Class 11 Maths · Chapter 4
Short answer:
Class 11 Maths Chapter 4 — Complex Numbers and Quadratic Equations wo chapter hai jahan pehli baar i = √-1 introduce hota hai, taaki negative number ka square root bhi define ho sake. Is chapter me complex numbers ki algebra (jod, ghata, guna, bhaag), modulus-argument, Argand plane, aur negative discriminant wali quadratic equations solve karna seekhte ho. Class 11 Maths NCERT solutions dhoondh rahe ho — to yahi hai poora step-by-step solved content, exam-ready working ke saath.
Class 10 me tumne quadratic equations solve kiye the — par sirf tab jab discriminant (D = b² − 4ac) zero ya positive ho. Jaise hi D negative aata hai, jaise x² + 3 = 0 (yaani x² = −3), Class 10 ka toolkit fail ho jaata hai kyunki koi bhi real number ka square negative nahi ho sakta. Chapter 4 isi gap ko bharta hai — ek naya number system, complex numbers, introduce karke.
Ye chapter conceptually algebra ki neev hai jo aage kai jagah kaam aayega — jaise Class 11 Maths Chapter 3 trigonometric functions NCERT solutions me argument/angle ka concept dobara milega, aur straight lines NCERT solutions, conic sections notes me coordinate geometry ke saath Argand plane ka connection dikhega. Agar tum poori tarah se Class 11 Maths NCERT solutions cover kar rahe ho — Chapter 1 Sets NCERT solutions se shuru karke, Chapter 8 sequences and series important questions tak — to is chapter ko halka mat lo, kyunki iske concepts trigonometry aur coordinate geometry dono chapters me dobara kaam aate hain.
Is guide me har NCERT-style question ka full step-by-step working diya gaya hai — sirf final answer nahi, balki wahi process jo exam me full marks paane ke liye zaroori hota hai. Neeche formulas ka table bhi hai (agar tum Class 11 Maths formulas PDF download karke revise kar rahe ho, to yahi core formulas hain), aur last me previous year style questions bhi.
Chapter 4 Summary — 5 Minute Revision
Complex Numbers and Quadratic Equations chapter ka core idea simple hai: i = √-1 define karke, har real aur "impossible" (negative sqrt wala) number ko ek hi system me le aana. Complex number z = a + ib ka real part a hai, imaginary part b hai. Ye numbers plus, minus, multiply, divide sab NCERT ke defined rules se hote hain (guna me FOIL jaisa hi hai, bas i² = −1 substitute karo).
Modulus |z| = √(a² + b²) origin se distance batata hai, aur argument θ, positive x-axis se banaya gaya angle. Ye dono milkar polar form z = r(cos θ + i sin θ) banate hain — jo Argand plane (complex plane) pe complex number ko geometrically represent karta hai.
Chapter ka doosra bada application: quadratic equation ax² + bx + c = 0 jab D = b² − 4ac negative ho, tab roots real nahi hote — woh complex conjugate pair me aate hain, x = (−b ± i√|D|) / 2a. Yehi wo tool hai jo Class 10 me missing tha.
