Mathematics 11 (Punjab Board)
Exercise 1.4 Solutions
Worked solutions for Punjab Board Class 11 Mathematics, Chapter 1: Complex Numbers.
Question 1: Find the three cube roots of: (i) 8 (ii) -8 (iii) 27 (iv) 64 (v) -125
Kisi number \(a\) ke teen cube roots \(a^{1/3}, a^{1/3}\omega, a^{1/3}\omega^2\) hote hain, jahan \(1, \omega, \omega^2\) cube roots of unity hain.
\[
\text{(i) } 2,\; 2\omega,\; 2\omega^2 \qquad \text{(ii) } -2,\; -2\omega,\; -2\omega^2 \qquad \text{(iii) } 3,\; 3\omega,\; 3\omega^2
\]
\[
\text{(iv) } 4,\; 4\omega,\; 4\omega^2 \qquad \text{(v) } -5,\; -5\omega,\; -5\omega^2
\]
Answer: (i) \(2, 2\omega, 2\omega^2\) (ii) \(-2, -2\omega, -2\omega^2\) (iii) \(3, 3\omega, 3\omega^2\) (iv) \(4, 4\omega, 4\omega^2\) (v) \(-5, -5\omega, -5\omega^2\)
Question 2: Find the four fourth roots of 16, 81, 625. Also show that their sum is zero in each case.
Positive real number \(a\) ke chaar fourth roots \(\pm a^{1/4}\) aur \(\pm i\,a^{1/4}\) hote hain. Inko jama karo to sab cancel ho kar zero banta hai.
(i) 16: roots \(2, -2, 2i, -2i\); sum \(= 2 - 2 + 2i - 2i = 0\).
(ii) 81: roots \(3, -3, 3i, -3i\); sum \(= 3 - 3 + 3i - 3i = 0\).
(iii) 625: roots \(5, -5, 5i, -5i\); sum \(= 5 - 5 + 5i - 5i = 0\).
Answer: Fourth roots are \(\pm a^{1/4}, \pm i\,a^{1/4}\); each sum is \(0\).
Question 3: If \(1, \omega, \omega^2\) are the cube roots of unity, show that \(1 + \omega^n + \omega^{2n} = 3\) where \(n\) is a multiple of 3.
\(n = 3k\) rakho, phir \(\omega^n = (\omega^3)^k = 1\) use karo.
\[
n = 3k \implies \omega^n = (\omega^3)^k = 1^k = 1, \qquad \omega^{2n} = (\omega^3)^{2k} = 1
\]
\[
1 + \omega^n + \omega^{2n} = 1 + 1 + 1 = 3
\]
Answer: Hence proved.
Question 4: Evaluate: (i) \(\left(\frac{-1+\sqrt{3}\,i}{2}\right)^3 + \left(\frac{-1-\sqrt{3}\,i}{2}\right)^3\) (ii) \((-1+\sqrt{3}\,i)^5 + (-1-\sqrt{3}\,i)^5\)
Pehchano ke \(\frac{-1+\sqrt{3}\,i}{2} = \omega\) aur \(\frac{-1-\sqrt{3}\,i}{2} = \omega^2\). Phir \(\omega^3 = 1\) aur \(1+\omega+\omega^2 = 0\) lagao.
(i)
\[
\omega^3 + (\omega^2)^3 = \omega^3 + \omega^6 = 1 + 1 = 2
\]
(ii) Note \(-1+\sqrt{3}\,i = 2\omega\) and \(-1-\sqrt{3}\,i = 2\omega^2\).
\[
(2\omega)^5 + (2\omega^2)^5 = 32\omega^5 + 32\omega^{10} = 32\omega^2 + 32\omega = 32(\omega + \omega^2) = 32(-1) = -32
\]
Answer: (i) \(2\) (ii) \(-32\)
Question 5: Show that \((1-\omega+\omega^2)(1-\omega^2+\omega^4)(1-\omega^4+\omega^8)\cdots\) to \(2n\) factors \(= 2^{2n}\).
Har factor ko simplify karo: \(1-\omega+\omega^2 = -2\omega\) aur \(1-\omega^2+\omega^4 = -2\omega^2\). Factors alternate hote hain, har pair ka product \(4\) banta hai.
Since \(\omega^2 = -1-\omega\): \(1-\omega+\omega^2 = 1-\omega-1-\omega = -2\omega\).
