Ask for a number whose square is −4 and the real numbers have no answer — squaring anything real gives a result that is zero or positive, never negative. Rather than stop there, mathematicians invented a new number to fill the gap, called i, defined by the one rule i² = −1. Build on that single idea and you get the complex numbers: a larger number system that contains the reals and finally lets every quadratic have a solution.
On the CSCA this topic is a reliable scorer — nearly every paper includes a complex-number question, and they cluster around a few predictable skills. This first page builds the foundation: what i means, how complex numbers are written, the picture that goes with them, and how to add, subtract, and (the key skill) multiply them. Practise the i² = −1 substitution until it is automatic, and the rest of the topic becomes much easier.
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The number i, and the form a + bi
Start from the definition. We let i be a number satisfying i² = −1 — equivalently, i = √−1. It is not a real number; it is a new building block. Multiply it by a real number and add a real number, and you get the general shape of every complex number.
Complex number. Any number of the form a + bi, where a and b are real and i² = −1. We call a the real part and b the imaginary part (note: b itself is a real number — it is the coefficient of i). A number with b = 0 is just a real number; a number with a = 0, like 3i, is called purely imaginary.
So 3 + 2i has real part 3 and imaginary part 2. The whole system is built to behave like ordinary algebra, with one extra rule: whenever an i² appears, replace it with −1. Every calculation on this page uses that rule.
Try it! · Check-in question
For the complex number 5 − 7i, the real part and imaginary part are ( ).
In a + bi the real part is a = 5 and the imaginary part is b = −7 (the coefficient of i, including its sign). The answer is C. Option A includes the i in the imaginary part — but the imaginary part is just the number −7, not −7i. Option B swaps the two parts. Option D drops the negative sign that belongs to the imaginary part.
The Argand plane
Complex numbers have a natural picture. Because a + bi is built from two real numbers, a and b, we can plot it as a point — or an arrow from the origin — on a plane: the real part along the horizontal axis, the imaginary part up the vertical axis. This picture is called the Argand plane (or complex plane), and it makes the later ideas of modulus and conjugate visual rather than abstract.
The complex number 3 + 2i plotted on the Argand plane: 3 along the real axis, 2 up the imaginary axis.
The Argand plane looks exactly like the coordinate plane, and that is the point: a complex number a + bi behaves much like the point (a, b) or the vector ⟨a, b⟩. Adding complex numbers will mirror adding vectors, and the modulus (next pages) will be the length of this arrow — the same Pythagoras you already know.
Adding and subtracting
Addition and subtraction are the easy operations: combine the real parts together and the imaginary parts together. It is exactly like collecting like terms in algebra — the real numbers are one kind of term, the i-terms another.
Addition and subtraction. (a + bi) ± (c + di) = (a ± c) + (b ± d)i. Real parts combine with real parts, imaginary with imaginary — they never mix.
Worked example · adding and subtracting
Given z = 3 + 2i and w = 1 − 5i, find z + w and z − w.
For the sum, combine real parts and imaginary parts: (3 + 1) + (2 + (−5))i.
That is 4 + (−3)i = 4 − 3i.
For the difference, subtract each part: (3 − 1) + (2 − (−5))i = 2 + 7i.
Watch the signs on subtraction: 2 − (−5) = 2 + 5 = 7. The imaginary parts are handled exactly like the real ones.
Try it! · Check-in question
If z + w = 5 + 2i and z = 3 − i, then w = ( ).
Since z + w = 5 + 2i, we get w = (5 + 2i) − z = (5 + 2i) − (3 − i). Subtract part by part: (5 − 3) + (2 − (−1))i = 2 + 3i. The answer is A. Check: (3 − i) + (2 + 3i) = 5 + 2i ✓. Option B adds z instead of subtracting it: (5 + 3) + (2 − 1)i = 8 + i. Option C loses the double minus: it works out 2 − 1 = 1 instead of 2 − (−1) = 3. Option D subtracts the wrong way round: z − (z + w) = (3 − 5) + (−1 − 2)i = −2 − 3i, which is −w, not w.
Try it! · Check-in question
The sum (5 − 2i) + (−3 + 6i) is plotted on the Argand plane. Its point is ( ).
Add part by part: (5 + (−3)) + (−2 + 6)i = 2 + 4i. On the Argand plane a + bi is the point (a, b), so the point is (2, 4). The answer is D. Option A subtracts the second number instead of adding it: (5 − (−3)) + (−2 − 6)i = 8 − 8i. Option B swaps the axes: it puts the imaginary part 4 across and the real part 2 up. Option C drops the minus sign on −2i, so it works out 2 + 6 = 8 instead of −2 + 6 = 4.
Multiplication: the central skill
This is the operation the exam tests most, and the one where the i² = −1 rule matters most. Multiply two complex numbers exactly as you would expand two brackets in algebra — every term times every term — and then convert any i² into −1. That conversion is what folds an imaginary piece back into the real part.
