The Scenario
Two drones are flying on a coordinate grid where positions are written as complex numbers. Drone A is at $z_1 = 3 + 2\mathrm{i}$ (3 km east, 2 km north). Drone B is at $z_2 = 1 + 4\mathrm{i}$.
To find the total displacement if both drones fly from the origin, you just add: $z_1 + z_2 = (3+1) + (2+4)\mathrm{i} = 4 + 6\mathrm{i}$. Addition is component-wise — easy, just like adding vectors.
But now the control tower sends a “rotate and scale” command: multiply the drone's position by $1 + \mathrm{i}$. A student tries to multiply component-by-component: $(3+2\mathrm{i})(1+\mathrm{i}) \stackrel{?}{=} (3 \cdot 1) + (2 \cdot 1)\mathrm{i} = 3 + 2\mathrm{i}$. But expanding properly with $\mathrm{i}^2 = -1$ gives a completely different answer: $3 + 3\mathrm{i} + 2\mathrm{i} + 2\mathrm{i}^2 = 1 + 5\mathrm{i}$. Multiplication is NOT component-wise!
The Question
We can add and subtract complex numbers component-by-component, just like vectors. But multiplication and division are different:
$\quad(3+2\mathrm{i})(1+\mathrm{i}) = 3 + 3\mathrm{i} + 2\mathrm{i} + 2\mathrm{i}^2 = 1 + 5\mathrm{i} \quad (\text{not } 3+2\mathrm{i}!)$
What are the correct algebraic rules for multiplying and dividing complex numbers in the form $x + \mathrm{i}y$? And how does the complex conjugate $z^* = x - \mathrm{i}y$ help us divide?
💡 The Bridge (click to reveal)
Addition and subtraction of complex numbers are component-wise: add real parts and imaginary parts separately.
Multiplication uses the distributive law (FOIL) with $\mathrm{i}^2 = -1$:
$$(a+b\mathrm{i})(c+d\mathrm{i}) = (ac - bd) + (ad + bc)\mathrm{i}$$
Division requires a clever trick: multiply numerator and denominator by the conjugate of the denominator. Since $z \cdot z^* = |z|^2$ is always real, this turns the denominator into a real number:
$$\frac{a+b\mathrm{i}}{c+d\mathrm{i}} = \frac{(a+b\mathrm{i})(c-d\mathrm{i})}{c^2+d^2}$$
Today we learn all four operations and the conjugate properties that make them work.
A complex number $z = x + \mathrm{i}y$ has a real part $\text{Re}(z) = x$ and an imaginary part $\text{Im}(z) = y$. Its complex conjugate is $z^* = x - \mathrm{i}y$ (reflect across the real axis). Today we learn the four arithmetic operations and the conjugate properties that power them.
Add (or subtract) the real and imaginary parts separately:
$$\boxed{\;(a + b\mathrm{i}) \pm (c + d\mathrm{i}) = (a \pm c) + (b \pm d)\mathrm{i}\;}$$Geometrically, this is vector addition on the Argand diagram — the parallelogram law.
Use the distributive law (FOIL) with $\mathrm{i}^2 = -1$:
$$\boxed{\;(a + b\mathrm{i})(c + d\mathrm{i}) = (ac - bd) + (ad + bc)\mathrm{i}\;}$$The $-bd$ term comes from $b\mathrm{i} \cdot d\mathrm{i} = bd\,\mathrm{i}^2 = -bd$. This is what makes multiplication not component-wise.
To divide, multiply numerator and denominator by the conjugate of the denominator. This makes the denominator real:
$$\boxed{\;\frac{a + b\mathrm{i}}{c + d\mathrm{i}} = \frac{(a + b\mathrm{i})(c - d\mathrm{i})}{c^2 + d^2} = \frac{(ac+bd) + (bc-ad)\mathrm{i}}{c^2+d^2}\;}$$The denominator becomes $c^2 + d^2$ because $(c+d\mathrm{i})(c-d\mathrm{i}) = c^2 - (d\mathrm{i})^2 = c^2 + d^2$ — always a positive real number (when $z \neq 0$).
