Resources For Teachers For Tutors For Students & Parents Pricing
Year 11 Specialist (Unit 1 & 2) Complex numbers

Starting to build the complex numbers

20 practice questions 0 video lessons Theory + worked examples

Start building the complex numbers for Year 11 Specialist Mathematics in Queensland (QCAA). This subtopic introduces the imaginary unit, the standard form of a complex number, and its real and imaginary parts.

You will learn to add and subtract complex numbers, simplify powers of i, take square roots of negative numbers, and solve equations by equating real and imaginary parts — the foundation for conjugates, the Argand plane and complex arithmetic later in the course.

Create a free accountTrack your progress and save your work as you go.
Create free account

Theory

A complex number extends the real numbers by introducing the imaginary unit \(i\), defined so that \(i^2=-1\). In Year 11 Specialist Mathematics (QCAA, Queensland) this page starts to build the complex numbers: the form \(z=a+bi\), real and imaginary parts, adding and subtracting, powers of \(i\), and square roots of negative numbers.

Some equations, such as \(x^2=-1\), have no real solution because no real number squares to a negative. To solve them we define the imaginary unit \(i\) as a number with \(i^2=-1\), equivalently \(i=\sqrt{-1}\).

A complex number is any number of the form \(z=a+bi\), where \(a\) and \(b\) are real. Here \(a\) is the real part \(\operatorname{Re}(z)=a\) and \(b\) is the imaginary part \(\operatorname{Im}(z)=b\). Note \(\operatorname{Im}(z)\) is the real coefficient \(b\), not \(bi\). The set of all complex numbers is written \(\mathbb{C}\).

Two complex numbers are equal exactly when their real parts match and their imaginary parts match. So \(a+bi=c+di\) means \(a=c\) and \(b=d\) — one complex equation gives two real equations.

You add and subtract complex numbers part by part: combine the real parts, then the imaginary parts. Because \(i^2=-1\), the powers of \(i\) run through a repeating cycle \(i,\,-1,\,-i,\,1\), and the square root of a negative number can be written using \(i\).

Argand plane point The complex plane with a horizontal real axis and a vertical imaginary axis. The number z equals minus 3 plus 2 i is plotted as a point 3 units left and 2 units up from the origin. Re Im z = -3 + 2i
Every \(z=a+bi\) is a point on the Argand plane: \(a\) along the real axis, \(b\) up the imaginary axis. Here \(z=-3+2i\).
Complex addition as vectors Adding z equals 2 plus i and w equals 1 plus 3 i. The two vectors from the origin form two sides of a parallelogram whose diagonal is z plus w equals 3 plus 4 i. Re Im z w z + w
Addition acts like vectors: \(z=2+i\) and \(w=1+3i\) give the diagonal \(z+w=3+4i\) of a parallelogram.

A complex number and its two parts:

\[ z=a+bi,\qquad \operatorname{Re}(z)=a,\qquad \operatorname{Im}(z)=b \]
z=a+bi

The defining property of the imaginary unit:

\[ i^2=-1,\qquad i=\sqrt{-1} \]
i2=1

Addition and subtraction combine like parts:

\[ (a+bi)\pm(c+di)=(a\pm c)+(b\pm d)i \]
(a+bi)±(c+di)

The powers of \(i\) repeat every four, and a root of a negative real splits off \(i\):

\[ i^4=1,\qquad \sqrt{-k}=i\sqrt{k}\ \ (k>0) \]
i4=1
Equate parts to solve. If two complex numbers are equal, set real part = real part and imaginary part = imaginary part. One complex equation becomes two real ones you can solve together.

Working with complex numbers

  1. Write in the form \(a+bi\): identify the real part \(a\) and the imaginary part \(b\) (the coefficient of \(i\), not \(bi\)).
  2. Add or subtract part by part: combine the real parts, then combine the imaginary parts, keeping any minus sign distributed across both parts.
  3. Simplify powers of \(i\) by peeling off \(i^4=1\): write \(i^n=(i^4)^q\times i^r\) where \(r\) is the remainder on dividing \(n\) by \(4\), then use \(i^0{=}1,\ i^1{=}i,\ i^2{=}-1,\ i^3{=}-i\).
  4. For a negative square root, write \(\sqrt{-k}=\sqrt{k}\times\sqrt{-1}=i\sqrt{k}\); to solve an equality, equate real and imaginary parts and solve.
Example 1 — Real and imaginary parts
For \(z=7-5i\), state \(\operatorname{Re}(z)\) and \(\operatorname{Im}(z)\).
Solution

Match \(z\) to the standard form \(a+bi\); the imaginary part is the coefficient of \(i\), including its sign:

\(z\)\(=\)\(a+bi\)
\(z\)\(=\)\(7+(-5)i\)
\(\operatorname{Re}(z)\)\(=\)\(7\)
\(\operatorname{Im}(z)\)\(=\)\(-5\)

\(\operatorname{Re}(z)=7\) and \(\operatorname{Im}(z)=-5\).

