[ALGEBRA] Complex arithmetic

This page introduces the complex numbers and their arithmetic. The real numbers contain no solution of $x^2+1=0$ . Adjoining a number $i$ with $i^2=-1$ supplies one while keeping the usual rules of addition and multiplication. A complex number then has a real component and a component along the new $i$ direction.

Cartesian form

[DEFINITION]

Complex number

A complex number is a number of the form

z=a+bi,\qquad a,b\in\mathbb R,\qquad i^2=-1.

The expression $a+bi$ is the Cartesian form of $z$ . Its real part is $\operatorname{Re}(z)=a$ and its imaginary part is $\operatorname{Im}(z)=b$ .

Every real number is a complex number with imaginary part zero. Equality is componentwise:

a+bi=c+di\quad\Longleftrightarrow\quad a=c\text{ and }b=d.

When simplifying powers of $i$ , use the four-term cycle

i^0=1,\qquad i^1=i,\qquad i^2=-1,\qquad i^3=-i,

which then repeats because $i^4=1$ .

[1]:Argand plane
[2]:Complex conjugate
[3]:Complex modulus
[4]:Add, multiply and divide complex numbers
8 more

The Argand plane

[DEFINITION]

Argand plane

The Argand plane represents $z=a+bi$ by the point $(a,b)$ , with the real axis horizontal and the imaginary axis vertical.

Under this identification, addition of complex numbers is vector addition, and $z-w$ is the displacement from $w$ to $z$ . So $|z-w|$ will measure the distance between the two points. The picture also makes modulus and argument geometric quantities in polar form.

The point represents the number but does not replace its algebra: complex multiplication still combines both components according to $i^2=-1$ .

[1]:Complex number
[2]:Read loci in the Argand plane

Conjugation and modulus

[DEFINITION]

Complex conjugate

If $z=a+bi$ , its complex conjugate is

\overline z=a-bi.

Conjugation changes the sign of the imaginary part and leaves the real part unchanged, so it reflects $z$ in the real axis. The product of $z$ and its conjugate is real:

z\overline z=(a+bi)(a-bi)=a^2+b^2.

[1]:Complex number
[2]:Rules for conjugation
[3]:Conjugate roots for real coefficients
[4]:A complex conjugate pair of roots

[DEFINITION]

Complex modulus

The modulus of $z=a+bi$ is

|z|=\sqrt{a^2+b^2}.

In the Argand plane, $|z|$ is the distance from $z$ to the origin. Division and polar form both use the identity $z\overline z=|z|^2$ .

Arithmetic and conjugation laws

[METHOD]

Add, multiply and divide complex numbers

For $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$ ,

z_1+z_2=(x_1+x_2)+i(y_1+y_2),

z_1z_2=(x_1x_2-y_1y_2)+i(x_1y_2+y_1x_2).

The additive inverse is $-z_1$ . For a non-zero $z=x+iy$ ,

z^{-1}=\frac{x}{x^2+y^2}-i\frac{y}{x^2+y^2}, \qquad zz^{-1}=1.

Division is defined using this inverse. In practice, when $z_2\ne0$ ,

\frac{z_1}{z_2} =\frac{z_1\overline{z_2}}{z_2\overline{z_2}}.

The multiplication formula follows by expanding and replacing $i^2$ by $-1$ . The conjugate in the division formula makes the denominator real:

z_2\overline{z_2}=|z_2|^2.

[1]:Complex number
[2]:Calculate with Cartesian forms

[EXAMPLE]

Calculate with Cartesian forms

Let $z_1=1+i$ and $z_2=\sqrt2-\pi i$ . Then

z_1+z_2=(1+\sqrt2)+(1-\pi)i,

and

\begin{aligned} z_1z_2 &=(1+i)(\sqrt2-\pi i)\\ &=(\sqrt2+\pi)+(\sqrt2-\pi)i. \end{aligned}

For subtraction,

z_1-z_2=(1-\sqrt2)+(1+\pi)i.

For division, multiply numerator and denominator by

\overline{z_2}=\sqrt2+\pi i:

\frac{z_1}{z_2} =\frac{(1+i)(\sqrt2+\pi i)}{(\sqrt2-\pi i)(\sqrt2+\pi i)} =\frac{(\sqrt2-\pi)+(\sqrt2+\pi)i}{2+\pi^2}.

[routine]

[1]:Add, multiply and divide complex numbers

[THEOREM]

Rules for conjugation

For $z,z_1,z_2\in\mathbb C$ with $z_2\ne0$ ,

\overline{\overline z}=z,\qquad \overline{z_1+z_2}=\overline{z_1}+\overline{z_2},\qquad \overline{z_1-z_2}=\overline{z_1}-\overline{z_2},\qquad \overline{z_1z_2}=\overline{z_1}\,\overline{z_2},

\overline{\frac{z_1}{z_2}} =\frac{\overline{z_1}}{\overline{z_2}}, \qquad \operatorname{Re}(z)=\frac{z+\overline z}{2}, \qquad \operatorname{Im}(z)=\frac{z-\overline z}{2i}.

Each identity can be checked by writing the numbers in Cartesian form. The last two recover the real and imaginary parts of $z$ from $z$ and $\overline z$ .

[1]:Complex conjugate

Fields and number systems

[DEFINITION]

Field

A field is a non-empty set equipped with addition and multiplication satisfying closure, associativity and commutativity for both operations, distinct additive and multiplicative identities $0\ne1$ , additive inverses, multiplicative inverses for non-zero elements and the two distributive laws.

In a field we can add, subtract, multiply and divide by any non-zero element without leaving the set. The real and complex numbers are fields. The natural numbers fail the additive inverse axiom: $1\in\mathbb N$ but $-1\notin\mathbb N$ . The integers fail the multiplicative inverse axiom: $2\in\mathbb Z$ , but $2^{-1}=1/2\notin\mathbb Z$ .

[1]:Vector spaces

The number systems used in the course form the chain

\mathbb N\subseteq\mathbb Z\subseteq\mathbb Q\subseteq\mathbb R\subseteq\mathbb C.

Here $\mathbb N=\{0,1,2,\ldots\}$ ; the convention matters when deciding which systems contain a solution. Each enlargement supplies solutions that the previous system lacks. For a quadratic with real coefficients, the last step is decided by the discriminant: a negative discriminant gives no real roots but does give complex roots.

[EXAMPLE]

Compare solutions across number systems

The equation $2x+4=8$ has the solution $x=2$ in every number system listed above. The equation $2x+4=-8$ first becomes solvable in $\mathbb Z$ , where $x=-6$ , while $2x+4=7$ first becomes solvable in $\mathbb Q$ , where $x=3/2$ . The equation $2x^2+4=7$ has real solutions $x=\pm\sqrt{3/2}$ , which are irrational, whereas

2x^2+4=3 \quad\Longrightarrow\quad x=\pm\frac{i}{\sqrt2}

requires $\mathbb C$ .

A solution in one system also belongs to every larger system in the chain, so each comparison identifies the smallest system that contains the solution.

[routine]

[1]:Complex number

Polar form describes the same numbers by distance from the origin and direction, which simplifies multiplication and roots.

Complex arithmetic

On this page