ATHENS ACADEMY · Mathematics · Grade 11 · Unit 1

ATHENS ACADEMY · Mathematics · Grade 11 · Unit 1 · Complex Numbers

Unit 1 — Complex Numbers

This unit is worth a reliable block of marks on the March 2027 paper, and almost every question in it is a method question — the examiner wants a fixed sequence of steps, not an insight. This note is built around those sequences. Part A covers the algebra; Part B covers polar form and the circuit application. Each section shows the method, the marks it typically carries, and the specific mistakes that lose them. Work the six examples with a pen, not by reading.

Part A Algebra of Complex Numbers The bread-and-butter half of this unit. Almost every question here is a method question — the examiner wants to see a fixed sequence of steps, not an insight. Learn the sequences and these marks are yours.

1. The Basics You Are Assumed to Know

Everything in this unit rests on one fact. Memorise it before anything else:

i=\sqrt{-1},\qquad\text{so}\qquad i^{2}=-1

Definition — write it exactly like this A complex number is

z=x+iy

, where

x

and

y

are real numbers.

x

is the real part,

Re(z)=x

y

is the imaginary part,

Im(z)=y

Definitions like this appear as one- and two-mark openers in Section B. Write the symbolic form and name the parts — the mark scheme wants both.

Guaranteed mark-loser

Im(7-11i)

-11

, not

-11i

. The imaginary part is a real number. Every year students lose this mark after doing the hard work correctly above it.

Two things carry this entire unit One fact:

i^{2}=-1

. One technique: multiply by the conjugate. If you can apply those two reliably, you can attempt every question in Part A. Nothing else here is difficult — it is bookkeeping.

2. Equality — The Marks Machine 4–6 marks

This is the most reliably examined idea in Part A, because it converts a hard-looking complex equation into two easy real ones.

x_{1}+iy_{1}=x_{2}+iy_{2}\quad\Longleftrightarrow\quad x_{1}=x_{2}\ \text{ and }\ y_{1}=y_{2}

Method — “find the real numbers

x

and

y

”

Clear any $i$ from a denominator first, using the conjugate.
Multiply everything out.
Group all real terms in one bracket, all imaginary terms in another.
Set real $=$ real, imaginary $=$ imaginary. You now have two ordinary equations.
Solve them simultaneously.

How the examiner phrases it “Find the values of the real numbers

x

and

y

…” — the words real numbers are your instruction to split into two equations. Whenever you see them, go straight to the method above.

3. The Four Operations 2–4 marks

3.1 Add and subtract

Real with real, imaginary with imaginary.

(x_{1}+iy_{1})\pm(x_{2}+iy_{2})=(x_{1}\pm x_{2})+i(y_{1}\pm y_{2})

$(2+3i)+(6+8i)=8+11i$ and $(4-3i)-(7+6i)=-3-9i$ .

Bracket trap Subtracting

-3i

adds

+3i

. When a minus sits in front of a bracket, change the sign of every term inside. Write the bracket-free line out in full before collecting — four seconds, saves the question.

3.2 Multiply

Expand as normal brackets, then replace $i^{2}$ with $-1$ . Do not memorise a formula for this — expanding is faster and safer under pressure.

$(2-6i)(4+9i)=8+18i-24i-54i^{2}=8-6i+54=62-6i$ .

The single most common error in the unit

-54i^{2}=+54

, not

-54

. The sign flips. Before writing any final answer, scan your working for a surviving

i^{2}

and kill it.

3.3 Powers of $i$ 2 marks

i^{1}=i,\qquad i^{2}=-1,\qquad i^{3}=-i,\qquad i^{4}=1

Speed method — do this in your head Divide the exponent by

4

and keep only the remainder:
remainder

1\rightarrow i

· remainder

2\rightarrow -1

· remainder

3\rightarrow -i

· remainder

0\rightarrow 1

.
So

i^{31}

31\div4

leaves

3

, answer

-i

. And

i^{58}

: leaves

2

, answer

-1

. A whole exercise takes under a minute.

3.4 Conjugate

Definition — conjugate The conjugate of

z=x+iy

\bar{z}=x-iy

. Change the sign of the imaginary part only; leave the real part alone.

