Why we need $i$

On the real number line, no number squares to a negative value. So we extend the number system by defining the imaginary unit $i$ as a number such that

That’s a definition. It is not something we prove from the real numbers.

A complex number is a number of the form

The set of all complex numbers is written $\mathbb{C}$.

Real numbers sit inside the complex numbers. If $b=0$, then $z=a+0i=a$. So complex numbers are not a separate universe from real numbers; they extend the system we already use.

Cartesian form and the named parts

We call $z=a+bi$ Cartesian form. It represents a complex number by giving its horizontal real coordinate and vertical imaginary coordinate directly.

For $z=a+bi$:

• The real part is the real number $a$ in the Cartesian form of $z$, written $\operatorname{Re}(z)=a$.
• The imaginary part is the real number $b$ multiplying $i$ in the Cartesian form of $z$, written $\operatorname{Im}(z)=b$. Be careful here: the imaginary part is $b$, not $bi$.
• The complex conjugate is the complex number found by changing the sign of the imaginary part while keeping the real part fixed, so $z^*=a-bi$.
• The modulus is the non-negative distance of the complex number from the origin in the complex plane, written $|z|=\sqrt{a^2+b^2}$.
• The argument is an angle from the positive real axis to the line segment joining the origin to the point representing $z$, written $\arg z$. The principal argument is usually taken in the interval $-\pi<\arg z\le \pi$.

The modulus formula comes straight from Pythagoras: the real and imaginary parts make the two perpendicular sides of a right triangle. For a non-zero complex number, $\tan(\arg z)=\frac{b}{a}$ when $a\ne0$, but you still have to check the quadrant. Since tangent repeats every $\pi$, the calculator value of $\tan^{-1}\left(\frac{b}{a}\right)$ can point to the wrong quadrant if you use it blindly.

The complex plane, or Argand diagram

The complex plane is a two-dimensional coordinate plane where the horizontal axis represents real parts and the vertical axis represents imaginary parts. The same plane is also called an Argand diagram, which is the IB term you will often see in questions.

A point with coordinates $(a,b)$ represents the complex number $a+bi$. That is why complex numbers link naturally with vectors: adding complex numbers in Cartesian form has the same geometry as adding displacement vectors in a plane.

Conjugation has a simple geometric meaning. If $z=a+bi$, then $z^*=a-bi$ is the reflection of $z$ in the real axis. The modulus stays the same, and the argument changes sign, apart from the usual care needed at $\pi$ and $-\pi$.

Language and modelling

The words “imaginary” and “complex” are historically unfortunate if they make the numbers sound fake or unnecessarily difficult. They are neither. Complex numbers give us a representation system: two real quantities are packaged into one object. In electrical engineering, for example, impedance can be represented as a quantity of the form $a+bi$, with one part modelling resistance and the other modelling reactance. The notation is compact, and the algebra carries useful geometric information.

Working in Cartesian form

Cartesian form is usually the easiest form for addition and subtraction: real parts go with real parts, and imaginary parts go with imaginary parts. Let $w=c+di$, where $w$ is another complex number and $c$ and $d$ are real numbers. Then

and

For multiplication, just expand as ordinary algebra and then use $i^2=-1$:

Division is the place where students most often try to make up a “divide the parts separately” rule. Don’t. To divide by $w=c+di$, multiply the numerator and denominator by the conjugate of the denominator:

This works because

which is real and positive when $w\ne0$. The denominator has turned into something we know how to divide by.

When two complex numbers are equal in Cartesian form, their real parts match and their imaginary parts match. In practice, one complex equation becomes two real equations, which is often the cleanest way into the problem.

Modulus-argument form

A modulus-argument form represents a complex number using its distance from the origin and its angle from the positive real axis. It is also called polar form. We write

where $r$ is the modulus of $z$ with $r\ge0$, and $\theta$ is an argument of $z$.

To convert between Cartesian and polar form, use

and

The quadrant matters here. A complex number in quadrant II or III can have the same tangent value as a number in another quadrant, so use its position on the Argand diagram to choose the correct argument.

