Complex Number Algebra and Roots

Overview

This branch develops the algebraic side of complex numbers: Cartesian form, equality, arithmetic, conjugates, division, square roots, and roots of real-coefficient polynomials. It supports the main hub by keeping the manipulation skills and conjugate-root reasoning in one reusable place.

Foundations

Why introduce $i$?

In the real number system, some equations have no solution. For example,

$$x^2=-1$$

has no real solution, because the square of any real number is always non-negative.

To extend the number system, we introduce a new number called the imaginary unit, denoted by $i$, defined by

$$i^2=-1.$$

With this definition, the equation

$$x^2+1=0$$

can now be solved:

$$x^2=-1,$$

so

$$x=i\text{or}x=-i.$$

Thus, introducing $i$ allows us to work with numbers beyond the real number system.

Square roots of negative numbers

Since

$$i^2=-1,$$

we write

$$\sqrt{-1}=i.$$

More generally, if $a>0$, then

$$\sqrt{-a}=i\sqrt{a}.$$

For example,

$$\sqrt{-9}=3i$$

and

$$\sqrt{-25}=5i.$$

Be careful with the difference between a square-root expression and solving a quadratic equation.

For example,

$$\sqrt{-9}=3i,$$

but the equation

$$x^2=-9$$

has two solutions:

$$x=3i\text{or}x=-3i.$$

So square roots of complex numbers often need attention to signs.

Cartesian form

A complex number is a number of the form

$$z=x+iy,$$

where

$$x,y\in \mathbb{R}.$$

This is called the Cartesian form of a complex number.

The number $x$ is called the real part of $z$, and the number $y$ is called the imaginary part of $z$:

$$\mathrm{Re}(z)=x,$$ $$\mathrm{Im}(z)=y.$$

A very common mistake is to think that the imaginary part is $iy$. It is not.

If

$$z=3-5i,$$

then

$$\mathrm{Re}(z)=3$$

and

$$\mathrm{Im}(z)=-5.$$

The imaginary part is the coefficient of $i$, not the whole term containing $i$.

Real, imaginary, and complex numbers

For

$$z=x+iy,$$

the values of $x$ and $y$ determine the type of number.

If $y=0$, then

$$z=x,$$

so $z$ is a real number.

If $x=0$ and $y\ne 0$, then

$$z=iy,$$

so $z$ is a purely imaginary number.

If both $x$ and $y$ are non-zero, then $z$ has both a real part and an imaginary part.

The number $0$ is a special case. It can be written as

$$0=0+0i.$$

In the Argand diagram, it lies at the intersection of the real axis and the imaginary axis.

Algebraic and geometric meanings

A complex number should be understood in two connected ways.

Algebraically,

$$z=x+iy$$

is an expression made from a real part and an imaginary part.

Geometrically,

$$z=x+iy$$

can be represented by the point

$$(x,y)$$

in the Argand diagram.

In this diagram:

• the horizontal coordinate represents the real part;
• the vertical coordinate represents the imaginary part.

So

$$z=x+iy$$

corresponds to the point

$$(x,y).$$

This dual viewpoint is the core of the topic. Many algebraic statements about complex numbers become geometric statements about points in a plane.

Equality of complex numbers

Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal.

So if

$$x_1+i{y}_1=x_2+i{y}_2,$$

then

$$x_1=x_2$$

and

$$y_1=y_2.$$

This is the standard method for solving simple equations involving complex numbers in Cartesian form.

For example, suppose

$$2+a+i(3-b)=5+4i.$$

Comparing real parts gives

$$2+a=5,$$

so

$$a=3.$$

Comparing imaginary parts gives

$$3-b=4,$$

so

$$b=-1.$$

Therefore,

$$a=3,b=-1.$$

The reason this works is that equality of complex numbers means equality of both coordinates in the Argand diagram.

A key restriction: no natural ordering

Real numbers can be ordered on a number line. For example, it is meaningful to say

$$3>2$$

because $3$ lies to the right of $2$ on the real number line.

Complex numbers are different. They are represented in a plane, not on a single line. Therefore, there is no natural meaning of “greater than” or “less than” for non-real complex numbers.

For example,

$$2+i>2-i$$

is not a meaningful statement.

Although $2+i$ lies above $2-i$ in the Argand diagram, being higher in the plane does not mean being greater. In this topic, complex numbers should not be compared using inequality signs unless the comparison is about a real-valued quantity such as modulus.

For example,

$$|2+i|=|2-i|$$

is meaningful, because both sides are real numbers.