In-Text Questions — Solutions
Express in the form a + ib: (5i)(−3i/5)
(5i) × (−3i/5) = −3i² = −3(−1) = 3
a + ib = 3 + 0i
Express in the form a + ib: i⁹ + i¹⁹
i⁹ = i⁽⁸⁺¹⁾ = i⁸·i = 1·i = i (kyunki i⁴ = 1, aur 8 is a multiple of 4)
i¹⁹ = i⁽¹⁶⁺³⁾ = i¹⁶·i³ = 1·(−i) = −i
i⁹ + i¹⁹ = i + (−i) = 0
Express in the form a + ib: i⁻³⁹
i⁴ = 1, isliye har exponent ko 4 se manage kar sakte hain. −39 ko 4 ka nearest multiple add karke positive banao: −39 + 40 = 1
i⁻³⁹ = i⁻³⁹⁺⁴⁰ = i¹ = i
Express in the form a + ib: 3(7 + i7) + i(7 + i7)
= 21 + 21i + 7i + 7i²
= 21 + 28i + 7(−1)
= 21 − 7 + 28i = 14 + 28i
Express in the form a + ib: (1 − i) − (−1 + i6)
= 1 − i + 1 − 6i
= 2 − 7i
Express in the form a + ib: (1/5 + i2/5) − (4 + i5/2)
Real part: 1/5 − 4 = 1/5 − 20/5 = −19/5
Imaginary part: 2/5 − 5/2 = 4/10 − 25/10 = −21/10
= −19/5 − i(21/10)
Express in the form a + ib: [(1/3 + i7/3) + (4 + i1/3)] − (−4/3 + i)
Pehle bracket ka sum: (1/3 + 4) + i(7/3 + 1/3) = 13/3 + i(8/3)
Ab (−4/3 + i) subtract karo: 13/3 − (−4/3) = 17/3; imaginary: 8/3 − 1 = 5/3
= 17/3 + i(5/3)
Express in the form a + ib: (1 − i)⁴
Pehle (1 − i)² nikalo: 1 − 2i + i² = 1 − 2i − 1 = −2i
Ab (1 − i)⁴ = (−2i)² = 4i² = 4(−1)
= −4
Express in the form a + ib: (1/3 + 3i)³
(a+b)³ = a³ + 3a²b + 3ab² + b³, jahan a = 1/3, b = 3i
a³ = 1/27
3a²b = 3(1/9)(3i) = i
3ab² = 3(1/3)(9i²) = 3(1/3)(−9) = −9
b³ = 27i³ = 27(−i) = −27i
Sum = 1/27 − 9 + i(1 − 27) = −242/27 − 26i
Express in the form a + ib: (−2 − 1/3 i)³
a = −2, b = −i/3
a³ = −8
3a²b = 3(4)(−i/3) = −4i
3ab² = 3(−2)(i²/9) = 3(−2)(−1/9) = 2/3
b³ = −i³/27 = −(−i)/27 = i/27
Sum = (−8 + 2/3) + i(−4 + 1/27) = −22/3 − (107/27)i
Find the multiplicative inverse of 4 − 3i.
z = 4 − 3i, so z̄ = 4 + 3i
|z|² = 4² + (−3)² = 16 + 9 = 25
z⁻¹ = z̄ / |z|² = (4 + 3i)/25 = 4/25 + i(3/25)
If z₁ = 2 − i, z₂ = 1 + i, find |(z₁ + z₂ + 1)/(z₁ − z₂ + 1)|.
z₁ + z₂ + 1 = (2 − i) + (1 + i) + 1 = 4
z₁ − z₂ + 1 = (2 − i) − (1 + i) + 1 = 2 − 2i
Ratio = 4/(2 − 2i). Conjugate se multiply karo: 4(2+2i) / [(2−2i)(2+2i)] = 4(2+2i)/8 = 1 + i
|1 + i| = √(1² + 1²) = √2
Find the conjugate of (3 − 2i)(2 + 3i)/(1 + 2i).
Numerator: (3−2i)(2+3i) = 6 + 9i − 4i − 6i² = 6 + 5i + 6 = 12 + 5i
Divide by (1+2i), multiply by conjugate (1−2i): (12+5i)(1−2i)/[(1)²+(2)²] = (12 − 24i + 5i − 10i²)/5 = (12 − 19i + 10)/5 = (22 − 19i)/5
Conjugate = 22/5 + i(19/5)
Find real numbers x and y such that (x − iy)(3 + 5i) is the conjugate of −6 − 24i.
Conjugate of −6 − 24i is −6 + 24i.
(x − iy)(3 + 5i) = 3x + 5xi − 3yi − 5yi² = (3x + 5y) + i(5x − 3y)
Equating: 3x + 5y = −6 ... (1) and 5x − 3y = 24 ... (2)
(1)×3: 9x + 15y = −18; (2)×5: 25x − 15y = 120. Adding: 34x = 102, x = 3
From (1): 9 + 5y = −6, so y = −3
x = 3, y = −3
Poore Class 11 Maths ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q16)
Find the modulus and argument of z = 1 − i.
|z| = √(1² + (−1)²) = √2
Point (1, −1) 4th quadrant me hai. tan α = 1/1 = 1, so α = π/4
arg(z) = −π/4
Find the modulus and argument of z = −1 + i.
