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Punjab Board Class 11 Mathematics

Chapter 1: Complex Numbers — Class 11 Maths Notes

Complex numbers from the ground up: modulus, conjugate and argument, operations, polar form and De Moivre's theorem, with fully worked solutions. This page has the theory, solved textbook examples, every exercise solution, MCQs and a formula sheet for Chapter 1.

Exercise solutions — Chapter 1

Every exercise of Complex Numbers solved step by step.

Mathematics 11 (Punjab Board)

Chapter 1: Complex Numbers — Theory

Definitions, key results and theory for Punjab Board Class 11 Mathematics, Chapter 1: Complex Numbers.

What is a complex number?

A complex number is an ordered pair \((a, b)\) of real numbers, written as \(z = a + ib\), where \(a\) is the real part, \(b\) is the imaginary part, and \(i = \sqrt{-1}\) with \(i^2 = -1\). Har woh equation jiska real numbers mein koi hal na ho, jaise \(x^2 + 1 = 0\), complex numbers mein hal ho jati hai.

\[ z = a + ib, \quad \text{Re}(z) = a, \quad \text{Im}(z) = b \]

Key point: \(\text{Im}(z)\) is the real number \(b\), not \(ib\). Is ghalti se bachein — board paper mein yeh aksar pucha jata hai.

Equality of two complex numbers

Two complex numbers are equal exactly when their real parts are equal and their imaginary parts are equal.

\[ a + ib = c + id \iff a = c \text{ and } b = d \]

Key point: Yahi rule Exercise 1.2 ke "find x and y" wale sawalon ki bunyad hai: pehle dono sides ko \(x + iy\) form mein lao, phir real aur imaginary parts compare karo.

Conjugate and modulus

The conjugate of \(z = a + ib\) is \(\bar{z} = a - ib\) (sign of the imaginary part flips). The modulus \(|z|\) is the distance of \(z\) from the origin in the Argand plane.

\[ \bar{z} = a - ib, \qquad |z| = \sqrt{a^2 + b^2}, \qquad |z|^2 = z \bar{z} \]

Useful properties: \(\overline{z_1 + z_2} = \bar{z_1} + \bar{z_2}\), \(\overline{z_1 z_2} = \bar{z_1}\,\bar{z_2}\), \(\overline{\left(\frac{z_1}{z_2}\right)} = \frac{\bar{z_1}}{\bar{z_2}}\), and \(|z_1 z_2| = |z_1|\,|z_2|\).

Key point: Division mein denominator ko real banane ke liye uske conjugate se multiply karte hain: \(\frac{1}{a+ib} = \frac{a-ib}{a^2+b^2}\).

Powers of \(i\)

Powers of \(i\) repeat every four steps, so any \(i^n\) reduces to one of \(i, -1, -i, 1\) by taking the remainder of \(n\) divided by 4.

\[ i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1, \quad i^n = i^r \]

Here \(r\) is the remainder when \(n\) is divided by 4 (division algorithm: \(n = 4q + r\)).

Key point: Four consecutive powers always add to zero: \(i^n + i^{n+1} + i^{n+2} + i^{n+3} = 0\).

Square root of a complex number

To find \(\sqrt{a + ib}\), set \(\sqrt{a+ib} = x + iy\) with real \(x, y\). Squaring gives the system \(x^2 - y^2 = a\) and \(2xy = b\). A shortcut uses \((x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2 = a^2 + b^2\), so \(x^2 + y^2 = \sqrt{a^2 + b^2}\).

\[ x^2 = \frac{\sqrt{a^2+b^2} + a}{2}, \qquad y^2 = \frac{\sqrt{a^2+b^2} - a}{2} \]

Key point: \(x\) aur \(y\) ke signs \(2xy = b\) se decide hote hain: agar \(b > 0\) to dono same sign, agar \(b < 0\) to opposite signs.

Complex polynomials as a product of linear factors

Over complex numbers, every polynomial of degree \(n\) splits completely into \(n\) linear factors (Fundamental Theorem of Algebra). A sum of squares, which never factors over real numbers, factors over complex numbers using \(i\).

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

Quadratic equations with negative discriminant give conjugate-pair roots: for \(az^2 + bz + c = 0\), \(z = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\), and \(\sqrt{\text{negative}} = i\sqrt{|\text{value}|}\).