Arithmetic in Cartesian Form

The basic principle

When working in Cartesian form,

$$z=x+iy,$$

treat $i$ like an algebraic symbol, but always remember the defining rule

$$i^2=-1.$$

Most calculations follow the same basic workflow:

1. expand or simplify using ordinary algebra;
2. replace $i^2$ by $-1$;
3. collect the real and imaginary parts;
4. write the final answer in Cartesian form $x+iy$.

This discipline is especially important in multiplication, division, and square-root problems.

Addition and subtraction

To add or subtract complex numbers, combine real parts with real parts, and imaginary parts with imaginary parts.

If

$$z_1=x_1+i{y}_1$$

and

$$z_2=x_2+i{y}_2,$$

then

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

and

$$z_1-z_2=(x_1-x_2)+i(y_1-y_2).$$

For example,

$$(2+3i)-(-1+2i)=2+3i+1-2i.$$

So

$$(2+3i)-(-1+2i)=3+i.$$

This is similar to adding or subtracting coordinates component by component:

$$x+iy⟷(x,y).$$

Multiplication

To multiply complex numbers, expand as in ordinary algebra, then use

$$i^2=-1.$$

For example,

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

Since

$$i^2=-1,$$

we have

$$-4i^2=4.$$

Therefore,

$$(2+4i)(1-i)=2+2i+4=6+2i.$$

In general,

$$(x_1+i{y}_1)(x_2+i{y}_2)$$

can be expanded as

$$x_1{x}_2+i{x}_1{y}_2+i{x}_2{y}_1+i^2{y}_1{y}_2.$$

Using

$$i^2=-1,$$

this becomes

$$x_1{x}_2-y_1{y}_2+i(x_1{y}_2+x_2{y}_1).$$

So

$$(x_1+i{y}_1)(x_2+i{y}_2)=(x_1{x}_2-y_1{y}_2)+i(x_1{y}_2+x_2{y}_1).$$

The formula is useful, but in most calculations it is safer to remember the method: expand normally, replace $i^2$ by $-1$, and collect terms.

Division

Division is slightly different because the denominator may contain $i$.

For example, in

$$\frac{5+10i}{4+3i},$$

the denominator is not real. To write the answer in Cartesian form, we multiply the numerator and denominator by the conjugate of the denominator.

The conjugate of

$$4+3i$$

is

$$4-3i.$$

The reason this works is that

$$(a+bi)(a-bi)=a^2+b^2,$$

which is real.

So

$$\frac{5+10i}{4+3i}=\frac{(5+10i)(4-3i)}{(4+3i)(4-3i)}.$$

The denominator becomes

$$(4+3i)(4-3i)=4^2+3^2=25.$$

The numerator is

$$(5+10i)(4-3i)=20-15i+40i-30i^2.$$

Since

$$i^2=-1,$$

we get

$$20-15i+40i-30i^2=20+25i+30=50+25i.$$

Therefore,

$$\frac{5+10i}{4+3i}=\frac{50+25i}{25}=2+i.$$

This process is often called realising the denominator, meaning that the denominator is changed into a real number.

Square roots in Cartesian form

To find the square roots of a complex number in Cartesian form, suppose one square root is

$$x+iy,$$

where $x$ and $y$ are real.

For example, to find the square roots of

$$3+4i,$$

let

$$\sqrt{3+4i}=x+iy.$$

Then

$$(x+iy{)}^2=3+4i.$$

Expanding the left-hand side gives

$$x^2+2xyi+i^2{y}^2=3+4i.$$

Since

$$i^2=-1,$$

we have

$$x^2-y^2+2xyi=3+4i.$$

Now compare real and imaginary parts:

$$x^2-y^2=3$$

and

$$2xy=4.$$

So

$$xy=2.$$

Solving these equations gives

$$x=2,y=1$$

or

$$x=-2,y=-1.$$

Therefore, the square roots of

$$3+4i$$

are

$$2+i$$

and

$$-2-i.$$

This pair is not accidental. If

$$w^2=z,$$

then

$$(-w{)}^2=z$$

as well.

So non-zero complex numbers have two square roots, and they are opposites of each other.

Common working discipline

When doing arithmetic in Cartesian form, keep the following habits:

• collect the final answer as $x+iy$;
• remember that $i^2=-1$, not $1$;
• do not leave a complex number in the denominator;
• compare real and imaginary parts only after both sides are written clearly in Cartesian form;
• in square-root problems, remember that roots usually come in opposite pairs.

Complex Conjugates

Definition

If

$$z=x+iy,$$

where $x,y\in \mathbb{R}$, then the complex conjugate of $z$ is

$$z^{\ast }=x-iy.$$

Here, $z^{\ast }$ denotes the conjugate of $z$. Some books use the notation $\overline{z}$ instead.

