MHT-CET Maths · Complex Numbers

Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity

Everything algebraic about z = x + iy: reduce powers of i modulo 4, multiply by the conjugate to clear a denominator, equate real and imaginary parts, and reduce powers of ω modulo 3.

All Complex Numbers notes MHT-CET Maths strategy

Why this matters

15 PYQs at 47% HARD — the chapter's largest and hardest page, though the difficulty is bookkeeping rather than ideas. The recurring stems are a polynomial evaluated at a complex x (answered by its minimal quadratic, never by substitution), a cube of a binomial, an equation solved for z by equating parts, and a determinant or power built on the cube roots of unity. Two of the fifteen carry stems the bank had garbled — a cube read as a division, an argument's denominator misprinted — both now repaired against the papers.

Concept 1 of 6

Powers of i and the Standard Form x + iy

Intuition

i^2 = -1

, so

i^3 = -i

and

i^4 = 1

: the powers of

i

cycle with period

4

. Any

i^n

is one of

1, i, -1, -i

, decided by

n \bmod 4

Definition

$i^{4k} = 1$ , $i^{4k+1} = i$ , $i^{4k+2} = -1$ , $i^{4k+3} = -i$ . So $i^{40} = 1$ , $i^{35} = i^{32}\cdot i^3 = -i$ , $i^{17} = i$ .
Collect real and imaginary parts separately: $4i^{40} - 5i^{35} + 6i^{17} + 2 = 4 + 5i + 6i + 2 = 6 + 11i$ .
Addition is componentwise; multiplication is FOIL with $i^2 = -1$ : $(a + ib)(c + id) = (ac - bd) + i(ad + bc)$ .
$\sqrt{-3} = i\sqrt3$ ; a stem written with $\sqrt{-1}$ means $i$ .
Two complex numbers are equal iff both real parts and both imaginary parts agree — the tool behind every 'find $x$ and $y$ ' stem.

Cycle of powers of i

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

Worked example

Simplify

3i^{22} + 2i^{15} - i^{9} + 7

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Complex NumbersMODERATE

Z_1=4i^{40}-5i^{35}+6i^{17}+2

Z_2=-1+i

, where

i=\sqrt{-1}

, then

|Z_1+Z_2|=

[Q110 · 9th May Shift 1 · 2024]

i³ = i, and other lapses in the cycle

i^3 = i^2\cdot i = -i

. Reading

i^{35}

i

instead of

-i

flips a sign in the imaginary part and lands on the wrong modulus, which is always offered.

Concept 2 of 6

The Conjugate: Rationalising a Denominator and Equating Conjugates

Intuition

z\bar z = |z|^2

is a real number, so multiplying top and bottom of a fraction by the conjugate of the denominator makes the denominator real — the single move that clears almost every fraction in the chapter.

Definition

$\bar z = x - iy$ ; $z\bar z = x^2 + y^2 = |z|^2$ ; $\overline{z_1/z_2} = \bar z_1/\bar z_2$ .
$\dfrac{2 - i}{3 + 4i} = \dfrac{(2 - i)(3 - 4i)}{25} = \dfrac{2 - 11i}{25}$ : then read off $a = \frac{2}{25}$ , $b = -\frac{11}{25}$ and compute the asked combination.
Conjugates of each other: $z_1 = \bar z_2$ means real parts equal and imaginary parts negatives. For $(3x + 2) - (5y - 3)i$ and $(6x + 3) + (2y - 4)i$ : $3x + 2 = 6x + 3$ and $5y - 3 = 2y - 4$ .
Trigonometric pairs: $\sin x + i\cos 2x$ and $\cos x - i\sin 2x$ are conjugates iff $\sin x = \cos x$ AND $\cos 2x = \sin 2x$ — two conditions that may have no common solution.

