HL · Complex numbers 1 · The number i and Cartesian form

Module 11 of 47Lesson 1 of 8 · HL · Complex numbers 1 · The number i and Cartesian form

Complex numbers · Essential understanding 1

The number i and Cartesian form

IB AI AHL 1.12with Prof. Kitso

💻 Using the TI-Nspire CX Student Software on a laptop

The screens in these notes are drawn exactly as the Student Software shows them. The keypad on the left of the software works like the handheld — click its keys with the mouse.
menu: click menu on the on-screen keypad (or Documents Toolbox ▸ 🔧). The menu numbers and letters are the same as in these notes.
enter = Enter · tab = Tab (next box) · esc = Esc · del = Backspace.
Powers: type ^ (Shift+6), the power, then press → to come back down before typing more. x² = ^2 then →.
Fraction: type / · √: type sqrt( · π: type pi · eˣ: type exp( · store: type :=. For EE (×10ⁿ) and i, click them on the keypad.
New page: Ctrl+I (or the Insert menu) · change page: Ctrl+← / Ctrl+→ · graph entry line: Ctrl+G · var: Ctrl+L.
Radian / Degree: click Settings in the status bar at the bottom of the window. The mode (RAD or DEG) shows in the top bar.
IB exam mode (Press-to-Test): everything in these notes works on the non-CAS TI-Nspire with the IB exam settings. In exam mode there are no dot-product, cross-product, unit-vector or eigenvalue keys and no Diff Eq graphs — the notes show the allowed way (type the components, use polyRoots on the characteristic equation, do Euler steps on a Calculator page).
If you type nDeriv(, the software shows it as centralDiff( — it is the same command.

The big idea: Some equations, like x² = −1, have no real answer. So we invent a new number, i, with i² = −1. A complex number is a real part plus a multiple of i, written z = a + bi, and you add, subtract, multiply and divide it with ordinary algebra.

By the end of this lesson you can:

use i² = −1 to simplify powers of i and square roots of negative numbers
write a complex number as z = a + bi and name its real and imaginary parts
add, subtract and multiply complex numbers
find the complex conjugate z* and use zz* = a² + b²
divide complex numbers by multiplying top and bottom by the conjugate
solve for unknowns by equating real parts and imaginary parts
do all of this on the TI-Nspire CX II

🧑🏾 Imagine this…

Long ago people said “you cannot take 5 from 3”, because there were no negative numbers. Then someone invented the number line to the left of zero, and suddenly 3 − 5 = −2 made sense. Today we do the same trick again. The equation x² = −1 has no answer on the number line, because a real number squared is never negative. So we say: “let there be a number called i whose square is −1”. It does not sit on the number line; it sits off it, at a right angle. Every complex number needs two measurements, like a map reference: how far along the real line, and how far in the i direction. Engineers use these numbers every day for electricity (Lesson 7).

The number i

the number with

i^2 = -1

. We write

i = \sqrt{-1}

Complex number

a number

z = a + bi

where

a

and

b

are real numbers. The set of all of them is written

\mathbb{C}

Real part

\text{Re}(z) = a

. For

z = 3 - 2i

\text{Re}(z) = 3

Imaginary part

\text{Im}(z) = b

(the number in front of i, not including the i). For

z = 3 - 2i

\text{Im}(z) = -2

Complex conjugate

z^* = a - bi

: same real part, opposite sign on the imaginary part.

Cartesian form

writing

z = a + bi

(the name comes from the x–y “Cartesian” coordinates of Lesson 3).

★ The rules you need

$i^2 = -1$ , so $i^3 = -i$ , $i^4 = 1$ , and then the pattern repeats every 4 powers: $i, -1, -i, 1, i, -1, \ldots$
For a positive number $k$ : $\sqrt{-k} = i\sqrt{k}$ . For example $\sqrt{-49} = 7i$ .
Add or subtract: do the real parts and the imaginary parts separately. $(a+bi) + (c+di) = (a+c) + (b+d)i$ .
Multiply: expand brackets like normal algebra, then replace $i^2$ by $-1$ .
Conjugate product: $zz^* = a^2 + b^2$ , which is always a real number.
Divide: $\dfrac{z}{w} = \dfrac{z \, w^*}{w \, w^*}$ . This is “rationalising the denominator”, just like clearing a surd.
Equal complex numbers: $a + bi = c + di$ means $a = c$ and $b = d$ .

⚠️ Watch out

$\text{Im}(3 - 2i)$ is −2, not $-2i$ . The imaginary part is a real number.
Do not forget $i^2 = -1$ when you multiply. The product $(2i)(3i)$ is $-6$ , not $6i^2$ and not $+6$ .
When you divide, multiply the top and the bottom by the conjugate of the bottom. Changing only the denominator changes the answer.
Keep your calculator in Rectangular complex mode (see the first GDC step), otherwise it may refuse to work with i.

