📚 Theory – Key Concepts & Formulas
Powers of i
Cycle of Powers of i
i¹ = i i² = −1 i³ = −i i⁴ = 1
i⁵ = i i⁶ = −1 i⁷ = −i i⁸ = 1 ...(repeats every 4)
Shortcut for iⁿ
| Divide n by 4 — Remainder | 0 | 1 | 2 | 3 |
|---|
| iⁿ = | 1 | i | −1 | −i |
💡 For negative power: i⁻ⁿ = 1/iⁿ = conjugate method
Operations on Complex Numbers
Basic Rules
- Equality: a+ib = c+id ⟺ a=c and b=d
- Addition: (a+ib)+(c+id) = (a+c) + i(b+d)
- Multiplication: (a+ib)(c+id) = (ac−bd) + i(ad+bc)
- Conjugate: If z=a+ib, then z̄ = a−ib
Division (Rationalising)
a+ib / c+id = (a+ib)(c−id) / (c²+d²)
Multiply numerator & denominator by conjugate of denominator
Square Root of Complex Number
Method
Let √(x+iy) = a+ib
Square both sides: x+iy = (a²−b²) + 2abi
Equate real: x = a²−b²
Equate imaginary: y = 2ab
Use: a²+b² = √(x²+y²)
Then solve for a and b
Quadratic Equations
Formula
ax²+bx+c = 0 → x = [−b ± √(b²−4ac)] / 2a
D = b²−4ac | If D < 0 → complex roots (conjugate pairs)
💡 If a,b,c ∈ R and p+iq is a root, then p−iq is also a root
Modulus, Argument & Polar Form
Formulas
Modulus: |z| = r = √(a²+b²)
Argument: θ = tan⁻¹(b/a) [adjust for quadrant]
Polar form: z = r(cosθ + i sinθ)
Exponential form: z = r·e^(iθ)
Argument by Quadrant
| Condition | Quadrant/Axis | arg(z) |
|---|
| a>0, b=0 | +X axis | 0 |
| a>0, b>0 | Q I | tan⁻¹(b/a) |
| a=0, b>0 | +Y axis | π/2 |
| a<0, b>0 | Q II | tan⁻¹(b/a) + π |
| a<0, b=0 | −X axis | π |
| a<0, b<0 | Q III | tan⁻¹(b/a) + π |
| a=0, b<0 | −Y axis | 3π/2 |
| a>0, b<0 | Q IV | tan⁻¹(b/a) + 2π |
🏆 Solved Examples (Textbook)
Ex. 1 – Write (1+2i)(1+3i)(2+i)⁻¹ in the form a+ib Solved Example
Given :(1+2i)(1+3i)(2+i)⁻¹
Step 1 :(1+2i)(1+3i) = 1+3i+2i+6i² = 1+5i+6(−1) = −5+5i
Step 2 := (−5+5i) / (2+i)
Step 3 :Multiply by conjugate (2−i)/(2−i) :
= (−5+5i)(2−i) / [(2+i)(2−i)]
= (−10+5i+10i−5i²) / (4+1)
= (−10+15i+5) / 5 [∵ i² = −1]
= (−5+15i) / 5
= −1 + 3i
∴ a = −1, b = 3 → z = −1 + 3i
Ex. 2 – Express 1/i + 2/i² + 3/i³ + 5/i⁴ in the form a+ib Activity
Known :i² = −1, i³ = −i, i⁴ = 1
Step 1 :1/i = i/i² = i/(−1) = −i
Step 2 :1/i² = 1/(−1) = −1
Step 3 :1/i³ = 1/(−i) = i (multiply by i/i → i/1 = i)
Step 4 :1/i⁴ = 1/1 = 1
∴ Sum := 1(−i) + 2(−1) + 3(i) + 5(1)
= −i − 2 + 3i + 5
= (5−2) + (3−1)i
= 3 + 2i
∴ a = 3, b = 2
Ex. 3 – If a,b ∈ R and (i⁴+3i)a + (i−1)b + 5i³ = 0, find a and b Solved Example
Given :(i⁴+3i)a + (i−1)b + 5i³ = 0+0i
Step 1 :Simplify powers: i⁴ = 1, i³ = −i
(1+3i)a + (i−1)b − 5i = 0+0i
Step 2 :Expand:
a + 3ai + bi − b − 5i = 0+0i
(a−b) + (3a+b−5)i = 0+0i
Step 3 :Equating real parts: a − b = 0 ∴ a = b
Step 4 :Equating imaginary parts: 3a + b − 5 = 0
