Mathematics Class 11
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🎓 Class 11 Mathematics CBSE Theory Ch 4 — Complex Numbers and Quadratic Equations ⏱ ~15 min
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Targeting Class 11 level in Algebra, with Advanced difficulty.

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End-of-Chapter Exercises

Exercise 4.1 — Express in standard form a + ib
1. \((5i)\!\left(-\dfrac{3}{5}i\right)\)
\(5i\cdot\bigl(-\tfrac{3}{5}i\bigr)=-3i^2=3\). So \(3+0i\).
2. \(i^9+i^{19}\)
\(i^9=i\) (9 mod 4 = 1); \(i^{19}=i^3=-i\) (19 mod 4 = 3). Sum \(=i+(-i)=0\).
3. \(i^{-39}\)
\(i^{-39}=i^{-39+40}=i^1=i\).
4. \(3(7+i7)+i(7+i7)\)
\(=21+21i+7i+7i^2=21+28i-7=14+28i\).
5. \((1-i)-(-1+i6)\)
\(=1-i+1-6i=2-7i\).
6. \(\left(\dfrac{1}{5}+i\dfrac{2}{5}\right)-\!\left(4+i\dfrac{5}{2}\right)\)
Real: \(1/5-4=-19/5\). Imag: \(2/5-5/2=4/10-25/10=-21/10\). Result: \(-\dfrac{19}{5}-\dfrac{21}{10}i\).
7. \(\left[\!\left(\dfrac{1}{3}+i\dfrac{7}{3}\right)+\!\left(4+i\dfrac{1}{3}\right)\right]-\!\left(-\dfrac{4}{3}+i\right)\)
Sum first two: \((1/3+4)+i(7/3+1/3)=13/3+i\,8/3\). Subtract third: \((13/3+4/3)+i(8/3-1)=17/3+i\,5/3\).
8. \((1-i)^4\)
\((1-i)^2=1-2i+i^2=-2i\). Then \((1-i)^4=(-2i)^2=4i^2=-4\). So \(-4+0i\).
9. \(\left(\dfrac{1}{3}+3i\right)^3\)
Use \((a+b)^3=a^3+3a^2b+3ab^2+b^3\) with \(a=1/3, b=3i\): \((1/3)^3+3(1/3)^2(3i)+3(1/3)(3i)^2+(3i)^3=1/27+i-9-27i=-242/27-26i\).
10. \(\left(-2-\dfrac{1}{3}i\right)^3\)
Let \(a=-2, b=-i/3\). \(a^3=-8\). \(3a^2b=3\cdot 4\cdot(-i/3)=-4i\). \(3ab^2=3(-2)(i^2/9)=2/3\). \(b^3=-i^3/27=i/27\). Sum: \(-8+2/3+i(-4+1/27)=-22/3-(107/27)i\).
Exercise 4.1 (continued) — Modulus, Conjugate, Standard form
11. Find the multiplicative inverse of \(4-3i\).
\(z^{-1}=\bar z/|z|^2=(4+3i)/(16+9)=(4+3i)/25=\dfrac{4}{25}+\dfrac{3}{25}i\).
12. Multiplicative inverse of \(\sqrt 5+3i\).
\(|z|^2=5+9=14\). \(z^{-1}=(\sqrt 5-3i)/14=\dfrac{\sqrt 5}{14}-\dfrac{3}{14}i\).
13. Multiplicative inverse of \(-i\).
\((-i)^{-1}=1/(-i)=i/(-i\cdot i)=i/1=i\) (since \(-i\cdot i=-i^2=1\)). So inverse is \(i\).
14. Express \((1-i)^4/(1+i)^4\) in standard form.
\(\left(\dfrac{1-i}{1+i}\right)^4\). Simplify base: multiply num and denom by \(1-i\): \(\dfrac{(1-i)^2}{(1+i)(1-i)}=\dfrac{-2i}{2}=-i\). So answer is \((-i)^4=i^4=1\).
Exercise 4.2 — Polar form
1. \(z=1-i\)
\(r=\sqrt 2\). Q4 (1, −1), reference \(\pi/4\), Arg \(=-\pi/4\). \(z=\sqrt 2(\cos(-\pi/4)+i\sin(-\pi/4))\).
2. \(z=-1+i\)
\(r=\sqrt 2\). Q2 (−1, 1), reference \(\pi/4\), Arg \(=3\pi/4\). \(z=\sqrt 2(\cos(3\pi/4)+i\sin(3\pi/4))\).
3. \(z=1\)
\(r=1\), Arg \(=0\). \(z=1(\cos 0+i\sin 0)\).
4. \(z=i\)
\(r=1\), Arg \(=\pi/2\). \(z=1(\cos(\pi/2)+i\sin(\pi/2))\).
5. \(z=-1-i\)
\(r=\sqrt 2\). Q3 (−1, −1), reference \(\pi/4\), Arg \(=-3\pi/4\). \(z=\sqrt 2(\cos(-3\pi/4)+i\sin(-3\pi/4))\).
6. \(z=-3\)
\(r=3\), Arg \(=\pi\). \(z=3(\cos\pi+i\sin\pi)\).
7. \(z=\sqrt 3+i\)
\(r=2\). Q1, \(\tan\theta=1/\sqrt 3\), Arg \(=\pi/6\). \(z=2(\cos(\pi/6)+i\sin(\pi/6))\).