Example 2 — A power of \(i\) and a negative root
Simplify \(i^{27}\) and write \(\sqrt{-50}\) in the form \(bi\).
Solution

Reduce \(i^{27}\) using \(i^4=1\): \(27=4\times6+3\), so keep \(i^3\):

\(i^{27}\)\(=\)\((i^4)^6\times i^3\)
\(=\)\(1^6\times i^3\)
\(=\)\(1\times(-i)\)
\(=\)\(-i\)

Split \(\sqrt{-1}=i\) off the root, then simplify the surd:

\(\sqrt{-50}\)\(=\)\(\sqrt{50}\times\sqrt{-1}\)
\(=\)\(\sqrt{25\times2}\times i\)
\(=\)\(5\sqrt{2}\,i\)

\(i^{27}=-i\) and \(\sqrt{-50}=5\sqrt{2}\,i\).

Example 3 — Add and subtract
Let \(z=6+i\) and \(w=2+4i\). Find \(z+w\) and \(z-w\).
Solution

Add real to real and imaginary to imaginary:

\(z+w\)\(=\)\((6+i)+(2+4i)\)
\(=\)\((6+2)+(1+4)i\)
\(=\)\(8+5i\)

Subtract part by part, distributing the minus sign across both parts of \(w\):

\(z-w\)\(=\)\((6+i)-(2+4i)\)
\(=\)\((6-2)+(1-4)i\)
\(=\)\(4-3i\)

\(z+w=8+5i\) and \(z-w=4-3i\).

Example 4 — Equate real and imaginary parts
Real numbers \(a\) and \(b\) satisfy \((a-b)+(a+b)i=8+2i\). Find \(a\) and \(b\).
Solution

Equality of complex numbers gives one equation from each part:

\(a-b\)\(=\)\(8\)
\(a+b\)\(=\)\(2\)

Add the equations to eliminate \(b\), then back-substitute:

\((a-b)+(a+b)\)\(=\)\(8+2\)
\(2a\)\(=\)\(10\)
\(a\)\(=\)\(5\)
\(5+b\)\(=\)\(2\)
\(b\)\(=\)\(-3\)

\(a=5\) and \(b=-3\).

Common pitfalls

Taking \(bi\) as the imaginary part. The imaginary part is the real coefficient \(b\), not \(bi\). For \(z=6i\), \(\operatorname{Im}(z)=6\), and for \(z=7-5i\), \(\operatorname{Im}(z)=-5\).
Dropping the minus sign when subtracting. The minus sign in \((a+bi)-(c+di)\) applies to both parts of the second number: it becomes \((a-c)+(b-d)i\).
Writing \(i^2=1\). By definition \(i^2=-1\). It follows that \(i^3=-i\) and \(i^4=1\); check by reducing the power modulo \(4\).
Forgetting \(\sqrt{-k}=i\sqrt{k}\). A negative under a square root is not "no answer" in \(\mathbb{C}\); peel off \(\sqrt{-1}=i\) first, e.g. \(\sqrt{-9}=3i\).

Frequently asked questions

What is the imaginary unit \(i\)?

It is the number defined so that \(i^2=-1\), equivalently \(i=\sqrt{-1}\). It lets us solve equations such as \(x^2=-1\) that have no real solution.

What are the real and imaginary parts of a complex number?

For \(z=a+bi\), the real part is \(\operatorname{Re}(z)=a\) and the imaginary part is \(\operatorname{Im}(z)=b\), the real coefficient of \(i\) (not \(bi\)).

How do you add or subtract complex numbers?

Combine like parts: \((a+bi)\pm(c+di)=(a\pm c)+(b\pm d)i\). Add or subtract the real parts, then the imaginary parts.

How do you simplify a power of \(i\)?

Use \(i^4=1\). Divide the exponent by \(4\) and keep the remainder: \(i^0=1\), \(i^1=i\), \(i^2=-1\), \(i^3=-i\). For example \(i^{27}=i^3=-i\).

What is the square root of a negative number?

Write \(\sqrt{-k}=i\sqrt{k}\) for \(k>0\). For example \(\sqrt{-36}=6i\) and \(\sqrt{-50}=5\sqrt{2}\,i\).

When are two complex numbers equal?

Exactly when their real parts are equal and their imaginary parts are equal. So one complex equation gives two real equations to solve.