For $z = x + \mathrm{i}y$ with conjugate $z^* = x - \mathrm{i}y$:
$$z \cdot z^* = x^2 + y^2 = |z|^2 \quad (\text{always real and } \geq 0)$$
$$(z^*)^* = z, \qquad (z_1 + z_2)^* = z_1^* + z_2^*, \qquad (z_1 z_2)^* = z_1^* \cdot z_2^*$$
$$\text{Re}(z) = \frac{z + z^*}{2}, \qquad \text{Im}(z) = \frac{z - z^*}{2\mathrm{i}}$$
The property $z \cdot z^* = |z|^2$ is the key that unlocks division — it turns a complex denominator into a real one.
The teacher demonstrates each operation on the board, pausing for student input at key moments. Students follow along and take notes.
Create your own mnemonic card for the complex division steps. Concise notes beat long sentences — use bullet points and a memorable phrase.
To compute $\dfrac{a+b\mathrm{i}}{c+d\mathrm{i}}$:
- Conjugate the denominator: write down $c - d\mathrm{i}$.
- Multiply top and bottom by $c - d\mathrm{i}$.
- Real denominator: the bottom becomes $c^2 + d^2$.
- Expand the numerator using FOIL (with $\mathrm{i}^2 = -1$).
- Simplify: split into $\dfrac{\text{real}}{c^2+d^2} + \dfrac{\text{imaginary}}{c^2+d^2}\mathrm{i}$.
- The five steps above (in your own words — shorter is better).
- A memorable phrase using the first letters C-M-R-E-S. Example: “Conjugate Makes Real Expressions Simple” — or invent your own!
- One worked example: $\dfrac{1}{3+\mathrm{i}}$ (answer: $\dfrac{3-\mathrm{i}}{10}$).
👁 Sample mnemonic card
Division of $(a+b\mathrm{i}) \div (c+d\mathrm{i})$:
Conjugate the bottom $\rightarrow c-d\mathrm{i}$
Multiply top & bottom by it
Real bottom $= c^2+d^2$
Expand top (FOIL, $\mathrm{i}^2=-1$)
Split into $x + y\mathrm{i}$
Phrase: “Cats Make Real Eggs Scrambled”
Example: $\dfrac{1}{3+\mathrm{i}} = \dfrac{1 \cdot (3-\mathrm{i})}{(3+\mathrm{i})(3-\mathrm{i})} = \dfrac{3-\mathrm{i}}{9+1} = \dfrac{3}{10} - \dfrac{1}{10}\mathrm{i}$
Express these complex numbers in the form $x + \mathrm{i}y$.
(i) $\dfrac{1}{3+\mathrm{i}}$ (ii) $\dfrac{1}{6-\mathrm{i}}$ (iii) $\dfrac{5\mathrm{i}}{6-2\mathrm{i}}$ (iv) $\dfrac{7+5\mathrm{i}}{6-2\mathrm{i}}$ (v) $\dfrac{3+2\mathrm{i}}{1+\mathrm{i}}$
(vi) $\dfrac{47-23\mathrm{i}}{6+\mathrm{i}}$ (vii) $\dfrac{2-3\mathrm{i}}{3+2\mathrm{i}}$ (viii) $\dfrac{5-3\mathrm{i}}{4+3\mathrm{i}}$ (ix) $\dfrac{6+\mathrm{i}}{2-5\mathrm{i}}$ (x) $\dfrac{12-8\mathrm{i}}{(2+2\mathrm{i})^2}$
