Mathematics 11 (Punjab Board)
Exercise 1.5 Solutions
Worked solutions for Punjab Board Class 11 Mathematics, Chapter 1: Complex Numbers.
Question 1: Plot the following points given in polar coordinates.
Polar point \((r, \theta)\) plot karne ke liye origin se \(\theta\) angle banao aur \(r\) distance lo. Agar \(r\) negative ho to opposite direction (angle \(+\ \pi\)) mein jao.
(i) \((2, 75^\circ)\): 2 units along \(75^\circ\) from the positive x-axis.
(ii) \((-3, 120^\circ)\): 3 units along \(120^\circ + 180^\circ = 300^\circ\).
(iii) \((2, \pi/6)\): 2 units along \(30^\circ\).
(iv) \((5, 5\pi/6)\): 5 units along \(150^\circ\).
(v) \((-5/2, \pi/3)\): 2.5 units along \(\pi/3 + \pi = 4\pi/3 = 240^\circ\).
(vi) \((-3, -2\pi/3)\): 3 units along \(-2\pi/3 + \pi = \pi/3 = 60^\circ\).
(vii) \((-9/2, -19\pi/12)\): 4.5 units along \(-19\pi/12 + \pi = -7\pi/12 = -105^\circ\).
(viii) \((-5/2, 5\pi/12)\): 2.5 units along \(5\pi/12 + \pi = 17\pi/12 = 255^\circ\).
Answer: Points plotted as described above on the Argand plane.
Question 2: Express the following complex numbers in polar form \(r(\cos\theta + i\sin\theta)\).
\(r = \sqrt{x^2+y^2}\) nikalo aur \(\theta\) ke liye quadrant dekho: pehle \(\tan^{-1}|y/x|\) lo, phir quadrant ke hisab se adjust karo.
(i) \(4 + 3i\): \(r = 5\), \(\theta = \tan^{-1}(3/4) \approx 36.87^\circ\).
\[
5(\cos 36.87^\circ + i\sin 36.87^\circ)
\]
(ii) \(1 + i\): \(r = \sqrt{2}\), \(\theta = \pi/4\).
\[
\sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)
\]
(iii) \(\frac{1}{2} + \frac{\sqrt{3}}{2}i\): \(r = 1\), \(\theta = \pi/3\).
\[
\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}
\]
(iv) \(\frac{5}{2} - \frac{5\sqrt{3}}{2}i\): \(r = 5\); the point is in the 4th quadrant, so \(\theta = -\pi/3\).
\[
5\left(\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right)
\]
(v) \(\frac{1-i}{1+i} = -i\): \(r = 1\), \(\theta = -\pi/2\).
\[
\cos\left(-\frac{\pi}{2}\right) + i\sin\left(-\frac{\pi}{2}\right)
\]
(vi) \(\frac{\sqrt{3}+i}{1+\sqrt{3}\,i} = \frac{\sqrt{3}-i}{2}\): \(r = 1\), \(\theta = -\pi/6\).
\[
\cos\left(-\frac{\pi}{6}\right) + i\sin\left(-\frac{\pi}{6}\right)
\]
Answer: Polar forms as shown above.
Common mistake: \(\tan^{-1}(y/x)\) blindly lagane se quadrant ghalat ho jata hai — hamesha \((x, y)\) ka quadrant check karo.
Question 3: Convert each complex number to rectangular form \(x+iy\).
\(x = r\cos\theta\) aur \(y = r\sin\theta\) use karo. Standard angles (jaise \(5\pi/3, 7\pi/6\)) ki values yaad rakho.