Key point: Pehle rational root ya grouping se real factors nikalo, phir bache hue quadratics ko quadratic formula se todo.

Cube roots of unity

The three cube roots of unity are \(1, \omega, \omega^2\), where \(\omega = \frac{-1 + \sqrt{3}\,i}{2}\). They satisfy two golden relations used in almost every proof of this section.

\[ \omega^3 = 1, \qquad 1 + \omega + \omega^2 = 0 \]

The three cube roots of any number \(a\) are \(a^{1/3}, a^{1/3}\omega, a^{1/3}\omega^2\). The four fourth roots of a positive real number \(a\) are \(\pm a^{1/4}, \pm i\,a^{1/4}\), and their sum is always zero.

Key point: \(1 + \omega + \omega^2 = 0\) se \(\omega + \omega^2 = -1\) milta hai — yeh substitution proofs mein bar bar kaam aata hai.

Polar form and the Argand plane

Every complex number \(z = x + iy\) can be written as \(z = r(\cos\theta + i\sin\theta)\), where \(r = |z| = \sqrt{x^2+y^2}\) is the modulus and \(\theta = \arg(z)\) is the argument — the angle from the positive real axis. The point \((x, y)\) plotted in the plane is the Argand diagram of \(z\).

\[ z = r(\cos\theta + i\sin\theta), \quad r = \sqrt{x^2+y^2}, \quad \tan\theta = \frac{y}{x} \]

Quadrant ka khayal rakhein: \(\tan^{-1}(y/x)\) hamesha sahi quadrant nahi deta. First quadrant mein \(\theta = \tan^{-1}(y/x)\), second mein \(\pi - \tan^{-1}|y/x|\), third mein \(\pi + \tan^{-1}|y/x|\), fourth mein \(-\tan^{-1}|y/x|\).

Key point: Polar form mein multiplication aur division asaan ho jate hain: moduli multiply/divide hote hain, arguments add/subtract hote hain.

Operations in polar form and argument properties

If \(z_1 = r_1(\cos\theta_1 + i\sin\theta_1)\) and \(z_2 = r_2(\cos\theta_2 + i\sin\theta_2)\), then:

\[ z_1 z_2 = r_1 r_2\big(\cos(\theta_1+\theta_2) + i\sin(\theta_1+\theta_2)\big) \]
\[ \frac{z_1}{z_2} = \frac{r_1}{r_2}\big(\cos(\theta_1-\theta_2) + i\sin(\theta_1-\theta_2)\big) \]

Hence \(\arg(z_1 z_2) = \arg(z_1) + \arg(z_2)\) and \(\arg(z_1/z_2) = \arg(z_1) - \arg(z_2)\) (up to multiples of \(2\pi\)).

Key point: \(|z - a| = r\) is a circle (centre \(a\), radius \(r\)), and \(|z - a| = |z - b|\) is the perpendicular bisector of the segment joining \(a\) and \(b\) — a straight line.

Complex numbers in the real world

In AC circuits, voltage, current and impedance are written as complex numbers (phasors). Ohm's law becomes \(V = IZ\), so current is \(I = V/Z\), computed exactly like complex division. The same multiply-by-a-complex-key idea is used in simple cryptographic ciphers: encrypt by multiplying with a key, decrypt by dividing (multiplying by the multiplicative inverse).

\[ V = IZ \quad\Longrightarrow\quad I = \frac{V}{Z} \]

Key point: Real-world sawalon mein pehle quantities ko \(a + ib\) ya polar form mein likho, phir normal complex operations apply karo.

Mathematics 11 (Punjab Board)

Solved Examples: Chapter 1 – Complex Numbers

Textbook solved examples for Punjab Board Class 11 Mathematics, Chapter 1: Complex Numbers, worked out step by step.

Example 1: If \(z_1 = (8, 9)\) and \(z_2 = (5, -6)\), find \(z_1 + z_2\), \(z_1 - z_2\) and \(z_1 \cdot z_2\).

Ordered pairs ki form mein addition aur subtraction mein corresponding parts add/subtract hote hain. Multiplication ke liye formula \((a, b)(c, d) = (ac - bd, ad + bc)\) use hota hai.