8. \(z=i\) (2nd part of textbook listing)
Same as Q4 — \(r=1\), Arg \(=\pi/2\).
Miscellaneous Exercise on Chapter 4
1. Evaluate \(\left[i^{18}+\left(\dfrac{1}{i}\right)^{25}\right]^3\).
\(i^{18}=i^2=-1\) (18 mod 4 = 2). \((1/i)^{25}=(-i)^{25}=(-1)^{25}i^{25}=-i\) (25 mod 4 = 1). Sum: \(-1-i\). Cube: \((-1-i)^3=-(1+i)^3\). \((1+i)^2=2i\), so \((1+i)^3=2i(1+i)=2i+2i^2=-2+2i\). Therefore \(-(1+i)^3=2-2i\).
2. For any two complex numbers \(z_1, z_2\), prove \(\text{Re}(z_1z_2)=\text{Re}\,z_1\,\text{Re}\,z_2-\text{Im}\,z_1\,\text{Im}\,z_2\).
Let \(z_1=a+ib, z_2=c+id\). Product \(=(ac-bd)+i(ad+bc)\). Real part \(=ac-bd=\text{Re}\,z_1\cdot\text{Re}\,z_2-\text{Im}\,z_1\cdot\text{Im}\,z_2\). \(\square\)
3. Reduce \(\left(\dfrac{1}{1-4i}-\dfrac{2}{1+i}\right)\left(\dfrac{3-4i}{5+i}\right)\) to standard form.
First: \(\dfrac{1}{1-4i}=\dfrac{1+4i}{17}\). Second: \(\dfrac{2}{1+i}=\dfrac{2(1-i)}{2}=1-i\). So the bracket = \(\dfrac{1+4i}{17}-1+i=\dfrac{1+4i-17+17i}{17}=\dfrac{-16+21i}{17}\). Now \(\dfrac{3-4i}{5+i}=\dfrac{(3-4i)(5-i)}{26}=\dfrac{15-3i-20i+4i^2}{26}=\dfrac{11-23i}{26}\). Product: \(\dfrac{(-16+21i)(11-23i)}{17\cdot 26}=\dfrac{-176+368i+231i-483i^2}{442}=\dfrac{307+599i}{442}\).
5. If \(z_1=2-i, z_2=1+i\), find \(\left|\dfrac{z_1+z_2+1}{z_1-z_2+1}\right|\).
\(z_1+z_2+1=4\); \(z_1-z_2+1=2-2i\). Ratio: \(\dfrac{4}{2-2i}=\dfrac{4(2+2i)}{8}=1+i\). Modulus: \(\sqrt 2\).
9. Find the modulus of \(\dfrac{1+i}{1-i}-\dfrac{1-i}{1+i}\).
\(\dfrac{1+i}{1-i}=\dfrac{(1+i)^2}{2}=i\). \(\dfrac{1-i}{1+i}=\dfrac{(1-i)^2}{2}=-i\). Difference: \(i-(-i)=2i\). Modulus = 2.
12. Find the number of non-zero integral solutions of \(|1-i|^x=2^x\).
\(|1-i|=\sqrt 2\). So equation is \((\sqrt 2)^x=2^x\), i.e. \(2^{x/2}=2^x\), so \(x/2=x\), giving \(x=0\). Excluding 0, there are 0 non-zero integral solutions.
14. If \(\left(\dfrac{1+i}{1-i}\right)^m=1\), find the least positive integer \(m\).
\(\dfrac{1+i}{1-i}=i\) (from Q9). So we need \(i^m=1\), least positive \(m=4\).
Activity: Conjugate Pairs of Quadratic Roots
L4 Analyse
Materials: Calculator, paper.
Predict: If \(ax^2+bx+c=0\) has real coefficients and one root is \(p+iq\) with \(q\ne 0\), what must the other root be?
  1. Solve \(x^2+1=0\). Roots: \(\pm i\) — a conjugate pair.
  2. Solve \(x^2+2x+5=0\). Use the quadratic formula: \(x=(-2\pm\sqrt{-16})/2=-1\pm 2i\) — a conjugate pair.
  3. Solve \(x^2+x+1=0\). Roots: \((-1\pm\sqrt{-3})/2\) — a conjugate pair.
  4. Why must complex roots of a real-coefficient polynomial come in conjugate pairs? Try \(P(\bar z)\) when \(P(z)=0\) and the coefficients are real.
If \(P(z)=0\) and the coefficients of \(P\) are all real, then taking conjugates of both sides: \(\overline{P(z)}=P(\bar z)\) (because conjugate distributes through real-coefficient polynomial), so \(P(\bar z)=\bar 0=0\). Hence \(\bar z\) is also a root. This means non-real roots of real-coefficient polynomials always come in conjugate pairs — a foundational fact for the entire chapter.