Its usefulness is that it kills the $i$ completely:

z\,\bar{z}=(x+iy)(x-iy)=x^{2}+y^{2}

The answer is always real and never negative. That is why division works. Two more results that appear as short questions:

\overline{z_{1}z_{2}}=\bar{z_{1}}\,\bar{z_{2}},\qquad\qquad \overline{\left(\frac{z_{1}}{z_{2}}\right)}=\frac{\bar{z_{1}}}{\bar{z_{2}}}

3.5 Divide — the technique everything depends on 4 marks

Method — division, every single time

Write down the conjugate of the denominator.
Multiply top and bottom by it.
Denominator: use $(a+b)(a-b)=a^{2}-b^{2}$ , then $i^{2}=-1$ . It becomes a real number.
Numerator: expand in full, collect terms.
Split into $x+iy$ and reduce the fractions.

Fatal error Multiplying by the denominator itself rather than its conjugate. For

2-5i

you need

2+5i

. Using

2-5i

again does not clear the

i

and the whole question is lost.

Worked Example 1 — Division 4 marks

Statement

Express

\dfrac{3+7i}{-4+6i}

in the form

x+iy

Working

Conjugate of

-4+6i

-4-6i

. Multiply top and bottom:

\frac{3+7i}{-4+6i}\times\frac{-4-6i}{-4-6i}=\frac{(3+7i)(-4-6i)}{(-4)^{2}-(6i)^{2}}

Numerator:

-12-18i-28i-42i^{2}

. Since

-42i^{2}=+42

, this is

30-46i

.
Denominator:

16-36i^{2}=16+36=52

=\frac{30-46i}{52}=\frac{30}{52}-\frac{46}{52}i

Divide both by

2

Answer

\dfrac{15}{26}-\dfrac{23}{26}i

Mark scheme

1 mark for the correct conjugate · 1 for the real denominator

52

· 1 for the expanded numerator · 1 for the reduced

x+iy

form. Reduce your fractions — an unreduced answer can cost the final mark.

4. Modulus 2–3 marks

Definition — modulus For

z=x+iy

\;|z|=\sqrt{x^{2}+y^{2}}

, a real number. It is the distance from the origin to the point

(x,\,y)

A complex number $x+iy$ is plotted as the point $(x,\,y)$ — real part across, imaginary part up. This picture is called the Argand plane, and you will need it again in Part B to identify quadrants.

Speed method — saves several minutes For a product or quotient, take the modulus of each piece separately:

|z_{1}z_{2}|=|z_{1}||z_{2}|,\qquad \left|\frac{z_{1}}{z_{2}}\right|=\frac{|z_{1}|}{|z_{2}|}

\left|\dfrac{(3+2i)(1-2i)}{4+3i}\right|=\dfrac{\sqrt{13}\cdot\sqrt{5}}{5}=\dfrac{\sqrt{65}}{5}

. One line — never expand the brackets.

Also examinable: $|z|=|-z|=|\bar{z}|$ , and $|z|\ge0$ always, with $|z|=0$ only when $z=0$ .

Two errors, both common

|3+4i|\ne 3+4

. It is

\sqrt{9+16}=5

. And there is never an

i

inside the square root — you square the real number

y

, not

iy

5. Real and Imaginary Parts of Powers

Your textbook lists eight formulas for this. Do not use them in the exam. They look alike, and one wrong subscript destroys the answer.

Use this instead — simplify first, read off second

Simplify the whole expression down to a single $x+iy$ . If it is a fraction, divide it out with the conjugate.
Apply the power by ordinary expansion.
Read $Re$ and $Im$ straight off the result.

Slightly longer, but it uses only skills you have already drilled and cannot fail through misremembering. The examiner marks the answer, not the route.

Example: for $\left(\dfrac{12+5i}{3+2i}\right)^{2}$ , divide first to get $\dfrac{46}{13}-\dfrac{9}{13}i$ , then square, then read off the parts.

Conceptual trap the examiner likes

Re(z_{1})\,Re(z_{2})\neq Re(z_{1}z_{2})

. You cannot take real parts first and multiply afterwards. Multiply the complex numbers properly, then take the real part.