Sum:

\[ z_1 + z_2 = (8 + 5, 9 + (-6)) = (13, 3) \]

Difference:

\[ z_1 - z_2 = (8 - 5, 9 - (-6)) = (3, 15) \]

Product (using \((ac-bd, ad+bc)\)):

\[ z_1 \cdot z_2 = (8 \times 5 - 9 \times (-6),\, 8 \times (-6) + 9 \times 5) = (40 + 54, -48 + 45) = (94, -3) \]
Answer: \(z_1 + z_2 = (13, 3)\), \(z_1 - z_2 = (3, 15)\), \(z_1 \cdot z_2 = (94, -3)\)

Example 2: Divide \(z_1 = (4, 2)\) by \(z_2 = (3, -1)\).

Division ke liye pehle ordered pairs ko \(a + ib\) form mein likhte hain, phir denominator ke conjugate se numerator aur denominator dono ko multiply karte hain.

\[ \frac{z_1}{z_2} = \frac{4 + 2i}{3 - i} \times \frac{3 + i}{3 + i} \]
\[ = \frac{12 + 4i + 6i + 2i^2}{9 - i^2} = \frac{12 + 10i - 2}{9 + 1} = \frac{10 + 10i}{10} = 1 + i \]
Answer: \(1 + i\)

Example 3: If \(z = \frac{(1+2i)^2}{2-i}\), evaluate \(|z|\).

Pehle \(z\) ko simplify karke \(a + ib\) form mein lao, phir modulus ka formula \(|z| = \sqrt{a^2 + b^2}\) lagao.

\[ z = \frac{(1+2i)^2}{2-i} = \frac{1 + 4i + 4i^2}{2-i} = \frac{-3 + 4i}{2-i} \times \frac{2+i}{2+i} \]
\[ = \frac{-6 - 3i + 8i + 4i^2}{4 + 1} = \frac{-10 + 5i}{5} = -2 + i \]
\[ |z| = \sqrt{(-2)^2 + 1^2} = \sqrt{5} \]
Answer: \(\sqrt{5}\)

Example 4: If \((3+2i)(x+iy) = 5+12i\), where \(x, y \in \mathbb{R}\), find \(x\) and \(y\).

Left side ko multiply karke \(a + ib\) form mein lao, phir dono sides ke real aur imaginary parts compare karke do equations banao (equality of complex numbers).

\[ 3x + 3iy + 2ix + 2i^2y = 5 + 12i \implies (3x - 2y) + (2x + 3y)i = 5 + 12i \]

Comparing real and imaginary parts:

\[ 3x - 2y = 5 \quad \text{(i)}, \qquad 2x + 3y = 12 \quad \text{(ii)} \]

Multiply (i) by 3 and (ii) by 2, then add:

\[ 9x - 6y = 15, \quad 4x + 6y = 24 \implies 13x = 39 \implies x = 3 \]

Put \(x = 3\) in (i): \(9 - 2y = 5 \implies y = 2\).

Answer: \(x = 3, \; y = 2\)

Example 5: Find the square root of \(5 + 12i\).

\(\sqrt{5+12i} = x + iy\) maan kar dono sides ka square lo. Phir \(x^2 - y^2 = 5\) aur \(2xy = 12\) milega. Shortcut: \((x^2+y^2)^2 = 5^2 + 12^2 = 169\).

Let \(\sqrt{5+12i} = x + iy\). Then \(x^2 - y^2 = 5\) and \(2xy = 12\).

\[ (x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2 = 25 + 144 = 169 \implies x^2 + y^2 = 13 \]
\[ 2x^2 = 18 \implies x = \pm 3, \qquad 2y^2 = 8 \implies y = \pm 2 \]

Since \(2xy = 12 > 0\), \(x\) and \(y\) have the same sign.

Answer: \(\pm(3 + 2i)\)

Example 6: Factorize the polynomial \(P(z) = z^2 + (i-3)z - 3i\).

