Mathematics 11 (Punjab Board)
Exercise 1.1 Solutions
Worked solutions for Punjab Board Class 11 Mathematics, Chapter 1: Complex Numbers.
Question 1: Find the multiplicative inverse of each complex number.
Kisi complex number \(z\) ka multiplicative inverse \(\frac{1}{z}\) hota hai. Ise nikalne ke liye numerator aur denominator dono ko denominator ke conjugate se multiply karo, phir \(a^2 + b^2\) se divide karo.
(i) \((-4, 7)\), i.e. \(z = -4 + 7i\)
\[
\frac{1}{-4+7i} = \frac{1}{-4+7i} \times \frac{-4-7i}{-4-7i} = \frac{-4-7i}{(-4)^2 + 7^2} = \frac{-4-7i}{65} = -\frac{4}{65} - \frac{7}{65}i
\]
(ii) \((\sqrt{2}, -\sqrt{5})\), i.e. \(z = \sqrt{2} - \sqrt{5}\,i\)
\[
\frac{1}{\sqrt{2}-\sqrt{5}\,i} = \frac{\sqrt{2}+\sqrt{5}\,i}{(\sqrt{2})^2 + (\sqrt{5})^2} = \frac{\sqrt{2}+\sqrt{5}\,i}{7} = \frac{\sqrt{2}}{7} + \frac{\sqrt{5}}{7}i
\]
(iii) \((1, 0)\), i.e. \(z = 1\)
\[
\frac{1}{1} = 1
\]
Answer: (i) \(\left(-\frac{4}{65}, -\frac{7}{65}\right)\) (ii) \(\left(\frac{\sqrt{2}}{7}, \frac{\sqrt{5}}{7}\right)\) (iii) \((1, 0)\)
Question 2: Separate into real and imaginary parts.
Fraction ko \(a + ib\) form mein lane ke liye denominator ke conjugate se multiply karo, phir \(i^2 = -1\) rakh kar real aur imaginary parts alag karo.
(i) \(\frac{2-7i}{4+5i}\)
\[
\frac{2-7i}{4+5i} \times \frac{4-5i}{4-5i} = \frac{8 - 10i - 28i + 35i^2}{16 + 25} = \frac{-27 - 38i}{41} = -\frac{27}{41} - \frac{38}{41}i
\]
(ii) \(\frac{(-2+3i)^2}{1+i}\)
\[
(-2+3i)^2 = 4 - 12i + 9i^2 = -5 - 12i
\]
\[
\frac{-5-12i}{1+i} \times \frac{1-i}{1-i} = \frac{-5 + 5i - 12i + 12i^2}{2} = \frac{-17 - 7i}{2} = -\frac{17}{2} - \frac{7}{2}i
\]
(iii) \(\frac{i}{1+i}\)
\[
\frac{i}{1+i} \times \frac{1-i}{1-i} = \frac{i - i^2}{2} = \frac{1 + i}{2} = \frac{1}{2} + \frac{1}{2}i
\]
(iv) \(\frac{(4+3i)^2}{4-3i}\)
\[
(4+3i)^2 = 16 + 24i + 9i^2 = 7 + 24i
\]
\[
\frac{7+24i}{4-3i} \times \frac{4+3i}{4+3i} = \frac{28 + 21i + 96i + 72i^2}{16 + 9} = \frac{-44 + 117i}{25} = -\frac{44}{25} + \frac{117}{25}i
\]
Answer: (i) \(-\frac{27}{41} - \frac{38}{41}i\) (ii) \(-\frac{17}{2} - \frac{7}{2}i\) (iii) \(\frac{1}{2} + \frac{1}{2}i\) (iv) \(-\frac{44}{25} + \frac{117}{25}i\)
Common mistake: Conjugate se multiply karte waqt numerator mein signs ka khayal rakho — \((a+ib)(a-ib) = a^2 + b^2\) hota hai, \(a^2 - b^2\) nahi.
Question 3: Prove that \(\bar{z} = z\) iff \(z\) is real.
Dono directions prove karni hain: pehle maan lo \(z\) real hai to \(\bar{z} = z\) dikhao, phir maan lo \(\bar{z} = z\) to \(z\) real dikhao. \(z = a + ib\) rakho.
Let \(z = a + ib\) with \(a, b \in \mathbb{R}\), so \(\bar{z} = a - ib\).
(⟹) If \(z\) is real, then \(b = 0\), so \(z = a\) and \(\bar{z} = a - 0i = a = z\).
(⟸) If \(\bar{z} = z\), then \(a - ib = a + ib\), so \(-ib = ib\), i.e. \(2ib = 0\). Since \(i \neq 0\), \(b = 0\), hence \(z = a\), a real number.
Answer: Hence proved — \(\bar{z} = z\) exactly when \(z\) is real.
Question 4: For \(z \in \mathbb{C}\), show that: (i) \(\frac{z + \bar{z}}{2} = \text{Re}(z)\) (ii) \(\frac{z - \bar{z}}{2i} = \text{Im}(z)\) (iii) \(|z|^2 = z \cdot \bar{z}\)
\(z = a + ib\) aur \(\bar{z} = a - ib\) rakh kar left side simplify karo — imaginary parts cancel ho jayengi.