Worked Example 2 — Find the real numbers

x

and

y

5 marks

Statement

Find the real numbers

x

and

y

\dfrac{x}{2+3i}-y(1+2i)=1+i

Working

Clear the

i

from the denominator using the conjugate

2-3i

\frac{x(2-3i)}{(2+3i)(2-3i)}=\frac{2x-3xi}{4+9}=\frac{2x-3xi}{13}

The equation becomes

\frac{2x}{13}-\frac{3x}{13}i-y-2yi=1+i

Group real terms and imaginary terms:

\left(\frac{2x}{13}-y\right)+\left(-\frac{3x}{13}-2y\right)i=1+i

Equate parts:

\frac{2x}{13}-y=1\ \text{...(1)}\qquad -\frac{3x}{13}-2y=1\ \text{...(2)}

Multiply (1) by

2

\dfrac{4x}{13}-2y=2

. Subtract (2) from this to remove

y

\frac{4x}{13}+\frac{3x}{13}=1 \;\Longrightarrow\; \frac{7x}{13}=1 \;\Longrightarrow\; x=\frac{13}{7}

Into (1):

\dfrac{2}{7}-y=1

, so

y=-\dfrac{5}{7}

Answer

x=\dfrac{13}{7},\qquad y=-\dfrac{5}{7}

Mark scheme

1 mark for rationalising · 1 for grouping into real and imaginary brackets · 1 for the two correct equations · 2 for solving. Both answers must be real — never leave an

i

attached. Watch the minus on bracket (2): both terms inside are negative.

6. Systems with Complex Coefficients 6–8 marks

The highest-mark question in Part A. The method is ordinary elimination — the only new thing is that a division at the end needs the conjugate.

Method — elimination

Multiply one or both equations so the coefficients of one variable match.
Subtract to eliminate that variable.
Solve for the remaining variable (this usually needs a conjugate division).
Substitute back to find the other.
Verify both values in both original equations.

Where the marks actually go Almost every lost mark here is a sign slip, not a misunderstanding. Keep real and imaginary parts in two vertical columns as you work down the page, and write

-(8-4i)\omega

on its own line before turning it into

(-8+4i)\omega

. Never do two sign steps in your head.

Worked Example 3 — System of equations 8 marks

Statement

Solve

\;2z-(1-3i)\omega=1+2i\;

...(1) and

\;(1+i)z+(2-i)\omega=2+i\;

...(2).

Working

Match the

z

coefficients: multiply (1) by

(1+i)

and (2) by

2

. Both then begin

2(1+i)z

. Subtracting removes

z

-\bigl[(1+i)(1-3i)+2(2-i)\bigr]\omega=(1+i)(1+2i)-2(2+i)

Left bracket:

(1+i)(1-3i)=4-2i

and

2(2-i)=4-2i

, sum

8-4i

.
Right side:

(1+i)(1+2i)=-1+3i

, and

2(2+i)=4+2i

, so

(-1+3i)-(4+2i)=-5+i

-(8-4i)\omega=-5+i \;\Longrightarrow\; (-8+4i)\omega=-5+i \;\Longrightarrow\; \omega=\frac{-5+i}{-8+4i}

Divide by the conjugate

-8-4i

\omega=\frac{44+12i}{80}=\frac{11}{20}+\frac{3}{20}i

Substitute into (1), writing

\omega=\dfrac{11+3i}{20}

and multiplying through by

20

40z-(1-3i)(11+3i)=20(1+2i)

Since

(1-3i)(11+3i)=20-30i

40z=20+40i+20-30i=40+10i \;\Longrightarrow\; z=1+\frac{1}{4}i

Answer

z=1+\dfrac{1}{4}i

\;\omega=\dfrac{11}{20}+\dfrac{3}{20}i

Mark scheme

2 marks for correct elimination · 2 for the expansions · 2 for the conjugate division · 2 for back-substitution. Check:

2z-(1-3i)\omega=1+2i

✓ and

(1+i)z+(2-i)\omega=2+i

✓. Twenty seconds, total certainty.

7. Factorising and Solving 3–5 marks

7.1 Sum of squares

Over the reals $z^{2}+a^{2}$ will not factorise. Over $\mathbb{C}$ it always does:

z^{2}+a^{2}=(z+ia)(z-ia)

So $z^{2}+169=(z+13i)(z-13i)$ . If there is a leading coefficient, take it out first: $2z^{2}+18=2(z+3i)(z-3i)$ .

Watch the wording “Cannot be factorised” is true only over the real numbers. If the question says over

\mathbb{C}

, it factorises — saying otherwise scores zero.

7.2 Cubics

Use the factor theorem: if $P(k)=0$ then $z-k$ is a factor. Try the factors of the constant term. For $P(z)=z^{3}+2z^{2}-5z-6$ , $P(-1)=0$ , so $z+1$ is a factor; dividing leaves $z^{2}+z-6=(z+3)(z-2)$ . Hence $P(z)=(z+1)(z+3)(z-2)$ .