👁 Reveal Solution
Multiply by $\dfrac{3-\mathrm{i}}{3-\mathrm{i}}$:
$\dfrac{1}{3+\mathrm{i}} = \dfrac{1 \cdot (3-\mathrm{i})}{(3+\mathrm{i})(3-\mathrm{i})} = \dfrac{3-\mathrm{i}}{9+1} = \dfrac{3-\mathrm{i}}{10}$
$\boxed{\dfrac{3}{10} - \dfrac{1}{10}\mathrm{i}}$
Multiply by $\dfrac{1-\mathrm{i}}{1-\mathrm{i}}$:
$\dfrac{(3+2\mathrm{i})(1-\mathrm{i})}{(1+\mathrm{i})(1-\mathrm{i})} = \dfrac{3 - 3\mathrm{i} + 2\mathrm{i} - 2\mathrm{i}^2}{1+1} = \dfrac{3 - \mathrm{i} + 2}{2} = \dfrac{5-\mathrm{i}}{2}$
$\boxed{\dfrac{5}{2} - \dfrac{1}{2}\mathrm{i}}$
Multiply by $\dfrac{3-2\mathrm{i}}{3-2\mathrm{i}}$:
$\dfrac{(2-3\mathrm{i})(3-2\mathrm{i})}{(3+2\mathrm{i})(3-2\mathrm{i})} = \dfrac{6 - 4\mathrm{i} - 9\mathrm{i} + 6\mathrm{i}^2}{9+4} = \dfrac{6 - 13\mathrm{i} - 6}{13} = \dfrac{-13\mathrm{i}}{13}$
$\boxed{-\mathrm{i}}$
First expand: $(2+2\mathrm{i})^2 = 4 + 8\mathrm{i} + 4\mathrm{i}^2 = 4 + 8\mathrm{i} - 4 = 8\mathrm{i}$.
Now divide: $\dfrac{12-8\mathrm{i}}{8\mathrm{i}}$. Multiply by $\dfrac{-8\mathrm{i}}{-8\mathrm{i}}$ (the conjugate of $8\mathrm{i}$ is $-8\mathrm{i}$):
$\dfrac{(12-8\mathrm{i})(-8\mathrm{i})}{(8\mathrm{i})(-8\mathrm{i})} = \dfrac{-96\mathrm{i} + 64\mathrm{i}^2}{-64\mathrm{i}^2} = \dfrac{-96\mathrm{i} - 64}{64} = \dfrac{-64 - 96\mathrm{i}}{64}$
$\boxed{-1 - \dfrac{3}{2}\mathrm{i}}$
Find real numbers $a$ and $b$ with $a > 0$ such that
(i) $(a + \mathrm{i}b)^2 = 21 + 20\mathrm{i}$ (ii) $(a + \mathrm{i}b)^2 = -40 - 42\mathrm{i}$ (iii) $(a + \mathrm{i}b)^2 = -5 - 12\mathrm{i}$
(iv) $(a + \mathrm{i}b)^2 = -9 + 40\mathrm{i}$ (v) $(a + \mathrm{i}b)^2 = 1 - 1.875\mathrm{i}$ (vi) $(a + \mathrm{i}b)^2 = \mathrm{i}$
👁 Reveal Solution
Expand: $a^2 - b^2 + 2ab\mathrm{i} = 21 + 20\mathrm{i}$.
Equate: $a^2 - b^2 = 21$ and $2ab = 20$, so $ab = 10$, giving $b = \dfrac{10}{a}$.
Substitute: $a^2 - \dfrac{100}{a^2} = 21 \Rightarrow a^4 - 21a^2 - 100 = 0 \Rightarrow (a^2 - 25)(a^2 + 4) = 0$.
So $a^2 = 25$ (since $a^2 + 4 > 0$), giving $a = 5$ (as $a > 0$) and $b = 2$.
$\boxed{a = 5, \; b = 2}$ (Check: $(5+2\mathrm{i})^2 = 25 + 20\mathrm{i} + 4\mathrm{i}^2 = 21 + 20\mathrm{i}$ ✓)
Equate: $a^2 - b^2 = -5$ and $2ab = -12$, so $ab = -6$, giving $b = \dfrac{-6}{a}$.
Substitute: $a^2 - \dfrac{36}{a^2} = -5 \Rightarrow a^4 + 5a^2 - 36 = 0 \Rightarrow (a^2 + 9)(a^2 - 4) = 0$.
