NCERT Plus One (Class 11) Mathematics • Chapter 4

Complex Numbers and Quadratic Equations

Imaginary unit i, powers of i, a + ib arithmetic, rationalisation, algebraic identities, modulus-conjugate & Argand plane.

📄 Official NCERT Textbook PDF

Topic 4.2: Imaginary Unit i = √(−1) & Integer Powers of i

Try solving x² + 1 = 0 using ordinary numbers: you need x² = −1, which is impossible on the real number line! So mathematician Euler created a superhero symbol i = √(−1) where i² = −1. Suddenly not only can we solve x² + 1 = 0, powers of i spin forever in a super-simple 4-step carousel: i, −1, −i, 1!
FOUNDATIONi = √(−1)  |  i² = −1
4-STEP CYCLEi¹ = i, i² = −1, i³ = −i, i⁴ = 1
ANY POWER iⁿDivide n by 4, keep remainder r: iⁿ = iʳ!
🧠 Memory Trick: Sum of ANY 4 consecutive powers of i is ALWAYS ZERO: iⁿ + iⁿ⁺¹ + iⁿ⁺² + iⁿ⁺³ = 0! And remember 1/i = −i!
💡 Simple Kid-Check: Think of powers of i like a 4-stop Ferris wheel returning to 1 every 4 clicks! Want i²⁷? Divide 27 ÷ 4 = 6 full spins with remainder 3, so i²⁷ = i³ = −i!
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    Negative Powers: i⁻³⁹ = 1i³⁹ = 1i³ = 1−i · ii = i−i² = +i!
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    Solving x² + k = 0: x² = −9 ⇒ x = ±3i. Negative discriminants now have neat roots!
⚠️ Square Root of Two Negatives: Compute √(−4) · √(−9):
Write each with i FIRST: (2i) · (3i) = 6i² = −6 (NEVER √((−4)(−9)) = √36 = +6)!
The #1 chapter trap: Writing √a · √b = √(ab) when BOTH a < 0 and b < 0! That law ONLY holds when at least one of a, b ≥ 0! Always convert √(−k) → i√k BEFORE multiplying!

Topic 4.3: Addition, Subtraction & Equality of Complex Numbers (a + ib)

Pack a Thrissur festival basket with a coconuts (Real part) and b mangoes (Imaginary part): written z = a + ib! When you add or subtract two baskets, coconuts ONLY combine with coconuts, and mangoes ONLY combine with mangoes — and two baskets are equal ONLY when both counts match!
STANDARD FORMz = a + ib (both a, b ∈ ℝ!). Re(z) = a, Im(z) = b
EQUALITYa + ib = c + id ⇔ a = c AND b = d
ADD & SUBTRACT(a + ib) ± (c + id) = (a ± c) + i(b ± d)
🧠 Memory Trick: Watch out — Im(3 + 4i) = 4 (NOT 4i)! Both Re(z) and Im(z) are pure REAL numbers!
💡 Simple Kid-Check: Every real number is secretly a complex number too (5 = 5 + 0i)! And a purely imaginary number like −7i has Re = 0 (0 − 7i).
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    Additive Identity: 0 = 0 + i0 leaves any z unchanged.
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    Additive Inverse (Negative): −z = −a − ib, because z + (−z) = 0.
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    No Inequality Order: You CANNOT compare 2 + 3i > 1 + 4i — complex numbers live on a 2D plane, not a 1D line!
🧩 Quick Equality Example: If 4x + i(3x − y) = 3 − 6i (where x, y ∈ ℝ), find x and y:
Match Real: 4x = 3 ⇒ x = 3/4. Match Imaginary: 3(3/4) − y = −6 ⇒ y = 9/4 + 6 = 33/4!
Two classic 1-mark traps: (1) Writing Im(5 − 2i) = −2i instead of −2 (never include i inside Im(z)!). (2) Trying to say 3 + 4i > 2 + 5i — order relations <, > do NOT exist for complex numbers!