(i) \(4\left(\cos\frac{5\pi}{3} + i\sin\frac{5\pi}{3}\right) = 4\left(\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) = 2 - 2\sqrt{3}\,i\)
(ii) \(\frac{3}{2}\left(\cos\frac{7\pi}{6} + i\sin\frac{7\pi}{6}\right) = \frac{3}{2}\left(-\frac{\sqrt{3}}{2} - i\frac{1}{2}\right) = -\frac{3\sqrt{3}}{4} - \frac{3}{4}i\)
(iii) \(|z| = 7, \arg(z) = \frac{23\pi}{12} = 345^\circ\): \(z = 7(\cos 345^\circ + i\sin 345^\circ) \approx 6.76 - 1.81i\)
(iv) \(|z| = 11, \arg(z) = -\frac{11\pi}{12} = -165^\circ\): \(z \approx -10.62 - 2.85i\)
(v) \(|z| = \frac{10}{3}, \arg(z) = -\frac{17\pi}{12} = 105^\circ\): \(z = \frac{10}{3}(\cos 105^\circ + i\sin 105^\circ) \approx -0.86 + 3.22i\)
(vi) \(2\cos(-33^\circ) + i\,2\sin(-33^\circ) \approx 1.68 - 1.09i\)
Answer: (i) \(2 - 2\sqrt{3}i\) (ii) \(-\frac{3\sqrt{3}}{4} - \frac{3}{4}i\) (iii) \(\approx 6.76 - 1.81i\) (iv) \(\approx -10.62 - 2.85i\) (v) \(\approx -0.86 + 3.22i\) (vi) \(\approx 1.68 - 1.09i\)
Question 4: Given \(z_1 = 9\left(\cos\frac{5\pi}{4} + i\sin\frac{5\pi}{4}\right)\) and \(z_2 = 5\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)\), find (i) \(z_1 + z_2\) (ii) \(z_1 - z_2\) (iii) \(z_1 z_2\) (iv) \(\frac{z_1}{z_2}\).
Addition/subtraction ke liye pehle rectangular form mein badlo. Multiplication/division seedha polar form mein: moduli multiply/divide, arguments add/subtract.
\(z_1 = 9\left(-\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}\right) \approx -6.36 - 6.36i\), \(z_2 = 5\left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = 2.5 + 4.33i\).
(i) \(z_1 + z_2 \approx -3.86 - 2.03i\)
(ii) \(z_1 - z_2 \approx -8.86 - 10.69i\)
(iii)
\[
z_1 z_2 = 45\left(\cos\frac{19\pi}{12} + i\sin\frac{19\pi}{12}\right) \approx 11.65 - 43.47i
\]
(iv)
\[
\frac{z_1}{z_2} = \frac{9}{5}\left(\cos\frac{11\pi}{12} + i\sin\frac{11\pi}{12}\right) \approx -1.74 + 0.47i
\]
Answer: (i) \(\approx -3.86 - 2.03i\) (ii) \(\approx -8.86 - 10.69i\) (iii) \(45\left(\cos\frac{19\pi}{12} + i\sin\frac{19\pi}{12}\right)\) (iv) \(\frac{9}{5}\left(\cos\frac{11\pi}{12} + i\sin\frac{11\pi}{12}\right)\)
Question 5: Given \(z_1 = 7\left(\cos\frac{23\pi}{12} + i\sin\frac{23\pi}{12}\right)\) and \(z_2 = 11\left(\cos\frac{11\pi}{12} + i\sin\frac{11\pi}{12}\right)\), find (i) \(z_1 + z_2\) (ii) \(z_1 - z_2\) (iii) \(z_1 z_2\) (iv) \(\frac{z_1}{z_2}\).
Wohi method: addition/subtraction rectangular form mein, multiplication/division polar form mein.
\(z_1 = 7(\cos 345^\circ + i\sin 345^\circ) \approx 6.76 - 1.81i\), \(z_2 = 11(\cos 165^\circ + i\sin 165^\circ) \approx -10.62 + 2.85i\).
(i) \(z_1 + z_2 \approx -3.86 + 1.03i\)
(ii) \(z_1 - z_2 \approx 17.38 - 4.65i\)
(iii)
\[
z_1 z_2 = 77\left(\cos\frac{34\pi}{12} + i\sin\frac{34\pi}{12}\right) = 77\left(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}\right) = -66.68 + 38.5i
\]
(iv)
\[
\frac{z_1}{z_2} = \frac{7}{11}(\cos\pi + i\sin\pi) = -\frac{7}{11}
\]
Answer: (i) \(\approx -3.86 + 1.03i\) (ii) \(\approx 17.38 - 4.65i\) (iii) \(-66.68 + 38.5i\) (iv) \(-\frac{7}{11}\)
Question 6: Prove that (i) \(\arg(z_1 z_2) = \arg(z_1) + \arg(z_2)\) (ii) \(\arg\left(\frac{z_1}{z_2}\right) = \arg(z_1) - \arg(z_2)\).
\(z_1 = r_1(\cos\theta_1 + i\sin\theta_1)\) aur \(z_2 = r_2(\cos\theta_2 + i\sin\theta_2)\) rakho, phir polar multiplication/division ke rules apply karo.
(i)
\[
z_1 z_2 = r_1 r_2\big(\cos(\theta_1+\theta_2) + i\sin(\theta_1+\theta_2)\big) \implies \arg(z_1 z_2) = \theta_1 + \theta_2
\]
(ii)
\[
\frac{z_1}{z_2} = \frac{r_1}{r_2}\big(\cos(\theta_1-\theta_2) + i\sin(\theta_1-\theta_2)\big) \implies \arg\left(\frac{z_1}{z_2}\right) = \theta_1 - \theta_2
\]
Answer: Hence proved (up to multiples of \(2\pi\)).