Quadratic formula \(z = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) se roots nikalo, phir \(P(z) = a(z - z_1)(z - z_2)\) likho.

\[ z = \frac{-(i-3) \pm \sqrt{(i-3)^2 - 4(1)(-3i)}}{2} = \frac{(3-i) \pm \sqrt{8 + 6i}}{2} \]

Now \(\sqrt{8+6i}\): with \(a^2 - b^2 = 8\), \(2ab = 6\), we get \((a^2+b^2)^2 = 64 + 36 = 100\), so \(a = \pm 3, b = \pm 1\) (same sign): \(\sqrt{8+6i} = \pm(3+i)\).

\[ z = \frac{(3-i) \pm (3+i)}{2} \implies z = 3 \text{ or } z = -i \]
Answer: \(P(z) = (z - 3)(z + i)\)

Example 7: Factorize the polynomial \(P(z) = z^2 - 4iz + 12\).

Yahan bhi quadratic formula lagao. Discriminant negative aayega, to \(\sqrt{-64} = 8i\) rakho.

\[ z = \frac{4i \pm \sqrt{(4i)^2 - 4(1)(12)}}{2} = \frac{4i \pm \sqrt{-16 - 48}}{2} = \frac{4i \pm 8i}{2} \]
\[ z = 6i \quad \text{or} \quad z = -2i \]
Answer: \(P(z) = (z - 6i)(z + 2i)\)

Example 8: Factorize the polynomial \(P(z) = z^3 + (1+i)z^2 + iz\).

Pehle common factor \(z\) bahar nikalo, phir bache hue quadratic ko factorize karo.

\[ P(z) = z\big(z^2 + (1+i)z + i\big) \]

For \(z^2 + (1+i)z + i = 0\): discriminant \(= (1+i)^2 - 4i = 2i - 4i = -2i = (1-i)^2\).

\[ z = \frac{-(1+i) \pm (1-i)}{2} \implies z = -i \text{ or } z = -1 \]
Answer: \(P(z) = z(z + 1)(z + i)\)

Example 9: Factorize the polynomial \(P(z) = z^3 - 3z^2 + z + 5\).

Pehle rational root test karo: \(z = -1\) rakhne par polynomial zero hota hai, is liye \((z+1)\) aik factor hai. Phir polynomial division se quadratic factor nikalo.

\[ P(-1) = -1 - 3 - 1 + 5 = 0 \implies (z+1) \text{ is a factor} \]
\[ z^3 - 3z^2 + z + 5 = (z+1)(z^2 - 4z + 5) \]

Roots of \(z^2 - 4z + 5 = 0\): \(z = \frac{4 \pm \sqrt{16-20}}{2} = \frac{4 \pm 2i}{2} = 2 \pm i\).

Answer: \(P(z) = (z + 1)(z - 2 - i)(z - 2 + i)\)

Example 10: Solve \(2z^2 - 12z + 50 = 0\) by completing the square, and express it as a product of linear factors.

Completing the square mein pehle \(2\) common lo, phir bracket ke andar perfect square banao. \((z-3)^2 = -16\) se \(z - 3 = \pm 4i\) milega.

\[ 2(z^2 - 6z) + 50 = 0 \implies 2\big[(z-3)^2 - 9\big] + 50 = 0 \]
\[ 2(z-3)^2 + 32 = 0 \implies (z-3)^2 = -16 \implies z - 3 = \pm 4i \]
\[ z = 3 + 4i \quad \text{or} \quad z = 3 - 4i \]
Answer: \(2z^2 - 12z + 50 = 2(z - 3 - 4i)(z - 3 + 4i)\)

Example 12: Express the complex number \(1 + i\sqrt{3}\) in polar form.

Modulus \(r = \sqrt{x^2+y^2}\) nikalo aur argument \(\theta = \tan^{-1}(y/x)\) — quadrant check karna na bhoolo (yahan first quadrant hai).

\[ r = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2, \qquad \theta = \tan^{-1}\!\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3} \]
Answer: \(2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)\)

Example 13: Find the product of \(5\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)\) and \(4\left(\cos\frac{3\pi}{2} + i\sin\frac{3\pi}{2}\right)\).

Polar form mein multiplication ka rule: moduli multiply karo, arguments add karo.

\[ = 20\left(\cos\left(\frac{\pi}{6} + \frac{3\pi}{2}\right) + i\sin\left(\frac{\pi}{6} + \frac{3\pi}{2}\right)\right) = 20\left(\cos\frac{5\pi}{3} + i\sin\frac{5\pi}{3}\right) \]
\[ = 20\left(\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) = 10 - 10\sqrt{3}\,i \]
Answer: \(10 - 10\sqrt{3}\,i\)

Example 14: Divide \(\frac{2}{7}\left(\cos\frac{7\pi}{6} + i\sin\frac{7\pi}{6}\right)\) by \(\frac{3}{5}\left(\cos\frac{-\pi}{2} + i\sin\frac{-\pi}{2}\right)\).