Conjugate identities

z\bar z = |z|^2 \qquad \frac{z_1}{z_2} = \frac{z_1\bar z_2}{|z_2|^2} \qquad z_1 = \bar z_2 \iff \operatorname{Re}z_1 = \operatorname{Re}z_2,\ \operatorname{Im}z_1 = -\operatorname{Im}z_2

Worked example

Write

\dfrac{3 + 2i}{1 - i}

in the form

a + ib

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Complex NumbersMODERATE

Z_1 = 2+i

and

Z_2 = 3-4i

and

\dfrac{\overline{Z_1}}{\overline{Z_2}} = a+bi

, then the value of

-7a + b

is (where

i = \sqrt{-1}

and

a,b \in \mathbb{R}

)

[Q132 · 9th May Shift 2 · 2024]

Conjugating only the numerator

\overline{Z_1}/\overline{Z_2}

needs BOTH conjugates before the division; and the division itself still needs a second conjugate multiplication to clear the denominator. Two conjugations, not one.

Drill 2 more on the conjugate: rationalising a denominator and equating conjugates

Concept 3 of 6

Purely Real or Purely Imaginary: Set the Other Part to Zero

Intuition

After rationalising, a complex expression has a real part and an imaginary part in the variable. 'Purely imaginary' means the real part is

0

; 'purely real' means the imaginary part is

0

. That equation is the whole question.

Definition

Rationalise first, then separate: $\dfrac{3 + 2i\sin\theta}{1 - 2i\sin\theta} = \dfrac{(3 - 4\sin^2\theta) + 8i\sin\theta}{1 + 4\sin^2\theta}$ .
Purely imaginary $\Rightarrow 3 - 4\sin^2\theta = 0 \Rightarrow \sin^2\theta = \dfrac34 \Rightarrow \theta = n\pi \pm \dfrac{\pi}{3}$ .
Purely real would instead need $8\sin\theta = 0$ , i.e. $\theta = n\pi$ .
Write the general solution in the form the options use: $\sin^2\theta = \sin^2\alpha \Rightarrow \theta = n\pi \pm \alpha$ .

Real and imaginary conditions

z \text{ purely imaginary} \iff \operatorname{Re}z = 0 \qquad z \text{ purely real} \iff \operatorname{Im}z = 0 \qquad \sin^2\theta = \sin^2\alpha \Rightarrow \theta = n\pi \pm \alpha

Worked example

For what real

a

\dfrac{a + 2i}{1 + i}

purely imaginary?

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Complex NumbersMODERATE

z =\frac{3 + 2isin\theta}{1 - 2isin\theta},(i =\sqrt{- 1})

will be purely imaginary if

\theta =

[Q115 · 22 April Shift II · 2025]

Setting the imaginary part to zero for 'purely imaginary'

Purely imaginary means NO real part. The condition is

\operatorname{Re} = 0

; solving

\operatorname{Im} = 0

gives

\theta = n\pi

, which is offered as option (D).

Concept 4 of 6

Solving for z: Put z = x + iy and Equate Parts

Intuition

An equation in

z

is two real equations in disguise — one from the real parts, one from the imaginary parts. With

z = x + iy

(and any given value of

x

y

substituted early), the two equations fix the unknowns.

Definition

$\dfrac{2z - n}{2z + n} = 2i - 1$ : cross-multiply, $2z - n = (2i - 1)(2z + n)$ , collect $z$ : $4z(1 - i) = 2in$ , so $z = \dfrac{in}{2(1 - i)} = \dfrac{in(1 + i)}{4} = -\dfrac{n}{4} + \dfrac{n}{4}i$ .
Given $\operatorname{Im}z = 10$ : $\dfrac{n}{4} = 10 \Rightarrow n = 40$ , and then $\operatorname{Re}z = -10$ .
Alternatively substitute $z = x + 10i$ at the start and equate parts of $(2x - n) + 20i = (2i - 1)(2x + n + 20i)$ .
Either way the answer is a pair $(n, \operatorname{Re}z)$ ; check both against the option.