So $a^2 = 4$, giving $a = 2$ (as $a > 0$) and $b = -3$.
$\boxed{a = 2, \; b = -3}$ (Check: $(2-3\mathrm{i})^2 = 4 - 12\mathrm{i} + 9\mathrm{i}^2 = -5 - 12\mathrm{i}$ ✓)
Equate: $a^2 - b^2 = 0$ and $2ab = 1$, so $ab = \dfrac{1}{2}$.
From $a^2 = b^2$: $b = \pm a$. Since $ab = \dfrac{1}{2} > 0$ and $a > 0$, we need $b > 0$, so $b = a$.
Then $2a^2 = 1$, giving $a = \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}$ and $b = \dfrac{\sqrt{2}}{2}$.
$\boxed{a = \dfrac{\sqrt{2}}{2}, \; b = \dfrac{\sqrt{2}}{2}}$ (Check: $\left(\dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2}\mathrm{i}\right)^2 = \dfrac{1}{2} + \mathrm{i} + \dfrac{1}{2}\mathrm{i}^2 = \mathrm{i}$ ✓)
Attempt each question on paper first, then click Reveal Solution to check your work. Questions are ordered from easiest to hardest.
Find real numbers $a$ and $b$ such that $\dfrac{a}{3+\mathrm{i}}+\dfrac{b}{1+2\mathrm{i}} = 1-\mathrm{i}$
💡 Hint
Rationalise each fraction first: $\dfrac{a}{3+\mathrm{i}} = \dfrac{a(3-\mathrm{i})}{10}$ and $\dfrac{b}{1+2\mathrm{i}} = \dfrac{b(1-2\mathrm{i})}{5}$.
Combine into a single fraction, then equate real and imaginary parts with $1 - \mathrm{i}$.
👁 Reveal Solution
$\dfrac{a}{3+\mathrm{i}} = \dfrac{a(3-\mathrm{i})}{(3+\mathrm{i})(3-\mathrm{i})} = \dfrac{a(3-\mathrm{i})}{10} = \dfrac{3a}{10} - \dfrac{a}{10}\mathrm{i}$
$\dfrac{b}{1+2\mathrm{i}} = \dfrac{b(1-2\mathrm{i})}{(1+2\mathrm{i})(1-2\mathrm{i})} = \dfrac{b(1-2\mathrm{i})}{5} = \dfrac{b}{5} - \dfrac{2b}{5}\mathrm{i}$
Combining over a common denominator of 10:
$\left(\dfrac{3a}{10} + \dfrac{2b}{10}\right) + \left(-\dfrac{a}{10} - \dfrac{4b}{10}\right)\mathrm{i} = 1 - \mathrm{i}$
Equating real parts: $\dfrac{3a + 2b}{10} = 1 \Rightarrow 3a + 2b = 10$
Equating imaginary parts: $\dfrac{-a - 4b}{10} = -1 \Rightarrow a + 4b = 10$
From the second equation: $a = 10 - 4b$.
Substitute into the first: $3(10 - 4b) + 2b = 10 \Rightarrow 30 - 12b + 2b = 10 \Rightarrow -10b = -20 \Rightarrow b = 2$.
Then $a = 10 - 4(2) = 2$.
$\boxed{a = 2, \; b = 2}$
Verification: $\dfrac{2}{3+\mathrm{i}} + \dfrac{2}{1+2\mathrm{i}} = \dfrac{2(3-\mathrm{i})}{10} + \dfrac{2(1-2\mathrm{i})}{5} = \dfrac{6-2\mathrm{i}}{10} + \dfrac{2-4\mathrm{i}}{5} = \dfrac{6-2\mathrm{i}+4-8\mathrm{i}}{10} = \dfrac{10-10\mathrm{i}}{10} = 1 - \mathrm{i}$ ✓
Solve these equations.