Polar form mein division ka rule: moduli divide karo, arguments subtract karo.

\[ = \frac{2/7}{3/5}\left(\cos\left(\frac{7\pi}{6} - \left(\frac{-\pi}{2}\right)\right) + i\sin\left(\frac{7\pi}{6} + \frac{\pi}{2}\right)\right) = \frac{10}{21}\left(\cos\frac{5\pi}{3} + i\sin\frac{5\pi}{3}\right) \]
\[ = \frac{10}{21}\left(\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) = \frac{5}{21} - \frac{5\sqrt{3}}{21}i \]
Answer: \(\frac{5}{21} - \frac{5\sqrt{3}}{21}i\)

Example 15: If \(z = x + iy\), write \(|3z - i| = |3z + 7|\) in terms of \(x\) and \(y\).

Dono sides ko modulus ki definition \(\sqrt{a^2+b^2}\) se likho, phir square karke simplify karo — yeh do points se barabar faslay wale points ka set hai (perpendicular bisector).

\[ |3x + i(3y-1)| = |(3x+7) + i(3y)| \]
\[ 9x^2 + (3y-1)^2 = (3x+7)^2 + 9y^2 \]
\[ 9x^2 + 9y^2 - 6y + 1 = 9x^2 + 42x + 49 + 9y^2 \implies 42x + 6y + 48 = 0 \]
Answer: \(7x + y + 8 = 0\)

Example 16: Show that \((x+2)^2 + y^2 = 8\) if \(\arg\left(\frac{z+2i}{z-2i}\right) = \frac{3\pi}{4}\) for \(z = x + iy\).

Fraction ko simplify karke real aur imaginary parts alag karo. \(\arg = 3\pi/4\) ka matlab \(\tan = -1\), yaani \(\frac{\text{Im}}{\text{Re}} = -1\).

\[ \frac{z+2i}{z-2i} = \frac{x + i(y+2)}{x + i(y-2)} \times \frac{x - i(y-2)}{x - i(y-2)} = \frac{(x^2 + y^2 - 4) + 4xi}{x^2 + (y-2)^2} \]

Since \(\tan\frac{3\pi}{4} = -1\):

\[ \frac{4x}{x^2 + y^2 - 4} = -1 \implies x^2 + y^2 + 4x - 4 = 0 \]
\[ (x+2)^2 + y^2 = 8 \]
Answer: \((x+2)^2 + y^2 = 8\)

Example 17: The impedance of a circuit is \(11(\cos 55.35^\circ + i\sin 55.35^\circ)\) ohms and the voltage is \(25(\cos 30^\circ + i\sin 30^\circ)\) V. Find the current.

Ohm's law \(V = IZ\) se \(I = V/Z\). Polar form mein divide karo: moduli ka quotient, arguments ka difference.

\[ I = \frac{25}{11}\big(\cos(30^\circ - 55.35^\circ) + i\sin(30^\circ - 55.35^\circ)\big) = \frac{25}{11}\big(\cos(-25.35^\circ) + i\sin(-25.35^\circ)\big) \]
\[ I \approx 2.27(0.9037 - 0.4281\,i) \approx 2.05 - 0.97\,i \]
Answer: \(I \approx 2.05 - 0.97i\) amperes

Example 18: Encrypt the word "MATH" by multiplying with the complex key \(K = 2 + 3i\), then decrypt it back using the multiplicative inverse.

Har letter ko uska number do (A=1, B=2, ...). Encryption: number \(\times\) key. Decryption: encrypted value \(\div\) key, yaani key ke multiplicative inverse se multiply.

Letters: M = 13, A = 1, T = 20, H = 8.