Consolidation Competency-Based Questions

Scenario: A student is preparing for the board exam on Chapter 4. He wants a quick decision tree: when faced with a complex-number problem, what's the right first move?
Q1. \(\left(\dfrac{1+i}{1-i}\right)^{2025}=\)
L3 Apply
  • (a) 1
  • (b) i
  • (c) −1
  • (d) −i
Answer: (b) i. Base = \(i\); 2025 mod 4 = 1, so \(i^{2025}=i\).
Q2. The argument of \(z=1+i\sqrt 3\) (in degrees) is:
L3 Apply
Answer: Q1, \(\tan\theta=\sqrt 3\), so \(\theta=60°\) (\(=\pi/3\) rad).
Q3. (T/F) "If \(|z|=|z-1|\), then z lies on the line Re(z) = 1/2."
L5 Evaluate
True. \(|z|=|z-1|\) means z is equidistant from 0 and 1 — geometrically, the perpendicular bisector of the segment from 0 to 1, which is the line Re(z) = 1/2.
Q4. If \(\alpha,\beta\) are roots of \(x^2-2x+5=0\), find \(|\alpha|\) and \(|\beta|\). Are they equal? Why?
L4 Analyse
Solution: \(x=(2\pm\sqrt{-16})/2=1\pm 2i\). \(|1\pm 2i|=\sqrt 5\). They are equal because the two roots are conjugates of each other, and conjugates always have the same modulus.
Q5. Design: a 2D rigid-body rotation by 90° anti-clockwise can be encoded as multiplication by \(i\). Express the rotation by 45° anti-clockwise as multiplication by some complex number \(c\). What is \(c\) and what is \(|c|\)?
L6 Create
Solution: Rotation by angle \(\alpha\) is multiplication by \(\cos\alpha+i\sin\alpha\). For \(\alpha=45°\): \(c=\cos 45°+i\sin 45°=\dfrac{1}{\sqrt 2}(1+i)\). Modulus \(|c|=1\) (rotations preserve length). Note \(c^2=i\) — two 45° rotations equal one 90° rotation, as expected.