|z| = √((−1)² + 1²) = √2
Point (−1, 1) 2nd quadrant me hai. tan α = 1, α = π/4
arg(z) = π − π/4 = 3π/4
Find the modulus and argument of z = −1 − i√3.
|z| = √(1 + 3) = 2
Point (−1, −√3) 3rd quadrant me hai. tan α = √3, α = π/3
arg(z) = −(π − π/3) = −2π/3
Find the modulus and argument of z = −3.
z = −3 + 0i, so |z| = 3
z negative real axis pe hai, isliye arg(z) = π
Find the modulus and argument of z = √3 + i.
|z| = √(3 + 1) = 2
Point (√3, 1) 1st quadrant me hai. tan α = 1/√3, α = π/6
arg(z) = π/6
Convert z = i into polar form.
z = 0 + i, so |z| = 1
z positive imaginary axis pe hai, arg(z) = π/2. Polar form: z = cos(π/2) + i sin(π/2)
Convert (1 + i)/(1 − i) into polar form.
Numerator aur denominator ko (1+i) se multiply karo: (1+i)²/[(1−i)(1+i)] = (1 + 2i + i²)/(1+1) = 2i/2 = i
|z| = 1
arg(z) = π/2, so z = cos(π/2) + i sin(π/2)
Convert 1/(1 + i) into polar form.
Conjugate se multiply karo: (1−i)/[(1+i)(1−i)] = (1 − i)/2 = 1/2 − i/2
|z| = √(1/4 + 1/4) = 1/√2
Point (1/2, −1/2) 4th quadrant, tan α = 1, α = π/4
arg(z) = −π/4
Solve x² + 3 = 0.
x² = −3
x = ±i√3
Solve 2x² + x + 1 = 0.
a = 2, b = 1, c = 1
D = b² − 4ac = 1 − 8 = −7
x = (−1 ± i√7)/4
Solve x² + 3x + 9 = 0.
D = 9 − 36 = −27
√(−27) = i√27 = 3i√3
x = (−3 ± 3i√3)/2
Solve x² − x + 2 = 0.
D = 1 − 8 = −7
x = (1 ± i√7)/2
Solve x² + 3x + 5 = 0.
D = 9 − 20 = −11
x = (−3 ± i√11)/2
Solve x² + x + 1 = 0.
D = 1 − 4 = −3
x = (−1 ± i√3)/2
Solve √2 x² + x + √2 = 0.
a = √2, b = 1, c = √2
D = 1 − 4(√2)(√2) = 1 − 8 = −7
x = (−1 ± i√7)/(2√2). Denominator ko rationalise karne ke liye √2/√2 se multiply karo
x = (−√2 ± i√14)/4
Solve x²/√2 + x + √2 = 0, i.e. x² + √2x + 2 = 0.
Poori equation ko √2 se multiply karke saral form me: x² + √2x + 2 = 0
a = 1, b = √2, c = 2
D = 2 − 8 = −6
x = (−√2 ± i√6)/2
Important Equations — Ek Nazar Me
| Concept | Formula / Rule |
|---|---|
| Imaginary unit | i = √−1, i² = −1 |
| Powers of i (cycle of 4) | i¹ = i, i² = −1, i³ = −i, i⁴ = 1 (fir repeat; n mod 4 use karo) |
| Complex number | z = a + ib, jahan Re(z) = a, Im(z) = b |
| Addition | (a+ib) + (c+id) = (a+c) + i(b+d) |
| Multiplication | (a+ib)(c+id) = (ac−bd) + i(ad+bc) |
| Conjugate | z̄ = a − ib |
| Modulus | |z| = √(a² + b²) |
| Multiplicative inverse | z⁻¹ = z̄ / |z|² |
| Division | z₁/z₂ = (z₁ · z̄₂) / |z₂|² |
| Polar (trigonometric) form | z = r(cos θ + i sin θ), r = |z|, θ = arg(z) |
| Principal argument range | −π < θ ≤ π |
| Quadratic formula | x = (−b ± √(b² − 4ac)) / 2a |
| Discriminant | D = b² − 4ac; D < 0 ⇒ roots complex conjugate pair |
| Square root of negative real | √−a = i√a (a > 0) |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- i² ko galti se +1 samajh lena — yaad rakho i² = −1 hamesha, ye poore chapter ki base hai.
