-2i)$ as $3-2i-8=-5-2i$ by treating $i^2$ as $+1$, or as $3-2i$ by dropping the term entirely. The correct value is $11-2i$, the $+8$ arising because $-8i^2=+8$. The modulus check catches it: $|11-2i|=\sqrt{125}$ matches $|z||w|=5\sqrt{5}$, while the wrong answers do not.

Comparing complex numbers. Writing or "the larger root". $\mathbb{C}$ carries no order compatible with its arithmetic, if then , so ; if then and squaring gives the same contradiction. Only moduli may be compared, and those are real.

Taking one root instead of $n$. Solving $z^3=1$ and answering $z=1$. There are three cube roots, at angles $0$, $2\pi /3$ and $4\pi /3$; the other two are $-\frac{1}{2}\pm \frac{\sqrt{3}}{2}i$. Every nonzero complex number has exactly $n$ distinct $n$th roots, and stopping at the first one found loses the rest.

Expecting an unpaired complex eigenvalue from a real matrix. Reporting that a real matrix has eigenvalues $2$ and $3+i$ and nothing else. The characteristic polynomial has real coefficients, so $3-i$ must also be a root. The pairing is forced. A stated spectrum that omits a conjugate is evidence of an arithmetic error rather than of an unusual matrix.

What unites them. The first two are arithmetic with a conceptual cause: not tracking what $i$ contributes. The middle one imports a property $\mathbb{R}$ has and $\mathbb{C}$ does not. The last two stop at the first answer when the structure guarantees more. One root of $n$, one half of a pair.