(i)
\[
\frac{z + \bar{z}}{2} = \frac{(a+ib) + (a-ib)}{2} = \frac{2a}{2} = a = \text{Re}(z)
\]
(ii)
\[
\frac{z - \bar{z}}{2i} = \frac{(a+ib) - (a-ib)}{2i} = \frac{2ib}{2i} = b = \text{Im}(z)
\]
(iii)
\[
|z|^2 = a^2 + b^2, \qquad z \cdot \bar{z} = (a+ib)(a-ib) = a^2 - i^2b^2 = a^2 + b^2
\]
Answer: All three identities hold, since both sides equal \(a\), \(b\), and \(a^2 + b^2\) respectively.
Question 5: Express \(\frac{\bar{z_1}\,\bar{z_2}}{z_3}\) in the form \(a+ib\), where \(z_1 = 2+i\), \(z_2 = 3-2i\), \(z_3 = 1+3i\).
Pehle dono conjugates nikalo aur multiply karo, phir \(z_3\) se divide karne ke liye uske conjugate se multiply karo.
\[
\bar{z_1}\,\bar{z_2} = (2-i)(3+2i) = 6 + 4i - 3i - 2i^2 = 8 + i
\]
\[
\frac{8+i}{1+3i} \times \frac{1-3i}{1-3i} = \frac{8 - 24i + i - 3i^2}{1 + 9} = \frac{11 - 23i}{10} = \frac{11}{10} - \frac{23}{10}i
\]
Answer: \(\frac{11}{10} - \frac{23}{10}i\)
Common mistake: \((2-i)(3+2i)\) mein \(-i \times 2i = -2i^2 = +2\) hota hai — \(i^2 = -1\) lagana na bhoolo.
Question 6: Evaluate for \(z_1 = 2+7i\) and \(z_2 = -5+3i\): (i) \(|2z_1 - 4z_2|\) (ii) \(|3z_1 + 2z_2|\) (iii) \(|-7z_2 + 2z_1|\) (iv) \(|(z_1 + z_2)^{-1}|\)
Pehle bracket ke andar complex arithmetic karo, phir modulus ka formula \(\sqrt{x^2+y^2}\) lagao. Part (iv) mein \(|z^{-1}| = 1/|z|\) use karo.
(i)
\[
2z_1 - 4z_2 = (4+14i) - (-20+12i) = 24 + 2i, \quad |24+2i| = \sqrt{576+4} = \sqrt{580} = 2\sqrt{145}
\]
(ii)
\[
3z_1 + 2z_2 = (6+21i) + (-10+6i) = -4 + 27i, \quad |-4+27i| = \sqrt{16+729} = \sqrt{745}
\]
(iii)
\[
-7z_2 + 2z_1 = (35-21i) + (4+14i) = 39 - 7i, \quad |39-7i| = \sqrt{1521+49} = \sqrt{1570}
\]
(iv)
\[
z_1 + z_2 = -3 + 10i, \quad |(z_1+z_2)^{-1}| = \frac{1}{|z_1+z_2|} = \frac{1}{\sqrt{9+100}} = \frac{1}{\sqrt{109}}
\]
Answer: (i) \(2\sqrt{145}\) (ii) \(\sqrt{745}\) (iii) \(\sqrt{1570}\) (iv) \(\frac{1}{\sqrt{109}}\)
Question 7: Show that \(i^{n} + i^{n+1} + i^{n+2} + i^{n+3} = 0\) for all \(n \in \mathbb{N}\).
\(i^n\) common factor bahar nikalo, phir bracket ke andar \(i^2 = -1\) aur \(i^3 = -i\) rakho — sab cancel ho jayega.
\[
\text{LHS} = i^n(1 + i + i^2 + i^3) = i^n(1 + i - 1 - i) = i^n \cdot 0 = 0
\]
Answer: Hence proved.
Question 8: Find the least positive value of \(n\) if \(\left(\frac{1+i}{1-i}\right)^{2n} = 1\).
Pehle bracket wali fraction simplify karo (conjugate se multiply karke) — yeh \(i\) ban jayegi. Phir \(i^{2n} = 1\) se \(n\) nikalo.
\[
\frac{1+i}{1-i} \times \frac{1+i}{1+i} = \frac{(1+i)^2}{1+1} = \frac{2i}{2} = i
\]
So \(i^{2n} = 1\). Since \(i^4 = 1\), the smallest positive \(2n\) giving 1 is \(2n = 4\).
\[
n = 2
\]
Answer: \(n = 2\)
Question 9: Show that the value of \(i^n\) for \(n \in \mathbb{N}\), \(n > 4\), is \(i^r\), where \(r\) is the remainder when \(n\) is divided by 4.
Division algorithm se \(n = 4q + r\) likho (\(0 \le r < 4\)), phir \(i^n = (i^4)^q \cdot i^r\) mein \(i^4 = 1\) rakho.
\[
n = 4q + r \implies i^n = i^{4q+r} = (i^4)^q \cdot i^r = 1^q \cdot i^r = i^r
\]
Answer: Hence proved.
