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Punjab Board Class 11 Mathematics

Exercise 1.3 Solutions — Complex Numbers

Complete worked solutions for Exercise 1.3 from Chapter 1 (Complex Numbers), Punjab Board Class 11 Mathematics. Each question shows every step and names the formula used.

Mathematics 11 (Punjab Board)

Exercise 1.3 Solutions

Worked solutions for Punjab Board Class 11 Mathematics, Chapter 1: Complex Numbers.

Question 1: Factorize each of the following into linear factors.

Sum of squares ko \(a^2 + b^2 = (a+ib)(a-ib)\) se toro. Quadratics ke liye quadratic formula se roots nikalo, phir \((z - r_1)(z - r_2)\) likho.

(i) \(a^2 + 4b^2\)

\[ a^2 + 4b^2 = a^2 - (2bi)^2 = (a + 2bi)(a - 2bi) \]

(ii) \(9a^2 + 16b^2\)

\[ 9a^2 + 16b^2 = (3a)^2 - (4bi)^2 = (3a + 4bi)(3a - 4bi) \]

(iii) \(3x^2 + 3y^2\)

\[ 3x^2 + 3y^2 = 3(x^2 + y^2) = 3(x + iy)(x - iy) \]

(iv) \(144x^2 + 225y^2\)

\[ 144x^2 + 225y^2 = (12x)^2 - (15yi)^2 = (12x + 15yi)(12x - 15yi) \]

(v) \(z^2 - 2z - 1\)

\[ z = \frac{2 \pm \sqrt{4 + 4}}{2} = 1 \pm \sqrt{2} \implies (z - 1 - \sqrt{2})(z - 1 + \sqrt{2}) \]

(vi) \(z^2 + 6z + 13\)

\[ z = \frac{-6 \pm \sqrt{36 - 52}}{2} = \frac{-6 \pm 4i}{2} = -3 \pm 2i \implies (z + 3 - 2i)(z + 3 + 2i) \]

(vii) \(z^2 + 4z + 5\)

\[ z = \frac{-4 \pm \sqrt{16 - 20}}{2} = \frac{-4 \pm 2i}{2} = -2 \pm i \implies (z + 2 - i)(z + 2 + i) \]

(viii) \(2z^2 - 22z + 65\)

\[ z = \frac{22 \pm \sqrt{484 - 520}}{4} = \frac{22 \pm 6i}{4} = \frac{11 \pm 3i}{2} \]
\[ 2\left(z - \frac{11+3i}{2}\right)\left(z - \frac{11-3i}{2}\right) \]
Answer: (i) \((a+2bi)(a-2bi)\) (ii) \((3a+4bi)(3a-4bi)\) (iii) \(3(x+iy)(x-iy)\) (iv) \((12x+15yi)(12x-15yi)\) (v) \((z-1-\sqrt{2})(z-1+\sqrt{2})\) (vi) \((z+3-2i)(z+3+2i)\) (vii) \((z+2-i)(z+2+i)\) (viii) \(2\left(z-\frac{11+3i}{2}\right)\left(z-\frac{11-3i}{2}\right)\)

Question 2: Factorize each of the following into linear factors.

Cube ke liye sum/difference of cubes ka formula yaad rakho: \(z^3 + a^3 = (z+a)(z^2-az+a^2)\). Biquadratics mein \(y = z^2\) substitute karo.

(i) \(z^3 + 8\)

\[ z^3 + 8 = (z+2)(z^2 - 2z + 4), \quad z^2-2z+4 = 0 \implies z = 1 \pm \sqrt{3}\,i \]
\[ (z+2)(z - 1 - \sqrt{3}\,i)(z - 1 + \sqrt{3}\,i) \]

(ii) \(z^3 + 27\)

\[ z^3 + 27 = (z+3)(z^2 - 3z + 9), \quad z^2-3z+9 = 0 \implies z = \frac{3 \pm 3\sqrt{3}\,i}{2} \]
\[ (z+3)\left(z - \frac{3+3\sqrt{3}\,i}{2}\right)\left(z - \frac{3-3\sqrt{3}\,i}{2}\right) \]

(iii) \(z^3 - 2z^2 + 16z - 32\)

\[ z^2(z-2) + 16(z-2) = (z-2)(z^2+16) = (z-2)(z+4i)(z-4i) \]

(iv) \(z^4 + 21z^2 - 100\)

\[ y = z^2: \; y^2 + 21y - 100 = 0 \implies y = \frac{-21 \pm 29}{2} = 4 \text{ or } -25 \]
\[ z^2 = 4 \implies z = \pm 2, \qquad z^2 = -25 \implies z = \pm 5i \]
\[ (z-2)(z+2)(z-5i)(z+5i) \]

(v) \(z^4 - 16\)

\[ (z^2-4)(z^2+4) = (z-2)(z+2)(z-2i)(z+2i) \]

(vi) \(z^4 + 3z^2 - 4\)

\[ y^2 + 3y - 4 = (y+4)(y-1) = 0 \implies z = \pm 1,\; \pm 2i \]
\[ (z-1)(z+1)(z-2i)(z+2i) \]

(vii) \(z^4 + 5z^2 + 6\)

\[ y^2 + 5y + 6 = (y+3)(y+2) = 0 \implies z^2 = -3 \text{ or } -2 \implies z = \pm i\sqrt{3},\; \pm i\sqrt{2} \]
\[ (z-i\sqrt{3})(z+i\sqrt{3})(z-i\sqrt{2})(z+i\sqrt{2}) \]