Conjugation changes only the sign of the imaginary part:

• the real part stays the same;
• the imaginary part changes sign.

For example, if

$$z=3-4i,$$

then

$$z^{\ast }=3+4i.$$

Geometric meaning

In the Argand diagram,

$$z=x+iy$$

is represented by the point

$$(x,y).$$

Its conjugate

$$z^{\ast }=x-iy$$

is represented by the point

$$(x,-y).$$

So conjugation reflects the point across the real axis.

This means that $z$ and $z^{\ast }$ have the same horizontal coordinate, but opposite vertical coordinates.

The key identity

The most important identity involving conjugates is

$$z{z}^{\ast }=x^2+y^2.$$

To see why, start from

$$z=x+iy$$

and

$$z^{\ast }=x-iy.$$

Then

$$z{z}^{\ast }=(x+iy)(x-iy).$$

Expanding gives

$$z{z}^{\ast }=x^2-ixy+ixy-i^2{y}^2.$$

The middle terms cancel, so

$$z{z}^{\ast }=x^2-i^2{y}^2.$$

Since

$$i^2=-1,$$

we get

$$z{z}^{\ast }=x^2+y^2.$$

This is always a real and non-negative number.

For example, if

$$z=3-4i,$$

then

$$z^{\ast }=3+4i.$$

So

$$z{z}^{\ast }=(3-4i)(3+4i)=3^2+4^2=25.$$

This identity is the reason conjugates are so useful: they can turn a complex product into a real number.

Extracting real and imaginary parts

Conjugates can also be used to isolate the real and imaginary parts of a complex number.

If

$$z=x+iy$$

and

$$z^{\ast }=x-iy,$$

then

$$z+z^{\ast }=(x+iy)+(x-iy)=2x.$$

Therefore,

$$z+z^{\ast }=2\mathrm{Re}(z).$$

Adding a complex number to its conjugate removes the imaginary part.

Similarly,

$$z-z^{\ast }=(x+iy)-(x-iy)=2iy.$$

Therefore,

$$z-z^{\ast }=2i\mathrm{Im}(z).$$

Subtracting the conjugate removes the real part and leaves the imaginary component.

Conjugates and division

Conjugates are especially useful for division in Cartesian form.

For example, suppose we want to simplify

$$\frac{1}{x+iy}.$$

We multiply the numerator and denominator by the conjugate of the denominator:

$$\frac{1}{x+iy}=\frac{1}{x+iy}\cdot \frac{x-iy}{x-iy}.$$

So

$$\frac{1}{x+iy}=\frac{x-iy}{(x+iy)(x-iy)}.$$

Using the key identity,

$$(x+iy)(x-iy)=x^2+y^2.$$

Therefore,

$$\frac{1}{x+iy}=\frac{x-iy}{x^2+y^2},$$

provided that

$$x+iy\ne 0.$$

The denominator is now real. This is why conjugates make division possible in Cartesian form.

Useful algebraic properties

Conjugation behaves naturally with the usual algebraic operations.

For any complex numbers $z$, $z_1$, and $z_2$,

$$(z^{\ast }{)}^{\ast }=z,$$ $$(z_1+z_2{)}^{\ast }=z_1^{\ast }+z_2^{\ast },$$

and

$$(z_1{z}_2{)}^{\ast }=z_1^{\ast }{z}_2^{\ast }.$$

If

$$z_2\ne 0,$$

then

$${\left(\frac{z_1}{z_2}\right)}^{\ast }=\frac{z_1^{\ast }}{z_2^{\ast }}.$$

These properties are useful because they allow conjugates to be handled algebraically without expanding every expression from first principles.

Why conjugates matter

Conjugates matter because they connect algebra and geometry.

Algebraically, they help us:

• turn products such as $z{z}^{\ast }$ into real numbers;
• divide complex numbers in Cartesian form;
• extract real and imaginary parts.

Geometrically, they represent reflection in the real axis.

Later, conjugates also explain an important symmetry of polynomial roots: if a polynomial has real coefficients, then its non-real complex roots occur in conjugate pairs.

Caption: Conjugation reflects a complex number across the real axis while preserving its modulus; the angle marker also shows how $\arg z$ is read from the positive real axis.

Complex Roots of Equations

Why complex roots matter

Some polynomial equations have no real roots.

For example,

$$x^2+1=0$$

gives

$$x^2=-1.$$

There is no real number whose square is $-1$. But in the complex number system,

$$i^2=-1,$$

so the equation has two complex roots:

$$x=i\text{or}x=-i.$$

This is one reason complex numbers are useful: they allow polynomial equations to have roots even when real numbers are not enough.