Topic 4.3: Multiplication, Multiplicative Inverse & Division by Rationalisation

Multiplying two complex numbers is just ordinary bracket expansion — except at the very end, (ib)(id) turns into −bd because i² = −1! And when dividing z₁ / z₂ leaves an ugly i in the denominator, we banish it downstairs by multiplying top and bottom by the denominator's mirror twin c − id!
MULTIPLY(a + ib)(c + id) = (ac − bd) + i(ad + bc)
INVERSE z⁻¹(a + ib)⁻¹ = a − iba² + b² = z̄|z|²
DIVISIONMultiply top & bottom of z₁z₂ by z₂̄ (conjugate!)
🧠 Memory Trick: Denominator trick (c + id)(c − id) = c² + d² (PLUS sign, not minus, because −i²d² = +d²)!
💡 Simple Kid-Check: Never memorize the big multiplication formula (ac − bd) + i(ad + bc) by heart — just FOIL expand the brackets normally and replace i² with −1!
🔎 Multiplicative Inverse of z = 4 − 3i:
z⁻¹ = 14 − 3i × 4 + 3i4 + 3i = 4 + 3i4² + 3² = 4 + 3i25 = 425 + i325!
Two super-common errors when dividing or inverting: (1) Writing (a + ib)(a − ib) = a² − b² instead of a² + b²! Remember −(ib)² = −(−b²) = +b²! (2) Leaving answers like (4 + 3i)/25 unsplit — Kerala HSE scoring keys prefer clear standard a + ib form: 4/25 + i(3/25)!

Topic 4.4: Identities (z₁ + z₂)² & Standard a + ib Simplifications

Want to expand (5 − 3i)³ or simplify 1 + i1 − i? Every algebra identity from Class 9 still works identically — PLUS you gain a cool new superpower: a² + b² now factors cleanly as (a + ib)(a − ib)!
POWER IDENTITIES(z₁ ± z₂)² = z₁² ± 2z₁z₂ + z₂²
z₁² − z₂² = (z₁ − z₂)(z₁ + z₂)
SUM OF SQUARESa² + b² = (a + ib)(a − ib) (impossible in ℝ, effortless in ℂ!)
VIP EXAM GEM1 + i1 − i = +i  |  1 − i1 + i = −i
🧠 Memory Trick: Check the numerator sign of (1 ± i)/(1 ∓ i) — the top sign wins! +i on top → +i; −i on top → −i!
💡 Simple Kid-Check: Notice how (1 + i)² = 1 + 2i + i² = 1 + 2i − 1 = 2i, and (1 − i)² = −2i! Memorizing those two tiny identities simplifies huge power problems in seconds!
Cube Identities (NCERT Fav):
(z₁ + z₂)³ = z₁³ + 3z₁²z₂ + 3z₁z₂² + z₂³
(z₁ − z₂)³ = z₁³ − 3z₁²z₂ + 3z₁z₂² − z₂³
🏆 Classic NCERT 2-Marker: Find least positive integer m such that (1 + i1 − i)ᵐ = 1.
Since 1 + i1 − i = i, we want iᵐ = 1 → least positive integer is m = 4!
When expanding (2 + 3i)², NEVER drop the middle term 2ab (giving 4 − 9 = −5)! Correct expansion: 2² + 2(2)(3i) + (3i)² = 4 + 12i − 9 = −5 + 12i! Also watch (i)³ = −i in cubic expansions!

Topic 4.4: Modulus |z| and Conjugate z̄ Properties

Imagine standing on the shore of Vembanad Lake! Your kite z = a + ib hovers above the water, and its reflection straight beneath the water is its Conjugate z̄ = a − ib. The straight diagonal length from you to the kite is its Modulus |z| = √(a² + b²) — and multiplying kite times reflection z · z̄ ALWAYS equals |z|²!
MODULUS |z||a + ib| = √(a² + b²) (always REAL and ≥ 0!)
CONJUGATE z̄overline{a + ib} = a − ib (flip sign of i term ONLY!)
GOLDEN LINKz · z̄ = |z|² = a² + b²  |  |z₁z₂| = |z₁||z₂|
🧠 Memory Trick: Need the modulus of a monster fraction like |(2+i)³ / (1−3i)|? NEVER expand first! Just take modulus of each piece separately: |z₁z₂| = |z₁|·|z₂| and |z₁/z₂| = |z₁|/|z₂|!
💡 Simple Kid-Check: Notice that |z| NEVER puts i inside the square root! Both z = 3 + 4i and its conjugate z̄ = 3 − 4i have the exact same modulus: √(3² + 4²) = √25 = 5!
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    Conjugate Distributes Cleanly: overline{z₁ ± z₂} = z̄₁ ± z̄₂, overline{z₁z₂} = z̄₁z̄₂, overline{(z₁/z₂)} = z̄₁/z̄₂, and overline{(z̄)} = z.
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    Modulus Rules: |zⁿ| = |z|ⁿ, |z̄| = |z|, and z⁻¹ = z̄ / |z|².
🚀 10-Second NCERT Shortcut: If x + iy = a + ibc + id, prove x² + y² = a² + b²c² + d²:
Take modulus squared on both sides! |x + iy|² = |a + ibc + id|² = |a + ib|²|c + id|² → done in TWO lines!
Watch two classic mistakes: (1) Including i inside modulus: writing |3 + 4i| = √(9 + (4i)²) = √(9 − 16) — WRONG! Modulus uses ONLY the real coefficients √(a² + b²)! (2) Flipping the sign of the Real part in Conjugate: overline{−2 + 5i} = −2 − 5i (Real part −2 NEVER flips!).