Question 7: Divide \(z_1 = 6(\cos 150^\circ + i\sin 150^\circ)\) by \(z_2 = 3(\cos 30^\circ + i\sin 30^\circ)\) and express the result in \(x+iy\) form.
Division rule: \(\frac{r_1}{r_2}\) aur angle \(\theta_1 - \theta_2\). Phir \(\cos 120^\circ = -1/2\), \(\sin 120^\circ = \sqrt{3}/2\) rakho.
\[
\frac{z_1}{z_2} = \frac{6}{3}\big(\cos(150^\circ-30^\circ) + i\sin(150^\circ-30^\circ)\big) = 2(\cos 120^\circ + i\sin 120^\circ)
\]
\[
= 2\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = -1 + \sqrt{3}\,i
\]
Answer: \(-1 + \sqrt{3}\,i\)
Question 8: Multiply \(z_1 = 2(\cos 60^\circ + i\sin 60^\circ)\) by \(z_2 = 5(\cos 90^\circ + i\sin 90^\circ)\) and express the result in \(x+iy\) form.
Multiplication rule: moduli multiply, angles add. \(\cos 150^\circ = -\sqrt{3}/2\), \(\sin 150^\circ = 1/2\).
\[
z_1 z_2 = 10\big(\cos(60^\circ+90^\circ) + i\sin(60^\circ+90^\circ)\big) = 10(\cos 150^\circ + i\sin 150^\circ)
\]
\[
= 10\left(-\frac{\sqrt{3}}{2} + i\frac{1}{2}\right) = -5\sqrt{3} + 5i
\]
Answer: \(-5\sqrt{3} + 5i\)
Question 9: Find the modulus and argument of \(z = -2 - 2i\).
Modulus \(\sqrt{x^2+y^2}\) se, aur argument ke liye quadrant dekho — \((-2,-2)\) third quadrant mein hai, is liye \(\theta = \pi + \tan^{-1}(1)\).
\[
|z| = \sqrt{(-2)^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2}
\]
\[
\arg(z) = \pi + \tan^{-1}\!\left(\frac{-2}{-2}\right) = \pi + \frac{\pi}{4} = \frac{5\pi}{4} \quad (+ 2n\pi)
\]
Answer: \(|z| = 2\sqrt{2}\), \(\arg(z) = \frac{5\pi}{4} + 2n\pi\)
Question 10: Write \(\arg(z - 2 + i) = \frac{2\pi}{3}\) in Cartesian form.
\(z = x+iy\) rakho, phir \(\tan(\arg) = \frac{\text{Imag}}{\text{Real}}\) use karo. Yaad rakho: yeh aik ray (half-line) hai, poori line nahi.
\[
z - 2 + i = (x-2) + i(y+1), \qquad \tan\frac{2\pi}{3} = \frac{y+1}{x-2}
\]
\[
-\sqrt{3} = \frac{y+1}{x-2} \implies y + 1 = -\sqrt{3}(x-2) \implies \sqrt{3}\,x + y - 2\sqrt{3} + 1 = 0
\]
Since \(\arg = 2\pi/3\) (second quadrant), we need \(x - 2 < 0\) and \(y + 1 > 0\).
Answer: \(\sqrt{3}\,x + y - 2\sqrt{3} + 1 = 0\), with \(x < 2\) and \(y > -1\)
Question 11: If \(\arg\left(\frac{z-1+2i}{z+1-2i}\right) = \frac{9\pi}{4}\) where \(z = x+iy\), show the Cartesian equation of the locus.
\(\frac{9\pi}{4} = 2\pi + \frac{\pi}{4}\), to effectively \(\arg = \pi/4\) aur \(\tan = 1\). Fraction ko simplify karke \(\frac{\text{Im}}{\text{Re}} = 1\) rakho.
\[
\frac{z-1+2i}{z+1-2i} = \frac{(x-1)+i(y+2)}{(x+1)+i(y-2)} \times \frac{(x+1)-i(y-2)}{(x+1)-i(y-2)}
\]
\[
= \frac{(x^2+y^2-5) + i(4x+2y)}{(x+1)^2+(y-2)^2}
\]
\[
\frac{4x+2y}{x^2+y^2-5} = \tan\frac{\pi}{4} = 1 \implies x^2 + y^2 - 4x - 2y - 5 = 0
\]
Answer: \(x^2 + y^2 - 4x - 2y - 5 = 0\) (the printed \(+2y\) does not follow; the correct sign is \(-2y\))
Question 12: If \(\arg(z-2-3i) - \arg(z+2+3i) = 2\pi\) where \(z = x+iy\), show that \(2y = 3x\).
