Mathematics 11 (Punjab Board)
Exercise 1.2 Solutions
Worked solutions for Punjab Board Class 11 Mathematics, Chapter 1: Complex Numbers.
Question 1: Find the real values of \(x\) and \(y\) in each of the following.
Har part mein pehle right side ko simplify karke \(a + ib\) form mein lao, phir equality of complex numbers se real aur imaginary parts compare karo.
(i) \(x + iy + 2 - 3i = i(5-i)(3+4i)\)
\[
\text{RHS} = (5i - i^2)(3+4i) = (1+5i)(3+4i) = 3 + 4i + 15i + 20i^2 = -17 + 19i
\]
\[
(x+2) + i(y-3) = -17 + 19i \implies x + 2 = -17,\; y - 3 = 19
\]
(ii) \((x+iy)(1-i) = (2-3i)(-5+5i)\left(-i+\frac{3}{5}\right)\)
\[
(2-3i)(-5+5i) = -10 + 10i + 15i - 15i^2 = 5 + 25i
\]
\[
(5+25i)\left(-i+\frac{3}{5}\right) = -5i + 3 - 25i^2 + 15i = 28 + 10i
\]
\[
(x+iy)(1-i) = (x+y) + i(y-x) = 28 + 10i \implies x + y = 28,\; y - x = 10
\]
Adding: \(2y = 38 \implies y = 19\), so \(x = 9\).
(iii) \(\frac{x}{2+i} + \frac{y}{3-i} = 4 + 5i\)
\[
\frac{x(2-i)}{5} + \frac{y(3+i)}{10} = 4 + 5i \implies \left(\frac{2x}{5} + \frac{3y}{10}\right) + i\left(\frac{-x}{5} + \frac{y}{10}\right) = 4 + 5i
\]
\[
4x + 3y = 40, \qquad -2x + y = 50
\]
From the second, \(y = 50 + 2x\); substituting: \(4x + 150 + 6x = 40 \implies x = -11\), \(y = 28\).
Answer: (i) \(x = -19, y = 22\) (ii) \(x = 9, y = 19\) (iii) \(x = -11, y = 28\)
Question 2: If \(z_1 = -13 + 24i\) and \(z_2 = x + yi\), find the real values of \(x\) and \(y\) such that \(z_1 - z_2 = -27 + 15i\).
\(z_1 - z_2\) ko expand karke \(a + ib\) form mein likho, phir real aur imaginary parts compare karo.
\[
(-13 - x) + (24 - y)i = -27 + 15i
\]
\[
-13 - x = -27 \implies x = 14, \qquad 24 - y = 15 \implies y = 9
\]
Answer: \(x = 14, \; y = 9\)
Question 3: Find the real values of \(x\) and \(y\) if:
\((x+iy)^2\) ko expand karo: \((x^2-y^2) + 2xyi\). Phir parts compare karo aur system \(x^2-y^2 = a\), \(2xy = b\) solve karo — \((x^2+y^2)^2 = a^2+b^2\) shortcut se.
(i) \((x+iy)^2 = 25 + 60i\)
\[
x^2 - y^2 = 25, \quad 2xy = 60; \qquad (x^2+y^2)^2 = 625 + 3600 = 4225 = 65^2
\]
\[
x^2 + y^2 = 65 \implies x^2 = 45,\; y^2 = 20
\]
Since \(xy = 30 > 0\), \(x\) and \(y\) have the same sign: \(x = \pm 3\sqrt{5}, y = \pm 2\sqrt{5}\).
(ii) \((x+iy)^2 = 64 + 48i\)
\[
x^2 - y^2 = 64, \quad 2xy = 48; \qquad (x^2+y^2)^2 = 4096 + 2304 = 6400 = 80^2
\]
\[
x^2 + y^2 = 80 \implies x^2 = 72,\; y^2 = 8
\]
Since \(xy = 24 > 0\): \(x = \pm 6\sqrt{2}, y = \pm 2\sqrt{2}\) (same sign).