Euler form

Euler form represents a complex number using the complex exponential:

where $e$ is Euler's number, the base of natural logarithms. Euler form is equivalent to polar form because

So the same complex number can be written in three equivalent ways:

The IB expects you to move between these forms fluently. Use Cartesian form for addition or subtraction; use polar or Euler form for multiplication, division, or powers.

Products, quotients and their geometry

Polar form makes multiplication very simple. Let $s$ be the modulus of $w$ with $s\ge0$, and let $\phi$ be an argument of $w$. If

then

Geometrically, multiplying by $w$ scales distances from the origin by factor $s$ and rotates arguments by angle $\phi$.

For quotients, provided $w\ne0$,

Geometrically, dividing by $w$ scales distances by factor $\frac{1}{s}$ and rotates arguments by angle $-\phi$.

Euler form gives a quick reason for these rules:

and

That is one reason the identity $e^{i\pi}+1=0$ is often called elegant: it links $e$, $i$, $\pi$, $1$ and $0$ in one short statement. Beauty in mathematics often comes from this kind of compression, where several ideas become one relationship.

Complex conjugate roots

A polynomial with real coefficients has all of its coefficients in the real numbers. When a polynomial like this has a non-real complex root, its conjugate must be a root as well.

One neat way to prove it is to take conjugates. Let

Real numbers do not change under conjugation, so conjugating $P(z)$ gives

If $P(z)=0$, then

That gives the conjugate-pair result. The phrase “real coefficients” matters. If the coefficients are not real, the result can fail, so don't assume roots come in conjugate pairs just because complex numbers appear in the equation.

For a quadratic or polynomial with real coefficients, any non-real roots therefore come in pairs such as $\alpha$ and $\alpha^*$

Multiplying the two linear factors produces a real quadratic factor:

Both $\alpha+\alpha^*$ and $\alpha\alpha^*$ are real. This ties in with the sum and product of roots: once one non-real root of a real polynomial is known, its conjugate is known straight away, and the remaining roots can often be found by factorising or comparing coefficients.

De Moivre's theorem

De Moivre's theorem gives powers of a complex number in polar form: raise the modulus to the power and multiply the argument by that power:

For positive integer powers, the result follows nicely by induction using the compound angle identities.

For $n=1$,

so the statement is true. Now assume it is true for $n=k$, where $k$ is a positive integer:

Then for $n=k+1$,

The final step uses the product rule for polar form, which comes from the compound angle identities for sine and cosine. Therefore, by mathematical induction, De Moivre's theorem is true for every positive integer $n$.

For negative integers, use reciprocals to get the same result, provided $r\ne0$. In practice, polar form is usually the most sensible form when powers are involved.

Extension to rational exponents

De Moivre's theorem also extends to rational exponents, but roots of complex numbers are usually multiple-valued.

This gives $q$ distinct values. The extra term $\frac{2\pi k}{q}$ is not just decoration; it accounts for all possible roots. Leave it out, and you usually find only one answer.

Powers and roots of complex numbers

For powers, write the number in polar or Euler form, apply De Moivre's theorem, then convert back if the question asks for Cartesian form. For instance, if $z=r\operatorname{cis}\theta$, then $z^n$ has modulus $r^n$ and argument $n\theta$.

For roots, begin by writing the target complex number in polar form. Suppose

The $q$ roots are

Geometrically, the $q$ roots sit equally spaced around a circle centred at the origin. The radius is $R^{\frac{1}{q}}$, and neighbouring roots are separated by an angle of $\frac{2\pi}{q}$. This picture is often the fastest way to notice a missing root.

When solving an equation such as $z^q=A$, where $A$ is a complex constant, keep track of which set each variable belongs to. If $z=r\operatorname{cis}\theta$, then $r$ is non-negative and real. So from $r^4=1$, for example, take $r=1$, not four complex values for $r$; the different complex roots come from the different allowed arguments.

The roots of unity are the standard model for this idea. Solving $z^q=1$ gives $q$ points equally spaced around the unit circle. In Cartesian form, some may look simple, such as $1$, $-1$, $i$ and $-i$ when $q=4$; in polar form, the full pattern is much easier to see.