Multiplying (a + bi)(c + di). (1) Expand all four products, just like FOIL: ac + adi + bci + bdi². (2) Replace i² with −1, so the bdi² term becomes −bd. (3) Collect: the real part is ac − bd, the imaginary part is ad + bc.
Worked example · multiplying two complex numbers
Find (2 + 3i)(1 − 2i).
Expand all four products: (2)(1) + (2)(−2i) + (3i)(1) + (3i)(−2i) = 2 − 4i + 3i − 6i².
Convert the i² term: −6i² = −6(−1) = +6.
Now collect real and imaginary parts: (2 + 6) + (−4 + 3)i.
That gives 8 − i.
The key step is the second one: −6i² became +6 and moved into the real part. Forget to convert i² and you would wrongly leave a −6i² sitting in the answer.
Always convert i². The single most common error is expanding the brackets correctly but leaving i² (or treating it as +1). Remember the rule: i² = −1, so a term like −6i² becomes +6, and +bdi² becomes −bd. Every i² must turn into a real number before you finish.
Try it! · Check-in question
i(4 − 3i) = ( ).
Multiply i into both terms: i · 4 − i · 3i = 4i − 3i². Since i² = −1, the term −3i² becomes +3. So the result is 3 + 4i. The answer is B. Option A treats i² as +1, so −3i² stays −3 and the result is −3 + 4i. Option C multiplies only the 3i by i and forgets the 4: 4 − 3i² = 4 + 3 = 7. Option D drops the −3i² term completely, as if i² were 0.
Worked example · squaring a complex number
Find (2 − i)².
Write it as a product and expand: (2 − i)(2 − i) = 4 − 2i − 2i + i².
Combine the middle terms and convert i²: 4 − 4i + (−1).
Collect: (4 − 1) − 4i = 3 − 4i.
Squaring is just multiplication by itself — the same expand-then-convert routine. Here i² became −1 and pulled the real part down from 4 to 3.
Try it! · Check-in question
If z = 1 + i, then z² − 2z = ( ).
First z² = (1 + i)(1 + i) = 1 + 2i + i² = 1 + 2i − 1 = 2i. Next 2z = 2 + 2i. So z² − 2z = 2i − (2 + 2i) = −2. The answer is D. Option A treats i² as +1, so z² = 2 + 2i and the difference becomes 0. Option B subtracts z instead of 2z: 2i − (1 + i) = −1 + i. Option C squares each part on its own, (1 + i)² = 1² + i² = 0, which loses the middle term 2i; then 0 − (2 + 2i) = −2 − 2i.
Try it! · Check-in question
For the real number k, the product (k + i)(2 − i) is purely imaginary. Then k = ( ).
Expand: (k + i)(2 − i) = 2k − ki + 2i − i². Since −i² = +1, this is (2k + 1) + (2 − k)i. Purely imaginary means the real part is 0, so 2k + 1 = 0 and k = −1/2. (The imaginary part is then 2 + 1/2 = 5/2, which is not 0.) The answer is C. Option A treats i² as +1, so the real part becomes 2k − 1 and k = 1/2. Option B sets the imaginary part 2 − k to 0 instead; that would make the product real, not purely imaginary. Option D drops the −i² term, so the real part is just 2k and k = 0.
Powers of i
The powers of i fall into a short repeating cycle, and recognising it turns a large power such as i²³ into a one-step answer. Build the first few from i² = −1: i¹ = i, i² = −1, then i³ = i²·i = −i, and i⁴ = i²·i² = (−1)(−1) = 1. After that it repeats, because i⁴ = 1 resets everything.
The cycle of i. i¹ = i, i² = −1, i³ = −i, i⁴ = 1, and then it loops: i⁵ = i, i⁶ = −1, … To find iⁿ, divide n by 4 and keep the remainder — the remainder (1, 2, 3, or 0) tells you which of i, −1, −i, 1 you land on. (Remainder 0 means iⁿ = 1.)
Worked example · a high power of i
Find i²³.
Divide the exponent by 4: 23 = 4 × 5 + 3, so the remainder is 3.
A remainder of 3 lands on i³.
And i³ = −i, so i²³ = −i.
The shortcut is just "remainder after dividing by 4." Twenty-three is three past a multiple of four, so i²³ matches i³ = −i. No need to multiply twenty-three i's together.
Try it! · Check-in question
The value of i⁵ · i⁹ is ( ).
When you multiply powers of the same base, you add the exponents: i⁵ · i⁹ = i¹⁴. Since 14 = 4 × 3 + 2, the remainder is 2, and i² = −1. The answer is A. You can also check it another way: i⁵ = i and i⁹ = i, so the product is i · i = i² = −1. Option B multiplies the exponents: 5 × 9 = 45, and 45 = 4 × 11 + 1 gives i. Option C takes i · i as +1, forgetting that i² = −1. Option D adds the two values, i + i = 2i, instead of multiplying them.