Encryption (multiply by \(2 + 3i\)):

\[ 13(2+3i) = 26 + 39i, \quad 1(2+3i) = 2 + 3i, \quad 20(2+3i) = 40 + 60i, \quad 8(2+3i) = 16 + 24i \]

Decryption (divide by \(2 + 3i\); note \(|2+3i|^2 = 13\)):

\[ \frac{26+39i}{2+3i} = \frac{(26+39i)(2-3i)}{13} = \frac{169}{13} = 13 \to \text{M} \]

Similarly the others give back 1, 20 and 8.

Answer: Encrypted: \(26+39i, 2+3i, 40+60i, 16+24i\); decrypted back to M, A, T, H

Mathematics 11 (Punjab Board)

MCQs: Chapter 1 – Complex Numbers

Choose the correct option. The answer key is checked against the worked solutions.

  1. If \(z = 3 - 4i\), the imaginary part of \(z\) is:

    1. \(3\)
    2. \(-4i\)
    3. \(-4\)
    4. \(4\)
    Imaginary part \(\text{Im}(z)\) sirf real number \(b\) hota hai, \(ib\) nahi — is liye \(-4\) sahi hai.
  2. \(i^{27}\) equals:

    1. \(-i\)
    2. \(i\)
    3. \(-1\)
    4. \(1\)
    27 ko 4 se divide karo: remainder 3, to \(i^{27} = i^3 = -i\).
  3. The multiplicative inverse of \(2 + 3i\) is:

    1. \(2 - 3i\)
    2. \(\frac{2}{13} - \frac{3}{13}i\)
    3. \(\frac{2}{13} + \frac{3}{13}i\)
    4. \(-2 - 3i\)
    \(\frac{1}{2+3i} = \frac{2-3i}{4+9} = \frac{2}{13} - \frac{3}{13}i\) — conjugate se multiply karo.
  4. If \(\bar{z} = z\), then \(z\) is:

    1. Purely imaginary
    2. Zero
    3. Any complex number
    4. A real number
    \(\bar{z} = z\) ka matlab \(a - ib = a + ib\), to \(b = 0\) — \(z\) real hai.
  5. \(|3 + 4i|\) equals:

    1. \(7\)
    2. \(5\)
    3. \(\sqrt{7}\)
    4. \(25\)
    \(|3+4i| = \sqrt{9+16} = \sqrt{25} = 5\) — yeh 3-4-5 triangle hai.
  6. The conjugate of \(-5 + 2i\) is:

    1. \(5 + 2i\)
    2. \(5 - 2i\)
    3. \(-5 - 2i\)
    4. \(-5 + 2i\)
    Conjugate mein sirf imaginary part ka sign flip hota hai: \(-5 - 2i\).
  7. \((2 + 3i) + (4 - 3i)\) equals:

    1. \(6\)
    2. \(6 + 6i\)
    3. \(-2\)
    4. \(6 - 6i\)
    Real parts: \(2+4 = 6\); imaginary parts: \(3-3 = 0\) — jawab sirf \(6\) hai.
  8. If \(1, \omega, \omega^2\) are cube roots of unity, then \(1 + \omega + \omega^2\) equals:

    1. \(1\)
    2. \(3\)
    3. \(\omega\)
    4. \(0\)
    Yeh cube roots of unity ki bunyadi identity hai — hamesha yaad rakho.
  9. \(\omega^3\) equals:

    1. \(\omega\)
    2. \(\omega^2\)
    3. \(1\)
    4. \(0\)
    \(\omega\) cube root of unity hai, is liye \(\omega^3 = 1\).
  10. The modulus of \(1 + i\) is:

    1. \(2\)
    2. \(\sqrt{2}\)
    3. \(1\)
    4. \(0\)
    \(|1+i| = \sqrt{1+1} = \sqrt{2}\).
  11. \(z \cdot \bar{z}\) equals:

    1. \(|z|^2\)
    2. \(|z|\)
    3. \(z^2\)
    4. \(0\)
    \((a+ib)(a-ib) = a^2 + b^2 = |z|^2\) — division mein yahi use hota hai.
  12. The argument of \(-1 - i\) (principal value) is:

    1. \(\frac{\pi}{4}\)
    2. \(-\frac{3\pi}{4}\)
    3. \(\frac{3\pi}{4}\)
    4. \(-\frac{\pi}{4}\)
    Point third quadrant mein hai, is liye \(\arg = -\pi + \pi/4 = -3\pi/4\) (principal value \(-\pi\) se \(\pi\) tak).
  13. \(\frac{1+i}{1-i}\) equals:

    1. \(1\)
    2. \(-1\)
    3. \(2i\)
    4. \(i\)
    Conjugate se multiply karo: \(\frac{(1+i)^2}{2} = \frac{2i}{2} = i\).
  14. One square root of \(-9\) is:

    1. \(3\)
    2. \(3i\)
    3. \(-3\)
    4. \(9i\)
    \(\sqrt{-9} = \sqrt{9}\sqrt{-1} = 3i\).
  15. If \(z_1 = 2(\cos\theta_1 + i\sin\theta_1)\) and \(z_2 = 3(\cos\theta_2 + i\sin\theta_2)\), then \(|z_1 z_2|\) equals:

    1. \(6\)
    2. \(5\)
    3. \(\theta_1 + \theta_2\)
    4. \(6(\cos(\theta_1+\theta_2) + i\sin(\theta_1+\theta_2))\)
    Polar multiplication mein moduli multiply hote hain: \(2 \times 3 = 6\).
  16. \(\arg(z_1 z_2)\) equals:

    1. \(\arg(z_1) \cdot \arg(z_2)\)
    2. \(\arg(z_1) - \arg(z_2)\)
    3. \(\frac{\arg(z_1)}{\arg(z_2)}\)
    4. \(\arg(z_1) + \arg(z_2)\)
    Product ka argument, arguments ka sum hota hai — polar form se sabit hota hai.
  17. The roots of \(z^2 + 1 = 0\) are:

    1. \(1, -1\)
    2. \(i, -i\)
    3. \(1, i\)
    4. No roots
    \(z^2 = -1\) se \(z = \pm i\) — complex numbers isi liye banaye gaye thay.
  18. \((1 + i)^2\) equals:

    1. \(2\)
    2. \(1 + 2i\)
    3. \(2i\)
    4. \(0\)
    \((1+i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i\).
  19. \(|z-2| = 3\) represents:

    1. A straight line
    2. A circle of radius 3 centred at \((2, 0)\)
    3. A circle of radius 2 centred at \((3, 0)\)
    4. A point
    \(|z - a| = r\) hamesha centre \(a\) aur radius \(r\) wala circle hota hai.
  20. \(i^n + i^{n+1} + i^{n+2} + i^{n+3}\) equals:

    1. \(0\)
    2. \(1\)
    3. \(4i^n\)
    4. \(2i^n\)
    \(i^n(1 + i + i^2 + i^3) = i^n(1 + i - 1 - i) = 0\).
  21. The polar form of \(1 + \sqrt{3}\,i\) is:

    1. \(2\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)\)
    2. \(2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)\)
    3. \(4\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)\)
    4. \(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\)
    \(r = \sqrt{1+3} = 2\) aur \(\theta = \tan^{-1}(\sqrt{3}) = \pi/3\).
  22. If \((x+iy)(1-i) = 4\), then \(x + y\) equals:

    1. \(2\)
    2. \(4\)
    3. \(0\)
    4. \(-4\)
    \((x+iy)(1-i) = (x+y) + i(y-x) = 4 + 0i\), to \(x + y = 4\).
  23. The four fourth roots of 16 are:

    1. \(2, -2\)
    2. \(2, -2, i, -i\)
    3. \(4, -4, 4i, -4i\)
    4. \(2, -2, 2i, -2i\)
    \(16^{1/4} = 2\), to roots \(\pm 2, \pm 2i\) hain.
  24. \(\overline{(3+2i)(1-i)}\) equals:

    1. \(5 - i\)
    2. \(5 + i\)
    3. \(-5 + i\)
    4. \(-5 - i\)
    \(\overline{z_1 z_2} = \bar{z_1}\bar{z_2} = (3-2i)(1+i) = 5 - i\).

Mathematics 11 (Punjab Board)

Chapter 1: Complex Numbers — Formula Sheet

Quick-revision formulas for Punjab Board Class 11 Mathematics, Chapter 1: Complex Numbers.