💡 Command terms

Write down: no working needed (for example, the conjugate). Find or simplify: show the expansion and the step with

i^2 = -1

. Express in the form a + bi means the final line must have the real part and the imaginary part clearly separated. Show that: every line must be visible, no calculator-only jumps.

🧪 Try it yourselfAdd, multiply and divide

z = 3 + 2i (blue) and w = 1 − 4i (orange) are the numbers from Example 1.2. Press z + w, z × w, z ÷ w and z* and compare with the working. Then drag the points and predict each result before you look.

Worked examples

Example 1.1Powers of i and square roots of negatives

Simplify (a) i⁵ (b) i²² (c) (4i)(5i) (d) √−49 (e) i²⁷

🧠 Think it through

Use $i^2 = -1$ . The powers of i go round a cycle of 4: $i, -1, -i, 1$ .
For a high power, divide the exponent by 4 and look only at the remainder: remainder 1 → i, 2 → −1, 3 → −i, 0 → 1.
For (c), multiply the numbers and the i’s separately: $20 \times i^2$ .
For (d), $\sqrt{-49} = \sqrt{49} \times \sqrt{-1}$ .

✍️ By hand
(a) 
(b) , so 
(c) 
(d) 
(e) , so 

🔢 On your TI-Nspire CX II

First set the calculator up. Open the document settings (home screen: doc ▾💻 File menu ▸ Settings ▸ Document Settings) and set Real or Complex to Rectangular and Angle to Radian. Make these the default for new documents if you can.
To type i, open the symbol palette with ctrl💻 Ctrl catalog💻 click 📖 on keypad and choose i. Do not type the letter i from the keyboard; it will be read as a variable.
Type the expressions and press enter ↵💻 Enter. For a high power it is quicker to let the GDC do the cycle for you.

✅ Answer (a) i (b) −1 (c) −20 (d) 7i (e) −i

Trick for (b) and (e): only the remainder when you divide the power by 4 matters. A power that is a multiple of 4 always gives 1.

Example 1.2Add, subtract and multiply

Let z = 3 + 2i and w = 1 − 4i. Find (a) z + w (b) z − w (c) zw (d) 2z − 3w. Then write down Re(zw) and Im(zw).

🧠 Think it through

Add and subtract the real parts together and the imaginary parts together.
For zw, expand all four terms of the double bracket (first × first, first × second, and so on).
The two i-terms multiply to $2i \times (-4i) = -8i^2 = +8$ .
For (d), multiply each number first, then subtract.

✍️ By hand
(a) 
(b) 
(c) 
(d) 

🔢 On your TI-Nspire CX II

Type the numbers in brackets. The GDC returns the answer in the form a + b·i.
To get a real or imaginary part use the Complex Number Tools menu (menu ▸ Number ▸ Complex Number Tools): real( and imag(.

✅ Answer (a) 4 − 2i (b) 2 + 6i (c) 11 − 10i (d) 3 + 16i. Re(zw) = 11 and Im(zw) = −10.

You can check this in the lab above: put z at (3, 2) and w at (1, −4) and press z × w.

Example 1.3The conjugate and zz*

Let z = 5 − 12i. (a) Write down z*. (b) Find zz*. (c) Find z + z*. (d) Show that z + z* is twice Re(z) and that z − z* is twice Im(z) times i.

🧠 Think it through

The conjugate flips the sign of the imaginary part only.
$zz^* = (a+bi)(a-bi) = a^2 - b^2 i^2 = a^2 + b^2$ . The middle terms cancel, so the answer is real.
For z + z* the imaginary parts cancel; for z − z* the real parts cancel.

✍️ By hand
(a) 
(b) 
(c) 
(d) 

🔢 On your TI-Nspire CX II

Use conj( from the Complex Number Tools menu (menu ▸ Number ▸ Complex Number Tools ▸ Complex Conjugate).

✅ Answer (a) 5 + 12i (b) 169 (c) 10 (d) z + z* = 10 = 2 × 5 and z − z* = −24i = 2 × (−12) × i.

Notice 169 = 13². The number 13 is the modulus of z, which you meet in Lesson 3.

Example 1.4Dividing complex numbers

Write

\dfrac{6 + 5i}{2 + i}

in the form a + bi.