Equating parts

a + ib = c + id \iff a = c \text{ and } b = d \qquad \frac{1}{1 - i} = \frac{1 + i}{2}

Worked example

Solve

\dfrac{z - 2}{z + 2} = i

for

z

Practice this conceptself-check

From the bank · past-year question

Example 4Complex NumbersHARD

Let z be the complex number with

Im(z) = 10

and satisfying

\frac{2z-n}{2z+n}= 2i- 1

, where

i=\sqrt{- 1}

, for some natural number '

n

' then

[Q129 · 25 April Shift I · 2025]

Dividing by 1 − i without rationalising

\dfrac{in}{2(1 - i)}

is not yet in standard form; multiply by

\dfrac{1 + i}{1 + i}

first. Reading the real part off the un-rationalised form gives

\operatorname{Re}z = 0

and a wrong sign on

n

Drill 1 more on solving for z: put z = x + iy and equate parts

Concept 5 of 6

A Polynomial at a Complex x: Use Its Minimal Quadratic, and Cube Expansions

Intuition

x = 1 + 2i

satisfies

x^2 - 2x + 5 = 0

. Divide the given polynomial by that quadratic; the remainder is the value, and substituting

1 + 2i

directly into a quartic is never necessary.

Definition

From $x = a + ib$ : $(x - a)^2 = -b^2$ , i.e. $x^2 - 2ax + (a^2 + b^2) = 0$ — the minimal quadratic. For $x = 1 + 2i$ : $x^2 - 2x + 5 = 0$ ; for $x = -2 + \sqrt3 i$ : $x^2 + 4x + 7 = 0$ .
Divide the polynomial by the quadratic (long division); $p(x) = q(x)\cdot(\text{quadratic}) + r(x)$ , and $p(x) = r(x)$ at the root. $2x^4 + 5x^3 + 7x^2 - x + 38 = (x^2 + 4x + 7)(2x^2 - 3x + 5) + 3$ , so the value is $3$ .
Cubes: $\left(-2 - \frac13 i\right)^3 = \dfrac{(-6 - i)^3}{27}$ and $(-6 - i)^3 = -216 - 108i + 18 + i = -198 - 107i$ by the binomial expansion with $i^2 = -1$ , $i^3 = -i$ .
Cube root written as $p + iq$ : $z = (p + iq)^3 = (p^3 - 3pq^2) + i(3p^2q - q^3)$ , so $\dfrac{x}{p} + \dfrac{y}{q} = (p^2 - 3q^2) + (3p^2 - q^2) = 4(p^2 - q^2)$ .

Minimal quadratic and the cube

x = a + ib \Rightarrow x^2 - 2ax + (a^2 + b^2) = 0 \qquad (p + iq)^3 = (p^3 - 3pq^2) + i(3p^2q - q^3)

Worked example

x = 2 + i

, find the value of

x^3 - 3x^2 + 4x + 1

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 5Complex NumbersMODERATE

x= - 2 +\sqrt{- 3}

, then the value of

2x^{4}+ 5x^{3}+ 7x^{2}-x+ 38

is equal to

[Q146 · 23 April Shift I · 2025]

Substituting the complex number directly

Raising

1 + 2i

to the fourth power by hand invites a sign error at every step. The minimal quadratic reduces the polynomial to its remainder in two lines of long division.

Drill 3 more on a polynomial at a complex x: use its minimal quadratic, and cube expansions

Concept 6 of 6

Cube Roots of Unity: ω³ = 1 and 1 + ω + ω² = 0

Intuition

z^3 = 1

has three roots:

1

\omega

and

\omega^2

, spaced

120^\circ

apart on the unit circle. Two facts —

\omega^3 = 1

and

1 + \omega + \omega^2 = 0

— reduce any expression in

\omega

to a number times

\omega

\omega^2

Definition

$\omega = \dfrac{-1 + i\sqrt3}{2}$ , $\omega^2 = \dfrac{-1 - i\sqrt3}{2} = \bar\omega$ ; either may be called $\omega$ in a stem.
$\omega^3 = 1$ : reduce every exponent modulo $3$ — $\omega^4 = \omega$ , $\omega^{100} = \omega$ . $1 + \omega + \omega^2 = 0$ : so $1 + \omega = -\omega^2$ , $\omega + \omega^2 = -1$ , $\dfrac1\omega = \omega^2$ .
$3 + \omega + 3\omega^2 = 3(1 + \omega^2) + \omega = -3\omega + \omega = -2\omega$ , so $(3 + \omega + 3\omega^2)^4 = 16\omega^4 = 16\omega$ .
$z^2 + z + 1 = 0$ means $z = \omega$ : then $z^3 + z^{-3} = 2$ and $z^4 + z^{-4} = \omega + \omega^2 = -1$ .
Determinants in $\omega$ : a row or column summing to $1 + \omega + \omega^2 = 0$ makes the determinant $0$ (add all columns into one); otherwise expand and reduce — $\begin{vmatrix} 1 & 1 & 1 \\ 1 & \omega & \omega^2 \\ 1 & \omega^2 & \omega \end{vmatrix} = 3\omega(\omega - 1)$ after replacing $-1 - \omega^2$ by $\omega$ and $\omega^4$ by $\omega$ .

The two facts

\omega^3 = 1, \qquad 1 + \omega + \omega^2 = 0, \qquad \omega^n = \omega^{\,n \bmod 3}, \qquad \bar\omega = \omega^2 = \frac{1}{\omega}

Worked example

Evaluate

(1 + \omega - \omega^2)^{3}

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 6Complex NumbersMODERATE

w=\frac{-1+i\sqrt{3}}{2}

, where

i=\sqrt{-1}

, then the value of

(3+w+3w^2)^4

[Q134 · 10th May Shift 1 · 2023]

Stopping at ω⁴

16\omega^4

is not an option;

16\omega

is. Reduce every exponent modulo

3

before matching, and remember

\omega^2

is a different option from

\omega

Drill 3 more on cube roots of unity: ω³ = 1 and 1 + ω + ω² = 0

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (6)

Powers of i and the Standard Form x + iy
Cycle of powers of i
$i^2 = -1,\quad i^3 = -i,\quad i^4 = 1,\qquad i^n = i^{\,n \bmod 4}$
The Conjugate: Rationalising a Denominator and Equating Conjugates
Conjugate identities
$z\bar z = |z|^2 \qquad \frac{z_1}{z_2} = \frac{z_1\bar z_2}{|z_2|^2} \qquad z_1 = \bar z_2 \iff \operatorname{Re}z_1 = \operatorname{Re}z_2,\ \operatorname{Im}z_1 = -\operatorname{Im}z_2$
Purely Real or Purely Imaginary: Set the Other Part to Zero
Real and imaginary conditions
$z \text{ purely imaginary} \iff \operatorname{Re}z = 0 \qquad z \text{ purely real} \iff \operatorname{Im}z = 0 \qquad \sin^2\theta = \sin^2\alpha \Rightarrow \theta = n\pi \pm \alpha$
Solving for z: Put z = x + iy and Equate Parts
Equating parts
$a + ib = c + id \iff a = c \text{ and } b = d \qquad \frac{1}{1 - i} = \frac{1 + i}{2}$
A Polynomial at a Complex x: Use Its Minimal Quadratic, and Cube Expansions
Minimal quadratic and the cube
$x = a + ib \Rightarrow x^2 - 2ax + (a^2 + b^2) = 0 \qquad (p + iq)^3 = (p^3 - 3pq^2) + i(3p^2q - q^3)$
Cube Roots of Unity: ω³ = 1 and 1 + ω + ω² = 0
The two facts
$\omega^3 = 1, \qquad 1 + \omega + \omega^2 = 0, \qquad \omega^n = \omega^{\,n \bmod 3}, \qquad \bar\omega = \omega^2 = \frac{1}{\omega}$

Watch out for (6)

i³ = i, and other lapses in the cycle
→ Powers of i and the Standard Form x + iy
Conjugating only the numerator
→ The Conjugate: Rationalising a Denominator and Equating Conjugates
Setting the imaginary part to zero for 'purely imaginary'
→ Purely Real or Purely Imaginary: Set the Other Part to Zero
Dividing by 1 − i without rationalising
→ Solving for z: Put z = x + iy and Equate Parts
Substituting the complex number directly
→ A Polynomial at a Complex x: Use Its Minimal Quadratic, and Cube Expansions
Stopping at ω⁴
→ Cube Roots of Unity: ω³ = 1 and 1 + ω + ω² = 0

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