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

30 PYQs at 30% 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 them 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

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

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 9th May Shift 1 · Q110Moderate

Example 1 · Complex Numbers · Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity

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

Z_2=-1+i

, where

i=\sqrt{-1}

, then

|Z_1+Z_2|=

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

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

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 9th May Shift 2 · Q132Moderate

Example 2 · Complex Numbers · Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity

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}

)

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

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?

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 22 April Shift II · Q115Moderate

Example 3 · Complex Numbers · Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity

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

will be purely imaginary if

\theta =

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

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

Practise this conceptself-check

The same idea in a real exam question:

MHT-CET · 2025 · 25 April Shift I · Q129Hard

Example 4 · Complex Numbers · Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity

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

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

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

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 23 April Shift I · Q146Moderate

Example 5 · Complex Numbers · Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity

x= - 2 +\sqrt{- 3}

, then the value of

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

is equal to

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

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}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 10th May Shift 1 · Q134Moderate

Example 6 · Complex Numbers · Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity

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

, where

i=\sqrt{-1}

, then the value of

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

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

Test yourself on Complex Numbers

15 past MHT-CET questions from this chapter, timed at 27 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Complex Numbers chapter test Drill all 30 questions

Related notes

Determinants — expanding a 3 × 3 determinant, for the ω determinants →