🧠 Think it through

We cannot leave i in the denominator. Make the denominator real by multiplying top and bottom by its conjugate, 2 − i.
The new denominator is $(2+i)(2-i) = 2^2 + 1^2 = 5$ , so it is real.
Expand the top carefully and use i² = −1.
Split the fraction into a real part and an imaginary part.

✍️ By hand

\dfrac{6+5i}{2+i} = \dfrac{(6+5i)(2-i)}{(2+i)(2-i)}

Top:

12 - 6i + 10i - 5i^2 = 12 + 4i + 5 = 17 + 4i

Bottom:

4 + 1 = 5

\dfrac{17+4i}{5} = \dfrac{17}{5} + \dfrac{4}{5}i = 3.4 + 0.8i

🔢 On your TI-Nspire CX II

Put the whole fraction in brackets. The GDC gives decimals; the exact fractions are $\dfrac{17}{5}$ and $\dfrac{4}{5}$ .

✅ Answer 3.4 + 0.8i (that is,

\dfrac{17}{5}

\dfrac{4}{5}

i).

Check by multiplying back: (3.4 + 0.8i)(2 + i) = 6.8 + 3.4i + 1.6i − 0.8 = 6 + 5i. ✓

Example 1.5Equating real and imaginary parts

(a) Find the complex number z such that 2z − z* = 3 + 9i. (b) Find the real numbers p and q such that (p + 2i)(3 − i) = q + 7i.

🧠 Think it through

Write $z = x + iy$ with $x$ and $y$ real, so $z^* = x - iy$ .
Two complex numbers are equal only if their real parts are equal and their imaginary parts are equal. That gives two ordinary equations.
For (b), expand the left side first, group the real and imaginary terms, then equate.

✍️ By hand
(a) 
, so 
(b) 
Real parts: . Imaginary parts: .
So  and then .

🔢 On your TI-Nspire CX II

Both parts are quick by hand. Use the calculator only to check your answer by substituting it back.
Check (b): put p = −1 into the left side. The result should be q + 7i with q = −1.

✅ Answer (a) z = 3 + 3i. (b) p = −1, q = −1.

Part (a) is the standard pattern: put z = x + iy, collect the real and imaginary parts, and you get two real equations. In (b) one of the equations only contained p, so it was quick.

Challenge questions

⭐ Challenge 1.6A square and a reciprocalPaper 1 style

Let z = 1 + 2i. Find (a) z² (b)

\dfrac{1}{z}

, giving each in the form a + bi.

✏️ Try it on paper and on your calculator first. Then check.

Show the full solution

🧠 Think it through

(a) Square the bracket: $(1+2i)^2 = 1 + 4i + 4i^2$ .
(b) Write 1 over z and multiply top and bottom by the conjugate 1 − 2i.

✍️ By hand

(a)

z^2 = 1 + 4i - 4 = -3 + 4i

(b)

\dfrac{1}{1+2i} = \dfrac{1-2i}{(1+2i)(1-2i)} = \dfrac{1-2i}{5}

= 0.2 - 0.4i

🔢 On your TI-Nspire CX II

Check both on the calculator.

✅ Answer (a) −3 + 4i (b) 0.2 − 0.4i (that is,

\dfrac{1}{5}

−

\dfrac{2}{5}

i).

⭐ Challenge 1.7Two complex numbersPaper 2 style

Let z = 3 + 4i and w = 1 − 2i.

(a) Find zw.
(b) Find $\dfrac{z}{w}$ in the form a + bi.
(c) Find the real number k such that z + kw is purely imaginary, and state the value of z + kw.
(d) Find the real numbers a and b such that az + bw = 8 + 6i.

✏️ Try it on paper and on your calculator first. Then check.

Show the full solution

🧠 Think it through

Expand zw with all four terms.
Multiply $\dfrac{z}{w}$ top and bottom by the conjugate of w, which is 1 + 2i.
“Purely imaginary” means the real part is zero. Write z + kw and set its real part to 0.
Expand az + bw, then equate real parts and imaginary parts to get two simultaneous equations.

✍️ By hand
(a) 
(b) 
(c) 
. Real part zero: 
, so 
(d) 
 and 
. Then , so 

🔢 On your TI-Nspire CX II

Check (a) and (b) on a Calculator page.
For (d) solve the two real equations with linSolve.

✅ Answer (a) 11 − 2i (b) −1 + 2i (c) k = −3 and z + kw = 10i (d) a = 2.2, b = 1.4.

Next: Lesson 2 · Complex roots of equations

IB DP Mathematics: Applications and Interpretation (AI) — SL and HL HL · Complex numbers 1 · The number i and Cartesian form

IB DP Mathematics: Applications and Interpretation (AI) — SL and HL

The number i and Cartesian form

Worked examples

Challenge questions