Topic 4.5: The Argand Plane Geometric Interpretation

How do you draw complex numbers when the regular X-axis is already full of Real numbers? Swiss mathematician Argand added a vertical street just for i! Walk x steps left–right on the Real Axis, then y steps up–down on the Imaginary Axis — every z = x + iy becomes point P(x, y) on the Argand Plane!
TWO AXESHorizontal = Real Axis (Re)
Vertical = Imaginary Axis (Im)
DISTANCE & POINTz = x + iy ↔ P(x, y)
Length OP = |z| = √(x² + y²)
GEOMETRY MOVESz̄ reflects across X-axis; × i rotates 90° anti-clockwise!
🧠 Memory Trick: Look at what multiplying by i does geometrically: 1 → i → −1 → −i is literally spinning a vector by 90° anti-clockwise on the Argand Plane!
💡 Simple Kid-Check: Purely real numbers like −4 = −4 + 0i sit right on the horizontal Real axis. Purely imaginary numbers like 3i = 0 + 3i sit straight on the vertical Imaginary axis!
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    Quadrant Matching: Just like 2D graphs! E.g. −3 + 2i → (−3, 2) lies in Quadrant II; 4 − i → (4, −1) lies in Quadrant IV.
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    Conjugate Reflection: Point z̄ = (x, −y) is the exact mirror image of z = (x, y) across the Real axis.
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    Distance Between Two Points: Distance from z₁ to z₂ is simply |z₁ − z₂|!
Don't mix up coordinate order! For z = 5i − 3, rewrite in standard x + iy order FIRST: −3 + 5i, so its Argand point is (−3, 5) (Quadrant II), NOT (5, −3)!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March • 1 Marks): Express i⁹ + i¹⁹ in the form a + ib.

i⁹ = i⁴⁽²⁾⁺¹ = i, and i¹⁹ = i⁴⁽⁴⁾⁺³ = i³ = −i.
Sum = i + (−i) = 0 + 0i.

Q2 (Kerala HSE Say • 2 Marks): Express (1 − i)⁴ in the form a + ib.

First compute (1 − i)² = 1 − 2i + i² = −2i.
Then (1 − i)⁴ = (−2i)² = 4i² = −4 + 0i.

Q3 (Kerala HSE March • 2 Marks): Find the multiplicative inverse of √5 + 3i.

z⁻¹ = √5 − 3i(√5)² + 3² = √5 − 3i5 + 9 = √514 − i314.

Q4 (Kerala HSE March • 2 Marks): Find real numbers x and y if (x − iy)(3 + 5i) is the conjugate of −6 − 24i.

Conjugate of −6 − 24i is −6 + 24i.
Expand LHS: (3x + 5y) + i(5x − 3y) = −6 + 24i.
Solve 3x + 5y = −6 and 5x − 3y = 24 ⇒ x = 3, y = −3.

Q5 (Kerala HSE Model • 2 Marks): If ((1 + i)/(1 − i))ᵐ = 1, find the least positive integral value of m.

Rationalise inside: (1 + i)(1 + i)1² + 1² = 2i2 = i.
So iᵐ = 1 ⇒ least positive integer m = 4.

Q6 (Kerala HSE March • 3 Marks): If a + ib = (x + i)² / (2x² + 1), prove that a² + b² = (x² + 1)² / (2x² + 1)².

Take modulus on both sides:
|a + ib| = |x + i|²|2x² + 1| = (√(x² + 1))²2x² + 1 = x² + 12x² + 1.
Square both sides → a² + b² = (x² + 1)²(2x² + 1)².

Q7 (Kerala HSE Say • 2 Marks): Find the modulus of (1 + i)/(1 − i) − (1 − i)/(1 + i).

Simplify using standard gems: i − (−i) = 2i = 0 + 2i.
Modulus = |2i| = √(0² + 2²) = 2.

Q8 (NCERT / DHSE Model • 3 Marks): Find the complex number z = (3 − 2i)(2 + 3i) / ((1 + 2i)(2 − i)) and state its quadrant on the Argand plane.

Numerator = (6 + 6) + i(9 − 4) = 12 + 5i.
Denominator = (2 + 2) + i(4 − 1) = 4 + 3i.
Multiply by 4 − 3i: (48 + 15) + i(20 − 36)16 + 9 = 6325 − i1625.
Point (63/25, −16/25) sits in Quadrant IV!