(i) $(1 + \mathrm{i})z = 3 + \mathrm{i}$ (ii) $(3 - 4\mathrm{i})(z - 1) = 10 - 5\mathrm{i}$ (iii) $(2 + \mathrm{i})(z - 7 + 3\mathrm{i}) = 15 - 10\mathrm{i}$ (iv) $(3 + 5\mathrm{i})(z + 2 - 5\mathrm{i}) = 6 + 3\mathrm{i}$
💡 Hint
For each part, isolate $z$ by dividing both sides by the complex coefficient.
For (i): $z = \dfrac{3+\mathrm{i}}{1+\mathrm{i}}$. Multiply by $\dfrac{1-\mathrm{i}}{1-\mathrm{i}}$.
For (ii)–(iv): first isolate the bracket, then divide.
👁 Reveal Solution
$z = \dfrac{3+\mathrm{i}}{1+\mathrm{i}} = \dfrac{(3+\mathrm{i})(1-\mathrm{i})}{(1+\mathrm{i})(1-\mathrm{i})} = \dfrac{3 - 3\mathrm{i} + \mathrm{i} - \mathrm{i}^2}{1+1} = \dfrac{3 - 2\mathrm{i} + 1}{2} = \dfrac{4 - 2\mathrm{i}}{2}$
$\boxed{z = 2 - \mathrm{i}}$
$z - 1 = \dfrac{10-5\mathrm{i}}{3-4\mathrm{i}} = \dfrac{(10-5\mathrm{i})(3+4\mathrm{i})}{(3-4\mathrm{i})(3+4\mathrm{i})} = \dfrac{30 + 40\mathrm{i} - 15\mathrm{i} - 20\mathrm{i}^2}{9+16} = \dfrac{30 + 25\mathrm{i} + 20}{25} = \dfrac{50 + 25\mathrm{i}}{25} = 2 + \mathrm{i}$
$z = 1 + 2 + \mathrm{i} = 3 + \mathrm{i}$
$\boxed{z = 3 + \mathrm{i}}$
$z - 7 + 3\mathrm{i} = \dfrac{15-10\mathrm{i}}{2+\mathrm{i}} = \dfrac{(15-10\mathrm{i})(2-\mathrm{i})}{(2+\mathrm{i})(2-\mathrm{i})} = \dfrac{30 - 15\mathrm{i} - 20\mathrm{i} + 10\mathrm{i}^2}{4+1} = \dfrac{30 - 35\mathrm{i} - 10}{5} = \dfrac{20 - 35\mathrm{i}}{5} = 4 - 7\mathrm{i}$
$z = 7 - 3\mathrm{i} + 4 - 7\mathrm{i} = 11 - 10\mathrm{i}$
$\boxed{z = 11 - 10\mathrm{i}}$
$z + 2 - 5\mathrm{i} = \dfrac{6+3\mathrm{i}}{3+5\mathrm{i}} = \dfrac{(6+3\mathrm{i})(3-5\mathrm{i})}{(3+5\mathrm{i})(3-5\mathrm{i})} = \dfrac{18 - 30\mathrm{i} + 9\mathrm{i} - 15\mathrm{i}^2}{9+25} = \dfrac{18 - 21\mathrm{i} + 15}{34} = \dfrac{33 - 21\mathrm{i}}{34}$
$z = -2 + 5\mathrm{i} + \dfrac{33 - 21\mathrm{i}}{34} = \dfrac{-68 + 33}{34} + \dfrac{170 - 21}{34}\mathrm{i} = \dfrac{-35}{34} + \dfrac{149}{34}\mathrm{i}$
$\boxed{z = -\dfrac{35}{34} + \dfrac{149}{34}\mathrm{i}}$
Find all the complex numbers $z$ for which $z^2 = 2z^*$
💡 Hint
Let $z = x + \mathrm{i}y$. Then $z^2 = (x^2 - y^2) + 2xy\mathrm{i}$ and $z^* = x - \mathrm{i}y$, so $2z^* = 2x - 2y\mathrm{i}$.
Equate real and imaginary parts to get two equations in $x$ and $y$.
From the imaginary part: $2xy = -2y$, so $2y(x+1) = 0$. This gives two cases: $y = 0$ or $x = -1$.