Basic Definitions

\[ z = a + ib, \quad \text{Re}(z) = a, \quad \text{Im}(z) = b, \quad i^2 = -1 \]
\[ a + ib = c + id \iff a = c \text{ and } b = d \]

Operations on Complex Numbers

\[ (a+ib) \pm (c+id) = (a \pm c) + i(b \pm d) \]
\[ (a+ib)(c+id) = (ac-bd) + i(ad+bc) \]
\[ \frac{a+ib}{c+id} = \frac{(a+ib)(c-id)}{c^2+d^2}, \qquad \frac{1}{a+ib} = \frac{a-ib}{a^2+b^2} \]

Conjugate and Modulus

\[ \bar{z} = a - ib, \qquad |z| = \sqrt{a^2+b^2}, \qquad |z|^2 = z\bar{z} \]
\[ \overline{z_1+z_2} = \bar{z_1}+\bar{z_2}, \quad \overline{z_1 z_2} = \bar{z_1}\,\bar{z_2}, \quad |z_1 z_2| = |z_1|\,|z_2| \]
\[ \frac{z+\bar{z}}{2} = \text{Re}(z), \qquad \frac{z-\bar{z}}{2i} = \text{Im}(z) \]

Powers of \(i\)

\[ i^2 = -1, \quad i^3 = -i, \quad i^4 = 1, \quad i^n = i^r \ (r = n \bmod 4) \]
\[ i^n + i^{n+1} + i^{n+2} + i^{n+3} = 0 \]

Square Root of a Complex Number

\[ \text{If } \sqrt{a+ib} = x+iy: \quad x^2-y^2 = a, \quad 2xy = b, \quad x^2+y^2 = \sqrt{a^2+b^2} \]

Quadratic Roots and Factorization

\[ az^2+bz+c = 0 \implies z = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]
\[ a^2+b^2 = (a+ib)(a-ib), \qquad z^3+a^3 = (z+a)(z^2-az+a^2) \]

Roots of Unity

\[ \omega = \frac{-1+\sqrt{3}\,i}{2}, \qquad \omega^3 = 1, \qquad 1+\omega+\omega^2 = 0 \]
\[ \text{Cube roots of } a: \ a^{1/3},\ a^{1/3}\omega,\ a^{1/3}\omega^2 \]
\[ \text{Fourth roots of } a > 0: \ \pm a^{1/4},\ \pm i\,a^{1/4} \]

Polar Form

\[ z = r(\cos\theta + i\sin\theta), \quad r = \sqrt{x^2+y^2}, \quad \tan\theta = \frac{y}{x} \]
\[ z_1 z_2 = r_1r_2\big(\cos(\theta_1+\theta_2) + i\sin(\theta_1+\theta_2)\big) \]
\[ \frac{z_1}{z_2} = \frac{r_1}{r_2}\big(\cos(\theta_1-\theta_2) + i\sin(\theta_1-\theta_2)\big) \]
\[ \arg(z_1 z_2) = \arg(z_1)+\arg(z_2), \quad \arg\left(\frac{z_1}{z_2}\right) = \arg(z_1)-\arg(z_2) \]

Loci in the Argand Plane

\[ |z-a| = r \ \text{ is a circle (centre } a, \text{ radius } r) \]
\[ |z-a| = |z-b| \ \text{ is the perpendicular bisector of } a \text{ and } b \]

Real-World Applications

\[ V = IZ \implies I = \frac{V}{Z} \]
\[ \text{Encrypt: } m \times k, \qquad \text{Decrypt: } \frac{m \times k}{k} \]

Chapter 1 FAQs

What is covered in Chapter 1 (Complex Numbers)?
This chapter covers Complex numbers, modulus, conjugate, argument, polar form, De Moivre's theorem. The page includes theory and definitions, solved textbook examples, complete exercise solutions, MCQs and a formula sheet.
How many marks does Chapter 1 carry in the Punjab Board paper?
Chapter 1 (Complex Numbers) carries approximately 7 marks in the Punjab Board Class 11 Mathematics paper.
Are these notes for the new PCTB syllabus?
Yes. These notes follow the PCTB 2025 new book for Class 11 Mathematics, used by BISE Lahore, Gujranwala, Rawalpindi, Faisalabad, Multan, Sahiwal, Sargodha, Bahawalpur and D.G. Khan.
Where are the exercise solutions for Chapter 1?
Use the exercise index on this page — every exercise of Chapter 1 has its own page with complete step-by-step solutions.