Consolidation Assertion–Reason

Assertion (A): Non-real roots of a polynomial with real coefficients occur in conjugate pairs.
Reason (R): The conjugate of a sum/product equals the sum/product of conjugates.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a). If \(P(z)=0\), taking conjugate gives \(P(\bar z)=\overline{P(z)}=0\), provided coefficients are real. The "provided" step uses R precisely.
Assertion (A): \(|i^{2025}|=1\).
Reason (R): Modulus is multiplicative: \(|z^n|=|z|^n\).
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a). \(|i|=1\), so \(|i^{2025}|=1^{2025}=1\). R is the rule that gives A.

Chapter Summary

Key formulas at a glance
  • Definition: \(i^2=-1\). A complex number \(z=a+ib\) has Re \(=a\), Im \(=b\).
  • Equality: \(a+ib=c+id\iff a=c\) and \(b=d\).
  • Operations: Add componentwise; multiply by expansion: \((a+ib)(c+id)=(ac-bd)+i(ad+bc)\).
  • Inverse: \(z^{-1}=\bar z/|z|^2\). Division: multiply by conjugate of denominator.
  • Powers of i: \(i^4=1\), so \(i^n\) depends only on \(n\bmod 4\) (\(1,i,-1,-i\)).
  • Square root of negatives: \(\sqrt{-a}=\sqrt a\,i\) for \(a>0\). The identity \(\sqrt a\sqrt b=\sqrt{ab}\) FAILS when both are negative — factor out \(i\) first.
  • Modulus: \(|z|=\sqrt{a^2+b^2}\); \(|z_1z_2|=|z_1|\,|z_2|\); \(z\bar z=|z|^2\).
  • Conjugate: \(\bar z=a-ib\); \(\overline{z_1+z_2}=\bar z_1+\bar z_2\); \(\overline{z_1z_2}=\bar z_1\bar z_2\); \(\overline{(\bar z)}=z\).
  • Argand plane: \(z=x+iy\) ↔ point \(P(x,y)\). \(|z|\) = distance OP. Conjugate = mirror across real axis.
  • Polar form: \(z=r(\cos\theta+i\sin\theta)\), \(r=|z|\), principal Arg \(\theta\in(-\pi,\pi]\).
  • Multiplication in polar: moduli multiply, arguments add. (Foundation of De Moivre's theorem.)

Historical Note

The fact that the square of a negative number does not exist in the real number system was recognised by the Greeks. But the credit goes to the Indian mathematician Mahavira (850 CE), who first stated this difficulty clearly. He mentions in his work "Ganitasara Sangraha" as in nature of things a negative (quantity) is not a square (quantity), it has, therefore, no square root.

Bhaskara II, another Indian mathematician, also writes in his work Bijaganita, written in 1150 CE: "There is no square root of a negative quantity, for it is not a square."

Cardan (1545) considered the problem of solving

\[x+y=10,\quad xy=40.\]

He obtained \(x=5+\sqrt{-15}\) and \(y=5-\sqrt{-15}\) as the solution, which was discarded by him by saying that these numbers are 'useless'. Albert Girard (about 1625) accepted square root of negative numbers and said that this will enable us to get as many roots as the degree of the polynomial equation. Euler was the first to introduce the symbol \(i\) for \(\sqrt{-1}\), and W. R. Hamilton (about 1830) regarded the complex number \(a+ib\) as an ordered pair of real numbers \((a,b)\) thus giving it a purely mathematical definition and avoiding use of the so called 'imaginary numbers'.

Frequently Asked Questions

How do you simplify i⁹ + i¹⁹?
i⁹ = i (since 9 mod 4 = 1) and i¹⁹ = i³ = −i (since 19 mod 4 = 3). So i⁹ + i¹⁹ = i + (−i) = 0.
What is the value of √(−25) × √(−16)?
Factor out i first: √(−25) = 5i and √(−16) = 4i. Product = 5i · 4i = 20i² = −20. Direct application of √a·√b = √(ab) would give the wrong answer +20.
What is the chapter summary of Class 11 Maths Chapter 4?
Complex numbers extend the real numbers by adding i with i² = −1. Every complex number z = a + ib has a real part a, an imaginary part b, a modulus |z| = √(a² + b²), a conjugate z̄ = a − ib, and a polar form r(cos θ + i sin θ). Powers of i cycle every 4. Conjugate, modulus, and quadratic-equation tools complete the algebra.
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