(iii) \((x+iy)^2 = \frac{2i-3}{3+i}\)
\[
\frac{-3+2i}{3+i} \times \frac{3-i}{3-i} = \frac{-9 + 3i + 6i - 2i^2}{10} = -\frac{7}{10} + \frac{9}{10}i
\]
\[
x^2 - y^2 = -\frac{7}{10}, \quad 2xy = \frac{9}{10}; \qquad (x^2+y^2)^2 = \frac{49}{100} + \frac{81}{100} = \frac{13}{10}
\]
\[
x^2 + y^2 = \frac{\sqrt{130}}{10} \implies x^2 = \frac{\sqrt{130}-7}{20},\; y^2 = \frac{\sqrt{130}+7}{20}
\]
Since \(xy = \frac{9}{20} > 0\), \(x\) and \(y\) take the same sign.
Answer: (i) \(x = \pm 3\sqrt{5}, y = \pm 2\sqrt{5}\) (same sign) (ii) \(x = \pm 6\sqrt{2}, y = \pm 2\sqrt{2}\) (same sign) (iii) \(x = \pm\sqrt{\frac{\sqrt{130}-7}{20}}, y = \pm\sqrt{\frac{\sqrt{130}+7}{20}}\) (same sign)
Question 4: If \(z_1 = 2 + 3i\) and \(z_2 = 1 - \alpha\), find the real value of \(\alpha\) such that \(\text{Im}(z_1 z_2) = 7\).
\(z_1 z_2\) multiply karo (note \(z_2\) real hai), phir imaginary part ko 7 ke barabar rakho.
\[
z_1 z_2 = (2+3i)(1-\alpha) = 2(1-\alpha) + 3(1-\alpha)i
\]
\[
\text{Im}(z_1 z_2) = 3(1-\alpha) = 7 \implies 1 - \alpha = \frac{7}{3} \implies \alpha = -\frac{4}{3}
\]
Answer: \(\alpha = -\frac{4}{3}\)
Question 5: If \(z_1 = x + yi\) and \(z_2 = a + bi\), find \(x, y, a, b\) such that \(z_1 + z_2 = 10 + 4i\) and \(z_1 - z_2 = 6 + 2i\).
Dono equations ko add karo to \(z_1\) milega, subtract karo to \(z_2\) milega — bilkul simultaneous equations ki tarah.
\[
2z_1 = (10+4i) + (6+2i) = 16 + 6i \implies z_1 = 8 + 3i
\]
\[
2z_2 = (10+4i) - (6+2i) = 4 + 2i \implies z_2 = 2 + i
\]
Answer: \(x = 8, y = 3, a = 2, b = 1\)
Question 6: Show that \(\forall z_1, z_2 \in \mathbb{C}\), \(\overline{z_1 z_2} = \bar{z_1} \cdot \bar{z_2}\).
\(z_1 = a+ib\), \(z_2 = c+id\) rakho. Pehle product ka conjugate nikalo, phir conjugates ka product — dono barabar aayenge.
\[
z_1 z_2 = (ac - bd) + (ad + bc)i \implies \overline{z_1 z_2} = (ac - bd) - (ad + bc)i
\]
\[
\bar{z_1} \cdot \bar{z_2} = (a-ib)(c-id) = (ac - bd) - (ad + bc)i
\]
Answer: Hence proved — both sides equal \((ac-bd) - (ad+bc)i\).
Question 7: Find the square root of the following complex numbers.
\(\sqrt{a+ib} = x+iy\) rakho, to \(x^2-y^2 = a\), \(2xy = b\), aur \((x^2+y^2)^2 = a^2+b^2\). Signs \(2xy = b\) se decide karo.
(i) \(-7 - 24i\)
\[
x^2 - y^2 = -7,\; 2xy = -24; \quad (x^2+y^2)^2 = 49 + 576 = 625 \implies x^2 + y^2 = 25
\]
\[
x^2 = 9,\; y^2 = 16; \quad xy = -12 < 0 \implies \text{opposite signs}
\]
(ii) \(8 - 6i\)
\[
x^2 - y^2 = 8,\; 2xy = -6; \quad (x^2+y^2)^2 = 64 + 36 = 100 \implies x^2 + y^2 = 10
\]
\[
x^2 = 9,\; y^2 = 1; \quad xy = -3 < 0 \implies \text{opposite signs}
\]
(iii) \(-15 - 36i\)
\[
x^2 - y^2 = -15,\; 2xy = -36; \quad (x^2+y^2)^2 = 225 + 1296 = 1521 = 39^2 \implies x^2 + y^2 = 39
\]
\[
x^2 = 6,\; y^2 = 27; \quad xy = -18 < 0 \implies \text{opposite signs}
\]
(iv) \(119 + 120i\)
\[
x^2 - y^2 = 119,\; 2xy = 120; \quad (x^2+y^2)^2 = 14161 + 14400 = 28561 = 169^2 \implies x^2 + y^2 = 169
\]
\[
x^2 = 144,\; y^2 = 25; \quad xy = 60 > 0 \implies \text{same sign}
\]
Answer: (i) \(\pm(3 - 4i)\) (ii) \(\pm(3 - i)\) (iii) \(\pm(\sqrt{6} - 3\sqrt{3}\,i)\) (iv) \(\pm(12 + 5i)\)
Common mistake: Sirf \(x^2\) aur \(y^2\) nikal kar ruk na jao — \(2xy = b\) ka sign check karna lazmi hai warna signs ghalat ho jayenge.