- Modulus |z| = √(a² + b²) nikalte waqt a ya b ka negative sign square karne se pehle hi drop kar dena — square karne se pehle sign matter nahi karta, ye galti se skip ho jaata hai.
- Argument nikalte waqt sirf tan θ = b/a se θ nikal lena, quadrant check kiye bina — isse 2nd aur 3rd quadrant wale complex numbers ka sign ulta aa jaata hai.
- Do complex numbers divide karte waqt denominator ka conjugate se multiply karna bhool jana — bina isके denominator real nahi banega aur answer galat form me rahega.
- Quadratic equation ka D negative aane par bhi √D ko directly real number treat kar lena, i factor out karna bhool jana — jaise √−7 ko galti se −√7 likh dena instead of i√7.
- i^n ki large power nikalte waqt n ko 4 se divide karke remainder use karna bhool jana — jaise i²³ ko seedha guna karne ki koshish karna instead of i²³ = i^(20+3) = i³ = −i.
Board-Style Important Questions
- Practice Question 1: Simplify i¹⁸ and express in the form a + ib.
- Practice Question 2: Find the modulus and argument of the complex number z = −√3 + i.
- Practice Question 3: Solve the quadratic equation x² + x + 1 = 0 and verify that the sum of the roots equals −1.
- Practice Question 4: Express (3 − 2i)/(1 + 2i) in the form a + ib, and hence find its conjugate.
- Practice Question 5: If z₁ = 2 − i and z₂ = 1 + i, find the modulus and argument of z₁z₂, showing all algebra steps.
Aksar Poochhe Jaane Wale Sawaal
Class 11 Maths Chapter 4 me kitne exercises hain?
NCERT ke current rationalised textbook (2026-27 session) me is chapter ka structure kaafi streamlined hai — sirf ek main Exercise 4.1 hai, jisme complex numbers ki algebra, modulus-conjugate, aur quadratic equations teeno se related sawaal cover hote hain, plus ek Miscellaneous Exercise chapter ke end me jisme mixed-difficulty questions hote hain. Agar tum poore Class 11 Maths NCERT solutions ka structure samajhna chahte ho, to har chapter (jaise sets, trigonometric functions, sequences and series) me bhi yahi kam-exercises + ek-miscellaneous-exercise wala rationalised pattern milega.
i ki value kya hoti hai aur uski powers kaise nikalte hain?
i = √−1 hai, aur i² = −1. Uski powers ek cycle of 4 me repeat karti hain: i, −1, −i, 1. Koi bhi bada exponent solve karne ke liye n ko 4 se divide karo aur remainder wali power use karo.
Modulus aur argument me kya fark hai?
Modulus |z| = √(a² + b²) batata hai complex number origin se kitni door hai (distance). Argument θ batata hai wo positive x-axis se kitne angle pe hai. Dono milkar polar form z = r(cos θ + i sin θ) banate hain, jo Argand plane pe complex number ko position karta hai.
Quadratic equation ka discriminant negative ho to roots kaise honge?
Agar D = b² − 4ac < 0, to roots real nahi honge — woh ek complex conjugate pair banayenge: x = (−b ± i√|D|)/2a. Ye exactly wahi case hai jo Class 10 me solve nahi hota tha.
Class 11 Maths ke saare chapters ke important questions aur formulas ek jagah kahan milenge?
Chapter-wise Class 11 Maths important questions with solutions aur Class 11 Maths formulas PDF download karke revise karna sabse efficient tarika hai — especially complex numbers, straight lines, limits and derivatives, aur probability jaise numerically-heavy chapters ke liye. Class 11 Maths NCERT book PDF download karke NCERT ke original questions ke saath match bhi kar sakte ho.
Ye chapter Class 11 Maths syllabus 2026-27 me kitna important hai?
Complex Numbers and Quadratic Equations algebra section ka core chapter hai aur direct-formula-based hone ki wajah se scoring bhi maana jaata hai. Exact marks weightage CBSE ke official curriculum document se hi verify karni chahiye — is guide me koi weightage number invent nahi kiya gaya hai. Class 11 Maths syllabus 2026-27 PDF ncert.nic.in se download karke chapter list cross-check kar sakte ho.
Class 11 Maths — Saare Chapters
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