Complex Arithmetic For Linear Algebra

Complex Numbers as Arithmetic Pairs

Complex numbers extend the real numbers by introducing a number \(i\) whose square is \(-1\). This makes it possible to solve equations such as \(x^2+1=0\), which have no real solutions. In linear algebra, complex arithmetic is needed when scalars, eigenvalues, or eigenvectors are complex. The calculations begin with a small set of rules, but it is important to apply them precisely: the imaginary unit behaves like a scalar in multiplication, and division requires a nonzero denominator.

The equality rule lets us compare complex numbers by matching their real and imaginary parts separately. Addition follows the same principle: collect real parts together and imaginary parts together. Multiplication uses distribution and the defining equation \(i^2=-1\). For \(z=a+bi\) and \(w=c+di\), these rules give

The minus sign in the real part of the product comes from \(bd\,i^2=-bd\). These operations agree with ordinary real arithmetic when the imaginary parts are zero. For example, \((a+0i)(c+0i)=ac+0i\).

Conjugation and Its Arithmetic Rules

The complex conjugate of \(z=a+bi\) is obtained by changing the sign of its imaginary part. It is written \(\overline{z}\). Conjugation keeps the real part and reverses the imaginary part. In particular, a real number is unchanged by conjugation.

Conjugation interacts predictably with addition and multiplication. The product identity is especially useful: rather than expand a quotient into unrelated real and imaginary terms, it allows us to turn the denominator into a real number. The identities follow directly from the arithmetic definitions.

Proof. Write \(z=a+bi\) and \(w=c+di\), with \(a,b,c,d\in\mathbb{R}\). Then \(z+w=(a+c)+(b+d)i\), so \(\overline{z+w}=(a+c)-(b+d)i=(a-bi)+(c-di)=\overline{z}+\overline{w}\). For multiplication, \(zw=(ac-bd)+(ad+bc)i\), hence \(\overline{zw}=(ac-bd)-(ad+bc)i\). Meanwhile, \(\overline{z}\,\overline{w}=(a-bi)(c-di)=(ac-bd)-(ad+bc)i\), so the two expressions agree. Finally, conjugating \(a-bi\) changes the sign of its imaginary part again, giving \(a+bi=z\). \(\square\)

For any real scalar \(r\), the same definition also gives \(\overline{rz}=r\overline{z}\): if \(z=a+bi\), then \(rz=ra+(rb)i\), whose conjugate is \(ra-(rb)i=r(a-bi)\). Two further useful identities come from multiplying a number by its conjugate and adding or subtracting them:

The first expression is real and nonnegative. It is zero only when \(a=b=0\), which is exactly when \(z=0\). These facts explain why conjugation is the key step in defining modulus and calculating reciprocals.

Modulus: The Size of a Complex Number

The formula treats \(a\) and \(b\) as the two real coordinates of \(z\). Thus \(|z|\) is zero exactly when \(z=0\), and \(|z|\) is always nonnegative. If \(z\) is real, so \(b=0\), then \(|z|=\sqrt{a^2}=|a|\), the ordinary absolute value. The squared-modulus identity follows because \(z\overline{z}=a^2+b^2\) and \(|z|^2=(\sqrt{a^2+b^2})^2=a^2+b^2\).

Proof. Since \(z\overline{z}=|z|^2\), the conjugation rule for products gives \((zw)\overline{(zw)}=(zw)(\overline{z}\,\overline{w})=(z\overline{z})(w\overline{w})=|z|^2|w|^2\). The left side is \(|zw|^2\), so \(|zw|^2=|z|^2|w|^2\). Both moduli are nonnegative; taking nonnegative square roots yields \(|zw|=|z||w|\). Now suppose \(w\ne0\). Then \(|w|>0\), and \(z=(z/w)w\). Applying the product identity gives \(|z|=|z/w||w|\). Division by the positive real number \(|w|\) proves \(\left|z/w\right|=|z|/|w|\). \(\square\)

Division by Using the Conjugate

To divide by a nonzero complex number, multiply the numerator and denominator by the denominator's conjugate. The denominator then becomes a positive real number: \(w\overline{w}=|w|^2\), which is nonzero when \(w\ne0\). This gives a quotient in the standard form \(a+bi\), with a real denominator that can be simplified.

This rule follows by multiplying a fraction by \(1=\overline{w}/\overline{w}\), which is permitted because \(\overline{w}\ne0\) whenever \(w\ne0\). More explicitly, if \(\overline{w}=0\), then \(w=\overline{\overline{w}}=0\), a contradiction. The denominator is \(w\overline{w}=|w|^2\), a positive real number.

A Quotient Identity for Conjugation

The addition and product identities extend to quotients, provided the denominator is nonzero. This matters in calculations where a quotient itself is conjugated. One must conjugate both the numerator and denominator; conjugating only one part does not preserve the quotient.

Proof. Since \(w\ne0\), also \(\overline{w}\ne0\). By the quotient rule, \(z/w=z\overline{w}/|w|^2\), where \(|w|^2\) is real. Conjugate this expression. Conjugation of a product, together with the fact that conjugation fixes real numbers, gives \(\overline{z/w}=\overline{z}\,\overline{\overline{w}}/|w|^2=\overline{z}w/|w|^2\). Applying the quotient rule to \(\overline{z}/\overline{w}\) gives \(\overline{z}/\overline{w}=\overline{z}\,\overline{\overline{w}}/|\overline{w}|^2=\overline{z}w/|w|^2\), because \(\overline{\overline{w}}=w\) and \(|\overline{w}|^2=|w|^2\). The two expressions are equal. \(\square\)

The equality \(|\overline{w}|=|w|\) used here follows directly from the definition: if \(w=c+di\), then \(\overline{w}=c-di\), and both moduli are \(\sqrt{c^2+d^2}\). The proof also shows why the nonzero-denominator hypothesis cannot be omitted: without it, the quotients in the proposition are not defined.

Common Pitfall: Conjugate the Denominator, Not Just the Numerator

A frequent error is to multiply only the numerator by the conjugate of the denominator and leave the denominator unchanged. For instance, an incorrect calculation might claim

This changes the value: the right-hand side is not the original fraction multiplied by \(1\). To preserve a quotient, multiply both numerator and denominator by the same nonzero number. The correct calculation is

Check: Multiplying the answer by \(1+i\) gives \(\left(\frac12-\frac12i\right)(1+i)=\frac12+\frac12i-\frac12i-\frac12i^2=1\). The corrected quotient therefore reproduces the numerator. The mistaken expression would instead give \((1-i)/(1+i)\), which is not \(1\) after multiplication by \(1+i\).

When checking complex arithmetic, use the defining representation \(a+bi\): verify real and imaginary parts separately, or multiply a proposed quotient by its denominator. For conjugation, test the sign change on the imaginary part. These checks catch most errors involving signs, distribution, or a missing conjugate.