Question 8: Find the square root of \(13 - 20\sqrt{3}\,i\) and represent it on an Argand diagram.
Wohi square-root method: \(x^2-y^2 = 13\), \(2xy = -20\sqrt{3}\). Argand diagram par dono roots origin ke gird opposite points honge.
\[
(x^2+y^2)^2 = 169 + 1200 = 1369 = 37^2 \implies x^2 + y^2 = 37
\]
\[
x^2 = 25,\; y^2 = 12; \quad xy = -10\sqrt{3} < 0 \implies x = \pm 5,\; y = \mp 2\sqrt{3}
\]
Argand diagram: Plot the points \((5, -2\sqrt{3})\) and \((-5, 2\sqrt{3})\). They lie on opposite sides of the origin, on the line \(y = -\frac{2\sqrt{3}}{5}x\), each at distance \(\sqrt{37}\) from the origin.
Answer: \(\pm(5 - 2\sqrt{3}\,i)\)
Question 9: Find the real values of \(x\) and \(y\) if \((-7+i)x + (y+iy) + (-1-5i) = i(11-i)\).
Right side simplify karo (\(i(11-i) = 1+11i\)), left side ko \((a) + (b)i\) form mein jama karo, phir parts compare karo.
\[
(-7x + y - 1) + i(x + y - 5) = 1 + 11i
\]
\[
-7x + y = 2, \qquad x + y = 16
\]
Subtracting: \(-8x = -14 \implies x = \frac{7}{4}\), then \(y = 16 - \frac{7}{4} = \frac{57}{4}\).
Answer: \(x = \frac{7}{4}, \; y = \frac{57}{4}\)
Question 10: Find the real values of \(x\) and \(y\) if \((5-2i)x + (iy) + 3 = i(11-i) - 4i\).
Right side: \(i(11-i) - 4i = 1 + 7i\). Left side ko real/imaginary mein divide karo.
\[
(5x + 3) + i(y - 2x) = 1 + 7i
\]
\[
5x + 3 = 1 \implies x = -\frac{2}{5}, \qquad y - 2x = 7 \implies y = 7 - \frac{4}{5} = \frac{31}{5}
\]
Answer: \(x = -\frac{2}{5}, \; y = \frac{31}{5}\)
Question 11: Find the real values of \(u\) and \(v\) if \(\frac{u-2}{2+i} + \frac{v-3}{2-i} = 4i\).
Har fraction ke denominator ko uske conjugate se multiply karke real banao, phir real aur imaginary parts compare karo.
\[
\frac{(u-2)(2-i)}{5} + \frac{(v-3)(2+i)}{5} = 4i
\]
\[
\text{Real: } \frac{2(u-2) + 2(v-3)}{5} = 0 \implies u + v = 5
\]
\[
\text{Imag: } \frac{-(u-2) + (v-3)}{5} = 4 \implies -u + v = 21
\]
Adding: \(2v = 26 \implies v = 13\), so \(u = -8\).
Answer: \(u = -8, \; v = 13\)
Question 12: If \(z_1 = 4 + 5i\) and \(z_2 = \alpha - 2i\), find the real value of \(\alpha\) such that \(\text{Re}(z_1 z_2) = 20\).
\(z_1 z_2\) multiply karke real part nikalo, use 20 ke barabar rakho.
\[
z_1 z_2 = (4+5i)(\alpha-2i) = (4\alpha + 10) + (5\alpha - 8)i
\]
\[
\text{Re}(z_1 z_2) = 4\alpha + 10 = 20 \implies \alpha = \frac{5}{2}
\]
Answer: \(\alpha = \frac{5}{2}\)