Try it! · Check-in question
The value of i + i² + i³ + i⁴ + i⁵ is ( ).
Replace each power using the cycle: i + (−1) + (−i) + 1 + i. The first four terms cancel: i − 1 − i + 1 = 0. That leaves i⁵ = i. The answer is B. Option A stops after the first four terms and forgets i⁵. Option C takes i³ as +i instead of −i: i − 1 + i + 1 + i = 3i. Option D treats i² as +1 all the way through, so i³ = i and i⁴ = 1: i + 1 + i + 1 + i = 2 + 3i. A useful fact: any four powers of i in a row add up to 0.
Your turn! · Practice questions
Practice · Question 1
For the complex number −2 + 3i, the real part and imaginary part are ( ).
In a + bi the real part is a = −2 and the imaginary part is b = 3. The answer is B. Option A swaps the parts. Option C includes the i in the imaginary part (it should be just 3). Option D drops the negative on the real part.
Practice · Question 2
(7 + 2i) − (3 + 5i) = ( ).
Subtract each part: (7 − 3) + (2 − 5)i = 4 + (−3)i = 4 − 3i. The answer is D. Option A adds the numbers instead of subtracting: (7 + 3) + (2 + 5)i. Option B subtracts the imaginary parts the wrong way round: 5 − 2 = 3. Option C removes the brackets without giving the minus sign to 5i: 7 + 2i − 3 + 5i = 4 + 7i. A minus sign in front of a bracket changes the sign of every term inside it.
Practice · Question 3
If z₁ = 4 + i and z₂ = 1 + 3i, the imaginary part of z₁z₂ is ( ).
Expand: (4 + i)(1 + 3i) = 4 + 12i + i + 3i². Since 3i² = −3, this is (4 − 3) + (12 + 1)i = 1 + 13i. The imaginary part is the coefficient of i, so it is 13. The answer is A. Option B keeps the i, but the imaginary part is the real number 13, not 13i. Option C multiplies part by part, as if it were addition: (4 × 1) + (1 × 3)i = 4 + 3i, whose imaginary part is 3. This misses the cross terms 12i and i. Option D gives the real part, 1, instead of the imaginary part.
Practice · Question 4
Which complex number has its square equal to 3 + 4i?
Test each option by squaring it. (2 + i)² = 4 + 4i + i² = 4 + 4i − 1 = 3 + 4i ✓. The answer is C. Option A has the wrong sign: (2 − i)² = 3 − 4i. Option B swaps the real and imaginary parts: (1 + 2i)² = 1 + 4i − 4 = −3 + 4i. Option D takes the square root of each part on its own (√3, and √4 = 2), but (√3 + 2i)² = 3 + 4√3 i − 4 = −1 + 4√3 i. Squaring does not work part by part, because of the middle term.
Practice · Question 5
The value of i¹⁰⁰ is ( ).
Divide 100 by 4: 100 = 4 × 25, so the remainder is 0. Remainder 0 means a whole number of cycles, so i¹⁰⁰ = (i⁴)²⁵ = 1²⁵ = 1. The answer is D. Option A assumes every even power of i is −1; that is true only when the remainder is 2. Option B uses the quotient 25 instead of the remainder: 25 = 4 × 6 + 1 gives i. Option C reads "remainder 0" as the value 0, but the remainder only tells you where you are in the cycle.
Practice · Question 6
(2 + 5i)(2 − 5i) = ( ).
Expand: (2)(2) + (2)(−5i) + (5i)(2) + (5i)(−5i) = 4 − 10i + 10i − 25i². The imaginary terms cancel, and −25i² = +25, leaving 4 + 25 = 29. The answer is A. Option B treats i² as +1, so it gets 4 − 25 = −21. Option C multiplies part by part, as if it were addition: 2 × 2 = 4 and 5 × (−5) = −25, giving 4 − 25i. Option D gets the sign of one cross term wrong: it adds 10i + 10i = 20i instead of −10i + 10i = 0. This is a preview of the next page: a number times its "conjugate" gives a real result.
Things to remember
Concept
What to know
The number i
i² = −1 (i = √−1). It is not a real number; it is a new building block.
Form a + bi
Real part a, imaginary part b (a real number — the coefficient of i, with its sign). b = 0 → real; a = 0 → purely imaginary.
Argand plane
Plot a + bi as the point/arrow (a, b): real part across, imaginary part up. Behaves like the vector ⟨a, b⟩.
Add / subtract
(a + bi) ± (c + di) = (a ± c) + (b ± d)i. Combine like parts; they never mix.
Multiply
Expand like FOIL, then convert i² = −1. Result: (ac − bd) + (ad + bc)i. The i² conversion is the step everyone forgets.
Powers of i
Cycle of 4: i, −1, −i, 1. For iⁿ, take n mod 4 (remainder 1→i, 2→−1, 3→−i, 0→1).