Fundamental Theorem of Algebra

The Fundamental Theorem of Algebra says that every polynomial of degree $n$ with complex coefficients has exactly $n$ roots in $\mathbb{C}$, counting repeated roots.

This means:

• a quadratic has $2$ complex roots, counting repeated roots;
• a cubic has $3$ complex roots, counting repeated roots;
• a polynomial of degree $n$ has $n$ complex roots, counting repeated roots.

The phrase “counting repeated roots” is important.

For example,

$$(x-1{)}^2=0$$

has only one distinct root,

$$x=1,$$

but this root is repeated twice. So it counts as two roots.

Similarly, a quadratic always has two roots in $\mathbb{C}$ when repeated roots are counted, even if it has no real roots.

For example,

$$x^2+1=0$$

has roots

$$i\text{and}-i.$$

Conjugate Root Theorem

The Conjugate Root Theorem says that if a polynomial has real coefficients, then any non-real root must occur together with its conjugate.

So if

$$a+ib$$

is a root, where $a,b\in \mathbb{R}$ and $b\ne 0$, then

$$a-ib$$

is also a root.

The condition that the polynomial has real coefficients is essential. The theorem does not necessarily apply if the polynomial has non-real coefficients.

For example, if a polynomial with real coefficients has root

$$1-i,$$

then it must also have root

$$1+i.$$

Why conjugate pairs give real factors

Suppose a polynomial with real coefficients has the non-real root

$$a+ib.$$

Then it must also have the conjugate root

$$a-ib.$$

The corresponding two linear factors are

$$x-(a+ib)$$

and

$$x-(a-ib).$$

Their product is

$$(x-(a+ib))(x-(a-ib)).$$

Rewrite each factor as

$$(x-a)-ib$$

and

$$(x-a)+ib.$$

Then

$$(x-(a+ib))(x-(a-ib))=((x-a)-ib)((x-a)+ib).$$

Using the difference of squares,

$$((x-a)-ib)((x-a)+ib)=(x-a{)}^2-(ib{)}^2.$$

Since

$$i^2=-1,$$

we have

$$(ib{)}^2=-b^2.$$

Therefore,

$$(x-a{)}^2-(ib{)}^2=(x-a{)}^2+b^2.$$

So

$$(x-(a+ib))(x-(a-ib))=(x-a{)}^2+b^2.$$

This is a quadratic factor with real coefficients.

This identity is very useful because it allows a conjugate pair of complex roots to be converted into a real quadratic factor.

Example: forming a real quadratic factor

Suppose a polynomial with real coefficients has root

$$1-i.$$

By the Conjugate Root Theorem, it also has root

$$1+i.$$

The corresponding quadratic factor is

$$(x-(1-i))(x-(1+i)).$$

Simplify each factor:

$$x-(1-i)=x-1+i,$$

and

$$x-(1+i)=x-1-i.$$

So

$$(x-(1-i))(x-(1+i))=(x-1+i)(x-1-i).$$

This is a product of conjugates, so

$$(x-1+i)(x-1-i)=(x-1{)}^2+1.$$

Therefore,

$$(x-(1-i))(x-(1+i))=x^2-2x+2.$$

So the conjugate pair

$$1-i,1+i$$

corresponds to the real quadratic factor

$$x^2-2x+2.$$

Solving polynomials with a given complex root

When a polynomial with real coefficients is given and one non-real root is known, use the conjugate root theorem as a problem-solving tool.

A reliable workflow is:

1. Write down the conjugate of the given non-real root.
2. Form the quadratic factor from the conjugate pair.
3. Divide the polynomial by this quadratic factor, or compare coefficients.
4. Solve the remaining factor.
5. Check that the total number of roots matches the degree of the polynomial.

For example, if a cubic polynomial with real coefficients has root

$$1-i,$$

then it must also have root

$$1+i.$$

These two roots already account for two roots of the cubic. Since a cubic has three roots in total, counting repeated roots, there is one remaining root to find.

Conceptual interpretation

The Conjugate Root Theorem expresses a symmetry.

A polynomial with real coefficients does not distinguish between the imaginary direction above the real axis and the imaginary direction below the real axis.

In the Argand diagram, the roots

$$a+ib$$

and

$$a-ib$$

are reflections of each other in the real axis.

That is why non-real roots of real-coefficient polynomials occur in conjugate pairs.

Related

• Complex Numbers
• Argand Diagram, Modulus, Argument, and Loci
• Polar Form, de Moivre, and Roots
• Content Map