Argument ka difference, quotient ka argument hota hai: \(\arg\frac{A}{B} = 2\pi\) ka matlab \(\frac{A}{B}\) positive real hai, yaani uska imaginary part zero.
\[
\arg\left(\frac{(x-2)+i(y-3)}{(x+2)+i(y+3)}\right) = 2\pi \implies \text{Im}\left(\frac{A}{B}\right) = 0
\]
\[
\text{Im numerator} = -(x-2)(y+3) + (y-3)(x+2) = -6x + 4y = 0
\]
\[
4y = 6x \implies 2y = 3x
\]
Answer: \(2y = 3x\)
Question 13: Solve \(|z-2i| = |z+2|\).
Yeh un points ka set hai jo \((0, 2)\) aur \((-2, 0)\) se barabar faslay par hain — perpendicular bisector, jo aik straight line hoti hai. Dono sides square karo.
\[
|x + i(y-2)| = |(x+2) + iy| \implies x^2 + (y-2)^2 = (x+2)^2 + y^2
\]
\[
x^2 + y^2 - 4y + 4 = x^2 + 4x + 4 + y^2 \implies -4y = 4x
\]
Answer: \(y = -x\) (the perpendicular bisector of the segment joining \((0,2)\) and \((-2,0)\))
Question 14: Solve \(|5z+4+i| = |5z-3+2i|\).
\(z = x+iy\) rakho, modulus expand karo, square karo — \(x^2\) aur \(y^2\) terms cancel ho jayenge aur aik line bachegi.
\[
|(5x+4) + i(5y+1)| = |(5x-3) + i(5y+2)|
\]
\[
(5x+4)^2 + (5y+1)^2 = (5x-3)^2 + (5y+2)^2
\]
\[
40x + 10y + 17 = -30x + 20y + 13 \implies 70x - 10y + 4 = 0
\]
Answer: \(35x - 5y + 2 = 0\)
Question 15: Determine the set of points satisfying \(|3z-2+i| = |3z+i|\).
Same perpendicular-bisector idea: dono sides ko \(|(3x-2)+i(3y+1)| = |3x+i(3y+1)|\) likho aur square karo.
\[
(3x-2)^2 + (3y+1)^2 = (3x)^2 + (3y+1)^2 \implies (3x-2)^2 = 9x^2
\]
\[
9x^2 - 12x + 4 = 9x^2 \implies x = \frac{1}{3}
\]
Answer: The vertical line \(x = \frac{1}{3}\)
Question 16: If \(z = x+iy\) and \(w = \frac{1-iz}{z-i}\), show that \(|w| = 1 \iff z\) is real.
\(|w|^2 = 1\) rakho aur numerator/denominator ke moduli compare karo — \(x^2\) terms cancel ho kar sirf \(y\) bachega.
\[
w = \frac{1 - i(x+iy)}{x+iy-i} = \frac{(1+y) - ix}{x + i(y-1)}, \qquad |w|^2 = \frac{(1+y)^2 + x^2}{x^2 + (y-1)^2}
\]
\[
|w| = 1 \iff (1+y)^2 + x^2 = x^2 + (y-1)^2 \iff 1 + 2y + y^2 = y^2 - 2y + 1 \iff 4y = 0
\]
\[
y = 0 \iff z = x \text{ is real}
\]
Answer: Hence proved — \(|w| = 1\) exactly when \(y = 0\), i.e. \(z\) is real.
Question 17: If \(z_1, z_2\) are different complex numbers with \(|z_2| = 1\), find \(\left|\frac{z_2-z_1}{1-\bar{z_1}z_2}\right|\).
\(|w|^2 = w\bar{w}\) likho aur \(|z_2| = 1\) (yaani \(z_2\bar{z_2} = 1\)) use karo — numerator aur denominator barabar niklenge.
\[
|w|^2 = \frac{(z_2-z_1)(\bar{z_2}-\bar{z_1})}{(1-\bar{z_1}z_2)(1-z_1\bar{z_2})} = \frac{1 - z_2\bar{z_1} - z_1\bar{z_2} + |z_1|^2}{1 - z_1\bar{z_2} - \bar{z_1}z_2 + |z_1|^2}
\]
Numerator and denominator are identical, so \(|w|^2 = 1\).