(viii) \(z^4 - 32z^2 - 3969\)

\[ y^2 - 32y - 3969 = 0 \implies y = \frac{32 \pm 130}{2} = 81 \text{ or } -49 \]
\[ z = \pm 9,\; \pm 7i \implies (z-9)(z+9)(z-7i)(z+7i) \]
Answer: (i) \((z+2)(z-1-\sqrt{3}i)(z-1+\sqrt{3}i)\) (ii) \((z+3)\left(z-\frac{3+3\sqrt{3}i}{2}\right)\left(z-\frac{3-3\sqrt{3}i}{2}\right)\) (iii) \((z-2)(z+4i)(z-4i)\) (iv) \((z-2)(z+2)(z-5i)(z+5i)\) (v) \((z-2)(z+2)(z-2i)(z+2i)\) (vi) \((z-1)(z+1)(z-2i)(z+2i)\) (vii) \((z-i\sqrt{3})(z+i\sqrt{3})(z-i\sqrt{2})(z+i\sqrt{2})\) (viii) \((z-9)(z+9)(z-7i)(z+7i)\)

Common mistake: \(z^2 = -25\) se \(z = \pm 5\) likh dena ghalat hai — negative ke square root mein \(i\) lazmi hai: \(z = \pm 5i\).

Question 3: Find the roots of \(z^4 + 7z^2 - 144 = 0\) and hence express it as a product of linear factors.

\(y = z^2\) substitution se biquadratic, quadratic ban jayegi. Usay factorize karo, phir har \(y\) se \(z\) ke do roots nikalo.

\[ y^2 + 7y - 144 = 0 \implies (y+16)(y-9) = 0 \implies y = 9 \text{ or } y = -16 \]
\[ z^2 = 9 \implies z = \pm 3, \qquad z^2 = -16 \implies z = \pm 4i \]
Answer: Roots \(3, -3, 4i, -4i\); \(z^4 + 7z^2 - 144 = (z-3)(z+3)(z-4i)(z+4i)\)

Question 4: Solve the following quadratic equations.

Quadratic formula \(z = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) lagao. Jab discriminant negative ho to \(\sqrt{-k} = i\sqrt{k}\) rakho.

(i) \(2z^2 - 3z + 4 = 0\)

\[ z = \frac{3 \pm \sqrt{9 - 32}}{4} = \frac{3 \pm \sqrt{23}\,i}{4} \]

(ii) \(z^2 - 6z + 30 = 0\)

\[ z = \frac{6 \pm \sqrt{36 - 120}}{2} = \frac{6 \pm \sqrt{84}\,i}{2} = 3 \pm \sqrt{21}\,i \]

(iii) \(3z^2 - 18z + 30 = 0\)

\[ z = \frac{18 \pm \sqrt{324 - 360}}{6} = \frac{18 \pm 6i}{6} = 3 \pm i \]

(iv) \(z^2 + 4z + 13 = 0\)

\[ z = \frac{-4 \pm \sqrt{16 - 52}}{2} = \frac{-4 \pm 6i}{2} = -2 \pm 3i \]

(v) \(2z^2 + 6z + 9 = 0\)

\[ z = \frac{-6 \pm \sqrt{36 - 72}}{4} = \frac{-6 \pm 6i}{4} = \frac{-3 \pm 3i}{2} \]

(vi) \(3z^2 - 5z + 7 = 0\)

\[ z = \frac{5 \pm \sqrt{25 - 84}}{6} = \frac{5 \pm \sqrt{59}\,i}{6} \]
Answer: (i) \(\frac{3 \pm \sqrt{23}\,i}{4}\) (ii) \(3 \pm \sqrt{21}\,i\) (iii) \(3 \pm i\) (iv) \(-2 \pm 3i\) (v) \(\frac{-3 \pm 3i}{2}\) (vi) \(\frac{5 \pm \sqrt{59}\,i}{6}\)

Question 5: Solve the following equations.

Common factor bahar nikalo jahan mumkin ho, phir biquadratic ko \(y = z^2\) se quadratic mein badlo.

(i) \(2z^4 - 32 = 0\)

\[ z^4 = 16 \implies z = \pm 2,\; \pm 2i \]

(ii) \(3z^5 - 243z = 0\)

\[ 3z(z^4 - 81) = 0 \implies z = 0 \text{ or } z^4 = 81 \implies z = 0,\; \pm 3,\; \pm 3i \]

(iii) \(5z^5 - 5z = 0\)

\[ 5z(z^4 - 1) = 0 \implies z = 0,\; \pm 1,\; \pm i \]

(iv) \(z^4 - 5z^2 + 4 = 0\)

\[ y^2 - 5y + 4 = (y-4)(y-1) = 0 \implies z = \pm 2,\; \pm 1 \]

(v) \(4z^4 - 25z^2 + 21 = 0\)

\[ y = \frac{25 \pm \sqrt{625 - 336}}{8} = \frac{25 \pm 17}{8} \implies y = \frac{21}{4} \text{ or } 1 \implies z = \pm\frac{\sqrt{21}}{2},\; \pm 1 \]

(vi) \(z^2 + z + 1 = 0\)

\[ z = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{3}\,i}{2} \]
Answer: (i) \(\pm 2, \pm 2i\) (ii) \(0, \pm 3, \pm 3i\) (iii) \(0, \pm 1, \pm i\) (iv) \(\pm 2, \pm 1\) (v) \(\pm\frac{\sqrt{21}}{2}, \pm 1\) (vi) \(\frac{-1 \pm \sqrt{3}\,i}{2}\)