7.3 Completing the square

Method

Divide through by the coefficient of $z^{2}$ .
Move the constant to the right.
Add $\left(\dfrac{\text{coefficient of }z}{2}\right)^{2}$ to both sides.
Square-root both sides, keeping $\pm$ .
If a negative appears under the root, pull out $i$ .

For $z^{2}-6z+25=0$ : $z^{2}-6z=-25$ ; add $9$ to both sides giving $(z-3)^{2}=-16$ ; so $z-3=\pm4i$ and $z=3\pm4i$ .

Five-second check A quadratic with real coefficients always gives complex roots as a conjugate pair. If your two answers are not conjugates, you have slipped — go back before writing anything else.

8. Part A — Formula Sheet

Formula / Rule	What it means in words	Common exam mistake
$i^{2}=-1$	The one fact the unit rests on.	Leaving $i^{2}$ in the answer, or treating it as $+1$ .
$z=x+iy$ ; $Re(z)=x$ , $Im(z)=y$	Both parts are real numbers.	Writing $Im(7-11i)=-11i$ instead of $-11$ .
$x_{1}+iy_{1}=x_{2}+iy_{2}\iff x_{1}=x_{2},\,y_{1}=y_{2}$	One complex equation $=$ two real equations.	Comparing a real part against an imaginary part.
$i^{1}=i,\ i^{2}=-1,\ i^{3}=-i,\ i^{4}=1$	Divide the exponent by 4, use the remainder.	Expanding every power the long way.
$\bar{z}=x-iy$	Flip the imaginary sign only.	Flipping the real part too.
$z\bar{z}=x^{2}+y^{2}=\|z\|^{2}$	Always real, never negative.	Writing $x^{2}-y^{2}$ .
$\dfrac{z_{1}}{z_{2}}=\dfrac{z_{1}\bar{z_{2}}}{z_{2}\bar{z_{2}}}$	Multiply top and bottom by the conjugate of the denominator.	Multiplying by the denominator itself.
$\|z\|=\sqrt{x^{2}+y^{2}}$	Distance from the origin.	$\|3+4i\|=7$ . It is $5$ .
$\|z_{1}z_{2}\|=\|z_{1}\|\|z_{2}\|$	Take moduli piece by piece.	Expanding brackets first and running out of time.
$z^{2}+a^{2}=(z+ia)(z-ia)$	Sums of squares factorise over $\mathbb{C}$ .	Answering “cannot be factorised.”
Real coefficients → conjugate roots	Complex roots come in pairs $p\pm qi$ .	Two answers that are not conjugates, unchecked.

Part B Polar Form and Applications The half students fear. In reality it is three procedures: convert to polar, multiply or divide in polar, and one circuit formula. The quadrant map in Section 10 is the single most valuable thing in this note — learn it and this part becomes routine.

9. Polar Form — What It Actually Means

You already describe a point by how far across and how far up: that is $x+iy$ . Polar form describes the same point by how far out and in which direction.

The idea in one sentence Nobody in a bazaar says “go 300 metres east and 400 metres north.” They say “go half a kilometre, that way.” Distance plus direction. That is polar form — the same shop, described differently.

The distance is $r=|z|$ . The direction is the angle $\theta$ , measured anticlockwise from the positive real axis, and called $\arg(z)$ . From the right-angled triangle:

x=r\cos\theta,\qquad y=r\sin\theta

Substituting into $z=x+iy$ gives the three forms you must recognise — all the same number:

z=r(\cos\theta+i\sin\theta)=r\,\text{cis}\,\theta=r\,e^{i\theta}

Know the vocabulary

x+iy

is rectangular (or algebraic) form.

r(\cos\theta+i\sin\theta)

is polar form.

re^{i\theta}

is the exponential form, and the step between the last two is Euler's formula. Questions switch between these names freely.

10. The Principal Argument — Where Most Marks Are Lost 4–6 marks

Here is the whole difficulty, stated plainly. Going round a full turn brings you back to the same point, so a point has infinitely many valid angles. To make the answer unique we insist on the one in a fixed range.

Definition — principal argument The value of

\arg(z)

lying in

-\pi\lt\theta\le\pi

is the principal argument, written

Arg(z)

Why students get this wrong: your calculator's $\tan^{-1}$ only ever returns angles in quadrants I and IV. It cannot tell $-\sqrt{3}+i$ from $\sqrt{3}-i$ , because $y/x$ has the same size for both. The calculator does not know the quadrant — you must supply it.