Since \(\omega^4 = \omega\): \(1-\omega^2+\omega^4 = 1-\omega^2+\omega = -2\omega^2\).
\[
\text{Product} = \underbrace{(-2\omega)(-2\omega^2)}_{=4\omega^3=4} \cdot \underbrace{(-2\omega)(-2\omega^2)}_{=4} \cdots \text{($n$ pairs)} = 4^n = 2^{2n}
\]
Answer: Hence proved — the product equals \(2^{2n}\).
Question 6: Evaluate \(\left(\frac{i+\sqrt{3}}{2}\right)^3 + \left(\frac{i-\sqrt{3}}{2}\right)^3\).
Dono numbers ko polar form mein likho: \(\frac{\sqrt{3}+i}{2} = \cos 30^\circ + i\sin 30^\circ\) aur \(\frac{-\sqrt{3}+i}{2} = \cos 150^\circ + i\sin 150^\circ\). Cube karne par dono \(i\) ban jate hain.
\[
\left(\frac{\sqrt{3}+i}{2}\right)^3 = \left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)^3 = \cos\frac{\pi}{2} + i\sin\frac{\pi}{2} = i
\]
\[
\left(\frac{-\sqrt{3}+i}{2}\right)^3 = \left(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}\right)^3 = \cos\frac{5\pi}{2} + i\sin\frac{5\pi}{2} = i
\]
\[
\text{Sum} = i + i = 2i
\]
Answer: \(2i\) (the printed statement's \(-1\) does not follow; direct computation gives \(2i\))
Question 7: Evaluate \(\sum_{k=0}^{5} \omega^{2k}\).
Terms likho aur \(\omega^3 = 1\) se powers reduce karo — phir \(1+\omega+\omega^2 = 0\) do dafa milega.
\[
\sum_{k=0}^{5} \omega^{2k} = 1 + \omega^2 + \omega^4 + \omega^6 + \omega^8 + \omega^{10} = 1 + \omega^2 + \omega + 1 + \omega^2 + \omega
\]
\[
= 2(1 + \omega + \omega^2) = 2 \cdot 0 = 0
\]
Answer: \(0\)
Question 8: If \(\omega\) is an imaginary cube root of unity, prove the standard identity \(\frac{a+b\omega+c\omega^2}{a\omega+b\omega^2+c} = \omega^2\).
Denominator ko \(\omega^2\) se multiply karo: \(\omega^2(a\omega+b\omega^2+c) = a\omega^3 + b\omega^4 + c\omega^2 = a + b\omega + c\omega^2\), jo numerator hai.
\[
\omega^2(a\omega + b\omega^2 + c) = a\omega^3 + b\omega^4 + c\omega^2 = a(1) + b(\omega) + c\omega^2 = a + b\omega + c\omega^2
\]
Hence \(\text{numerator} = \omega^2 \times \text{denominator}\).
\[
\frac{a+b\omega+c\omega^2}{a\omega+b\omega^2+c} = \omega^2
\]
Answer: Hence proved: \(\frac{a+b\omega+c\omega^2}{a\omega+b\omega^2+c} = \omega^2\)
Common mistake: As printed (denominator \(a\omega^2+b\omega+c\), RHS \(\omega\)), the identity holds only when \(b = c\); the standard correct form proved above has denominator \(a\omega+b\omega^2+c\) and value \(\omega^2\).
Question 9: Reduce \(\frac{a\omega^3+b\omega^7+c\omega^{19}}{a\omega^4+b\omega^6+c\omega^{20}}\) using \(\omega^3 = 1\).
Har power ko 3 se divide karke remainder nikalo: \(\omega^7 = \omega\), \(\omega^{19} = \omega\), \(\omega^4 = \omega\), \(\omega^6 = 1\), \(\omega^{20} = \omega^2\).
\[
\text{Numerator} = a(1) + b\omega + c\omega = a + (b+c)\omega
\]
\[
\text{Denominator} = a\omega + b(1) + c\omega^2 = a\omega + b + c\omega^2
\]
\[
\frac{a\omega^3+b\omega^7+c\omega^{19}}{a\omega^4+b\omega^6+c\omega^{20}} = \frac{a+(b+c)\omega}{a\omega+b+c\omega^2}
\]
This reduced fraction equals \(\omega\) only under extra conditions on \(a, b, c\) — e.g. with \(a = 1, b = c = 0\) it equals \(\omega^3/\omega^4 = \omega^2 \neq \omega\). So the claim " \(= \omega\) for all \(a,b,c\)" does not hold as printed.
Answer: Reduced form \(\frac{a+(b+c)\omega}{a\omega+b+c\omega^2}\); not identically equal to \(\omega\)