Answer: \(1\)
Question 18: An AC source has voltage \(V = 120\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)\) V and impedance \(Z = \frac{1+i\sqrt{3}}{2}\) ohms. Find the current.
Ohm's law \(I = V/Z\). Pehle \(Z\) ko polar form mein likho: \(Z = \cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\) (kyunke \(r = 1\)). Phir divide karo.
\[
I = \frac{120\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)}{\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}} = 120\left(\cos\left(\frac{\pi}{4}-\frac{\pi}{3}\right) + i\sin\left(\frac{\pi}{4}-\frac{\pi}{3}\right)\right)
\]
\[
I = 120\left(\cos\left(-\frac{\pi}{12}\right) + i\sin\left(-\frac{\pi}{12}\right)\right)
\]
Answer: \(I = 120\left(\cos\left(-\frac{\pi}{12}\right) + i\sin\left(-\frac{\pi}{12}\right)\right)\) amperes
Question 19: An AC circuit has voltage \(V = 90 + 30i\) V and impedance \(Z = 3 - 6i\) ohms. Find the current in rectangular and polar form.
\(I = V/Z\) — normal complex division (conjugate se multiply). Phir \(r = \sqrt{x^2+y^2}\), \(\theta = \tan^{-1}(y/x)\).
\[
I = \frac{90+30i}{3-6i} \times \frac{3+6i}{3+6i} = \frac{270 + 540i + 90i + 180i^2}{45} = \frac{90 + 630i}{45} = 2 + 14i
\]
\[
r = \sqrt{4 + 196} = \sqrt{200} = 10\sqrt{2}, \qquad \theta = \tan^{-1}(7) \approx 81.87^\circ
\]
Answer: \(I = 2 + 14i = 10\sqrt{2}\,(\cos 81.87^\circ + i\sin 81.87^\circ)\) amperes
Question 20: Encrypt the word "CODE" using the complex key \(k = 2 - i\), then decrypt it back.
Letters ko numbers do (A=1, B=2, ...). Encryption: number \(\times\) key. Decryption: key se divide karo.
C = 3, O = 15, D = 4, E = 5.
Encrypt: \(3(2-i) = 6-3i\); \(15(2-i) = 30-15i\); \(4(2-i) = 8-4i\); \(5(2-i) = 10-5i\).
Decrypt (divide by \(2-i\), multiply by conjugate):
\[
\frac{6-3i}{2-i} = \frac{(6-3i)(2+i)}{5} = \frac{15}{5} = 3 \to \text{C}
\]
Similarly \(30-15i \to 15\) (O), \(8-4i \to 4\) (D), \(10-5i \to 5\) (E).
Answer: Encrypted: \(6-3i, 30-15i, 8-4i, 10-5i\); decrypted: C, O, D, E
Question 21: Encrypt the word "QUIZ" using the complex key \(k = 3 - 3i\), then decrypt it back.
Wohi method: multiply for encryption, divide for decryption.
Q = 17, U = 21, I = 9, Z = 26.
Encrypt: \(17(3-3i) = 51-51i\); \(21(3-3i) = 63-63i\); \(9(3-3i) = 27-27i\); \(26(3-3i) = 78-78i\).
Decrypt: \(\frac{51-51i}{3-3i} = 17 \to\) Q; similarly \(63-63i \to 21\) (U), \(27-27i \to 9\) (I), \(78-78i \to 26\) (Z).
Answer: Encrypted: \(51-51i, 63-63i, 27-27i, 78-78i\); decrypted: Q, U, I, Z
Question 22: Encrypt the word "CLASS" using the complex key \(k = -3 + 4i\), then decrypt it back.
Wohi method: multiply for encryption, divide for decryption (note \(|-3+4i|^2 = 25\)).
C = 3, L = 12, A = 1, S = 19, S = 19.
Encrypt: \(3(-3+4i) = -9+12i\); \(12(-3+4i) = -36+48i\); \(1(-3+4i) = -3+4i\); \(19(-3+4i) = -57+76i\) (twice).
Decrypt: \(\frac{-9+12i}{-3+4i} = \frac{(-9+12i)(-3-4i)}{25} = \frac{75}{25} = 3 \to\) C; similarly \(-36+48i \to 12\) (L), \(-3+4i \to 1\) (A), \(-57+76i \to 19\) (S).
Answer: Encrypted: \(-9+12i, -36+48i, -3+4i, -57+76i, -57+76i\); decrypted: C, L, A, S, S