👁 Reveal Solution
Let $z = x + \mathrm{i}y$. Then $z^2 = (x^2 - y^2) + 2xy\mathrm{i}$ and $2z^* = 2x - 2y\mathrm{i}$.
Equating real parts: $x^2 - y^2 = 2x$ (1)
Equating imaginary parts: $2xy = -2y$ (2)
$2xy + 2y = 0 \Rightarrow 2y(x + 1) = 0$
So either $y = 0$ or $x = -1$.
Substituting into (1): $x^2 = 2x \Rightarrow x(x-2) = 0 \Rightarrow x = 0$ or $x = 2$.
This gives $z = 0$ or $z = 2$.
Substituting into (1): $1 - y^2 = -2 \Rightarrow y^2 = 3 \Rightarrow y = \pm\sqrt{3}$.
This gives $z = -1 + \sqrt{3}\mathrm{i}$ or $z = -1 - \sqrt{3}\mathrm{i}$.
$\boxed{z = 0, \quad z = 2, \quad z = -1 + \sqrt{3}\,\mathrm{i}, \quad z = -1 - \sqrt{3}\,\mathrm{i}}$
Verification for $z = -1 + \sqrt{3}\mathrm{i}$: $z^2 = 1 - 2\sqrt{3}\mathrm{i} + 3\mathrm{i}^2 = 1 - 3 - 2\sqrt{3}\mathrm{i} = -2 - 2\sqrt{3}\mathrm{i}$. And $2z^* = 2(-1 - \sqrt{3}\mathrm{i}) = -2 - 2\sqrt{3}\mathrm{i}$ ✓
For $z = x + \mathrm{i}y$, find $\dfrac{1}{z} + \dfrac{1}{z^*}$ in terms of $x$ and $y$.
💡 Hint 1
Start by writing $\dfrac{1}{z}$ and $\dfrac{1}{z^*}$ separately.
Recall: $\dfrac{1}{z} = \dfrac{z^*}{z \cdot z^*} = \dfrac{z^*}{|z|^2}$ and $\dfrac{1}{z^*} = \dfrac{z}{z \cdot z^*} = \dfrac{z}{|z|^2}$.
💡 Hint 2
Adding: $\dfrac{1}{z} + \dfrac{1}{z^*} = \dfrac{z^* + z}{|z|^2}$.
Now substitute $z = x + \mathrm{i}y$, $z^* = x - \mathrm{i}y$, and $|z|^2 = x^2 + y^2$.
What is $z + z^*$? (The imaginary parts cancel!)
👁 Reveal Solution
$\dfrac{1}{z} = \dfrac{1}{x + \mathrm{i}y} = \dfrac{x - \mathrm{i}y}{(x+\mathrm{i}y)(x-\mathrm{i}y)} = \dfrac{x - \mathrm{i}y}{x^2 + y^2}$
$\dfrac{1}{z^*} = \dfrac{1}{x - \mathrm{i}y} = \dfrac{x + \mathrm{i}y}{(x-\mathrm{i}y)(x+\mathrm{i}y)} = \dfrac{x + \mathrm{i}y}{x^2 + y^2}$
Since both fractions have the same denominator $x^2 + y^2$:
$\dfrac{1}{z} + \dfrac{1}{z^*} = \dfrac{(x - \mathrm{i}y) + (x + \mathrm{i}y)}{x^2 + y^2} = \dfrac{2x}{x^2 + y^2}$
The imaginary parts $-\mathrm{i}y$ and $+\mathrm{i}y$ cancel, leaving a purely real result.
$\boxed{\dfrac{1}{z} + \dfrac{1}{z^*} = \dfrac{2x}{x^2 + y^2}}$
Note: this equals $\dfrac{2\,\text{Re}(z)}{|z|^2}$, which is always real. The result is real because $\dfrac{1}{z^*} = \left(\dfrac{1}{z}\right)^*$, so the sum of a complex number and its conjugate is always real: $w + w^* = 2\,\text{Re}(w)$.
Exit Ticket
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