Method — converting

x+iy

to polar form

$r=\sqrt{x^{2}+y^{2}}$ . Always positive.
Sketch the point. Ten seconds. Which quadrant?
Reference angle $\alpha=\tan^{-1}\left|\dfrac{y}{x}\right|$ — use the absolute value, so $\alpha$ is a positive acute angle.
Apply the quadrant rule: I $\rightarrow\alpha$ · II $\rightarrow\pi-\alpha$ · III $\rightarrow\alpha-\pi$ · IV $\rightarrow-\alpha$ .
Check the answer sits in $-\pi\lt\theta\le\pi$ , then write $z=r(\cos\theta+i\sin\theta)$ .

The habit that fixes this topic Sketch before you calculate. Every time. Students who draw a quick pair of axes get these right; students who trust the calculator get quadrants II and III wrong nearly every time. It is the cheapest set of marks in the unit.

Two errors that dominate One: a negative

r

. It is a distance — it cannot be negative.
Two: an answer outside the range.

\frac{7\pi}{6}

is a valid argument but not the principal one; subtract

2\pi

to get

-\frac{5\pi}{6}

. If the question says “principal,” the range is compulsory.

Worked Example 4 — Modulus and principal argument, all four quadrants 6 marks

Statement

Find the modulus and principal argument of

\sqrt{3}+i

-\sqrt{3}+i

-\sqrt{3}-i

and

\sqrt{3}-i

Working

All four have

|x|=\sqrt{3}

and

|y|=1

, so the modulus and the reference angle are the same for all four:

r=\sqrt{3+1}=2,\qquad \alpha=\tan^{-1}\left|\frac{1}{\sqrt{3}}\right|=\frac{\pi}{6}

Only the quadrant differs, so only the argument differs.

(i)

\sqrt{3}+i

: quadrant I

\rightarrow\ Arg=\alpha=\dfrac{\pi}{6}

(ii)

-\sqrt{3}+i

: quadrant II

\rightarrow\ Arg=\pi-\dfrac{\pi}{6}=\dfrac{5\pi}{6}

(iii)

-\sqrt{3}-i

: quadrant III

\rightarrow\ Arg=\dfrac{\pi}{6}-\pi=-\dfrac{5\pi}{6}

(iv)

\sqrt{3}-i

: quadrant IV

\rightarrow\ Arg=-\dfrac{\pi}{6}

Answer

Modulus

2

for all four. Arguments

\dfrac{\pi}{6}

\dfrac{5\pi}{6}

-\dfrac{5\pi}{6}

-\dfrac{\pi}{6}

Mark scheme

1 mark for

r=2

· 1 for

\alpha=\frac{\pi}{6}

· 1 for each correct quadrant rule. Notice what this question is really testing: the modulus and reference angle carry no information about direction — the quadrant rule does all the work. All four answers lie in

-\pi\lt\theta\le\pi

. ✓

11. Multiplying and Dividing in Polar Form 4–6 marks

This is why polar form exists. It turns multiplication into addition.

z_{1}z_{2}=r_{1}r_{2}\,\text{cis}(\theta_{1}+\theta_{2}),\qquad \frac{z_{1}}{z_{2}}=\frac{r_{1}}{r_{2}}\,\text{cis}(\theta_{1}-\theta_{2})

Say it as a sentence Multiply: moduli multiply, arguments add.
Divide: moduli divide, arguments subtract.
No brackets to expand, no

i^{2}

to chase. Four times faster than the rectangular method.

Geometrically, multiplying by $z_{2}$ does two things at once: it stretches the arrow by $r_{2}$ and rotates it through $\theta_{2}$ . Also examinable:

z^{-1}=\frac{1}{r}(\cos\theta-i\sin\theta),\qquad \arg(z_{1}z_{2})=\arg z_{1}+\arg z_{2}

Finish the job After adding two arguments the total may leave the principal range. If the question asks for

Arg

of the product, add or subtract

2\pi

to bring it back. Stopping at

\frac{7\pi}{6}

is a half-finished answer.

Worked Example 5 — Multiply in polar form 5 marks

Statement

Find

\dfrac{2}{3}\left(\cos\dfrac{\pi}{3}+i\sin\dfrac{\pi}{3}\right)\times 6\left(\cos\dfrac{5\pi}{6}+i\sin\dfrac{5\pi}{6}\right)

, giving the answer in rectangular form.

Working

Multiply the moduli, add the arguments:

r=\frac{2}{3}\times6=4,\qquad \theta=\frac{\pi}{3}+\frac{5\pi}{6}=\frac{7\pi}{6}

So the product is

4\left(\cos\dfrac{7\pi}{6}+i\sin\dfrac{7\pi}{6}\right)

.
Convert to rectangular form. Since

\dfrac{7\pi}{6}=\pi+\dfrac{\pi}{6}

lies in quadrant III, both cosine and sine are negative:

\cos\frac{7\pi}{6}=-\frac{\sqrt{3}}{2},\qquad \sin\frac{7\pi}{6}=-\frac{1}{2}

4\left(-\frac{\sqrt{3}}{2}-\frac{1}{2}i\right)=-2\sqrt{3}-2i

Answer

-2\sqrt{3}-2i

Mark scheme

1 mark for

r=4

· 1 for

\theta=\frac{7\pi}{6}

· 2 for the correct signs of cosine and sine in quadrant III · 1 for the final form. Had the question asked for

Arg

, the answer would be

\frac{7\pi}{6}-2\pi=-\frac{5\pi}{6}

12. Loci — Turning a Condition into an Equation 4 marks

A question may give a condition such as $|z-2i|=|\bar{z}+3|$ and ask what it describes in $x$ and $y$ . The method never varies.

Method

Put $z=x+iy$ , so $\bar{z}=x-iy$ .
Substitute into the condition.
Collect each side into a single $\text{real}+i(\text{real})$ .
Apply $|a+ib|=\sqrt{a^{2}+b^{2}}$ to each side and square both sides.
Simplify. The $x^{2}$ and $y^{2}$ terms usually cancel, leaving a straight line.

For $|z-2i|=|\bar{z}+3|$ : squaring gives $x^{2}+(y-2)^{2}=(x+3)^{2}+y^{2}$ , so $-4y+4=6x+9$ , that is $\;6x+4y+5=0$ — a straight line.

For an inequality using $Re$ or $Im$ there is no squaring at all: $Re(z-3)\le2$ becomes $x-3\le2$ , so $x\le5$ .

13. Applications 5–8 marks

13.1 Alternating-current circuits — the examinable one

One formula links current $I$ , impedance $Z$ and voltage $E$ :

E=IZ\qquad\Longrightarrow\qquad Z=\frac{E}{I},\qquad I=\frac{E}{Z}

Notation warning — this is not a new topic Electrical engineers write the imaginary unit as

J

, because

I

already means current. So you will see

(-6-2J)

amperes. Nothing changes mathematically:

J^{2}=-1

, and the question is an ordinary conjugate division wearing a different letter. The AC supply WAPDA delivers at 220 V, 50 Hz is described with exactly this arithmetic.

Worked Example 6 — Impedance of a circuit 6 marks

Understand

In an AC circuit the voltage is

(-50+100J)

volts and the current is

(-6-2J)

amperes. Find the impedance

Z

Set up

From

E=IZ

Z=\frac{E}{I}=\frac{-50+100J}{-6-2J}

An ordinary division. The conjugate of

-6-2J

-6+2J

Solve

Z=\frac{-50+100J}{-6-2J}\times\frac{-6+2J}{-6+2J}

Numerator:

300-100J-600J+200J^{2}

. Since

200J^{2}=-200

, this is

100-700J

Unit 1 — Complex Numbers

1. The Basics You Are Assumed to Know

2. Equality — The Marks Machine 4–6 marks

3. The Four Operations 2–4 marks

3.1 Add and subtract

3.2 Multiply

3.3 Powers of iii 2 marks

3.4 Conjugate

3.5 Divide — the technique everything depends on 4 marks

4. Modulus 2–3 marks

5. Real and Imaginary Parts of Powers

6. Systems with Complex Coefficients 6–8 marks

7. Factorising and Solving 3–5 marks

7.1 Sum of squares

7.2 Cubics

7.3 Completing the square

8. Part A — Formula Sheet

9. Polar Form — What It Actually Means

10. The Principal Argument — Where Most Marks Are Lost 4–6 marks

11. Multiplying and Dividing in Polar Form 4–6 marks

12. Loci — Turning a Condition into an Equation 4 marks

13. Applications 5–8 marks

13.1 Alternating-current circuits — the examinable one

3.3 Powers of $i$ 2 marks

📐 Chapter 1 · Complex Numbers