∴ 3a + a = 5 (since a = b)
∴ 4a = 5
∴ a = 5/4
∴ a = b = 5/4
Ex. 4 – x + 2i + 15i⁶y = 7x + i³(y+4). Find x+y, x,y ∈ R Solved Example
Given :x + 2i + 15i⁶y = 7x + i³(y+4)
Step 1 :Simplify: i⁶ = −1, i³ = −i
∴ x + 2i − 15y = 7x − (y+4)i
∴ (x−15y) + 2i = 7x − (y+4)i
Step 2 :Equating real parts:
x − 15y = 7x
∴ −6x − 15y = 0 ... (i)
Step 3 :Equating imaginary parts:
2 = −(y+4)
∴ y + 6 = 0
∴ y = −6 ... (ii)
Step 4 :Substituting y = −6 in (i):
−6x − 15(−6) = 0
−6x + 90 = 0 ∴ x = 15
∴ x = 15, y = −6 → x + y = 15 + (−6) = 9
Ex. 5 – Show that (√3/2 + i/2)³ = i Proof
To prove :(√3/2 + i/2)³ = i
Step 1 :Write as ((√3+i)/2)³ = (√3+i)³ / 8
Step 2 :Expand (√3+i)³ using (a+b)³ = a³+3a²b+3ab²+b³:
= (√3)³ + 3(√3)²(i) + 3(√3)(i²) + i³
= 3√3 + 9i + 3√3(−1) + (−i)
= 3√3 + 9i − 3√3 − i
= (3√3 − 3√3) + (9i − i)
= 0 + 8i = 8i
Step 3 :∴ (√3+i)³ / 8 = 8i / 8 = i = R.H.S.
∴ L.H.S. = i = R.H.S. Proved ✓
Sq. Root Ex. 1 – Find √(6+8i) Square Root
Let :√(6+8i) = a+ib, a,b ∈ R
Step 1 :Squaring both sides:
6 + 8i = (a+ib)² = (a²−b²) + 2abi
Step 2 :Equating real parts: a² − b² = 6 ... (1)
Step 3 :Equating imaginary parts: 2ab = 8 ∴ ab = 4 ∴ a = 4/b ... (2)
Step 4 :Substituting (2) in (1):
(4/b)² − b² = 6
16/b² − b² = 6
∴ b⁴ + 6b² − 16 = 0
Step 5 :Put m = b²:
m² + 6m − 16 = 0
(m+8)(m−2) = 0
∴ m = −8 or m = 2
Step 6 :Since b is real, b² ≠ −8
∴ b² = 2 ∴ b = ±√2
Step 7 :For b = √2 : a = 4/√2 = 2√2
For b = −√2 : a = 4/(−√2) = −2√2
∴ √(6+8i) = 2√2 + √2·i or −2√2 − √2·i = ±√2(2+i)
Sq. Root Ex. 2 – Find √(3−4i) Square Root
Let :√(3−4i) = a+ib, a,b ∈ R
Step 1 :Squaring: (a+ib)² = (a²−b²) + 2abi
∴ a²−b² = 3 ... (1) and 2ab = −4 ∴ ab = −2 ... (2)
Step 2 :Use (a²+b²)² = (a²−b²)² + (2ab)²
= 9 + 16 = 25
∴ a² + b² = 5 ... (3)
Step 3 :Adding (1) and (3):
2a² = 8 ∴ a² = 4 ∴ a = ±2
Step 4 :For a = 2 : b = −4/(2×2) = −1
For a = −2 : b = −4/(2×−2) = 1
∴ √(3−4i) = 2−i or −2+i = ±(2−i)
✏️ Exercise 1.1 – Complete Solutions
Q.1 – Simplify
i) √(−16) + 3√(−25) + √(−36) − √(−625)
Step 1 :√(−16) = 4i, 3√(−25) = 3×5i = 15i, √(−36) = 6i, √(−625) = 25i
= 4i + 15i + 6i − 25i
= (4+15+6−25)i
= 0·i
Answer: 0
ii) 4√(−4) + 5√(−9) − 3√(−16)
Step 1 :4√(−4) = 4×2i = 8i, 5√(−9) = 5×3i = 15i, 3√(−16) = 3×4i = 12i
= 8i + 15i − 12i = 11i
Answer: 11i (a=0, b=11)
Q.3 – Find a and b
i) a + 2b + 2ai = 4 + 6i
Step 1 :Equate real parts: a + 2b = 4 ... (1)
Step 2 :Equate imaginary parts: 2a = 6 ∴ a = 3
Step 3 :Substitute in (1): 3 + 2b = 4 ∴ b = 1/2
∴ a = 3, b = 1/2
ii) (a−b) + (a+b)i = a + 5i
Step 1 :Equate real parts: a − b = a ∴ −b = 0 ∴ b = 0
Step 2 :Equate imaginary parts: a + b = 5 ∴ a + 0 = 5 ∴ a = 5
∴ a = 5, b = 0
iii) (a+b)(2+i) = b+1 + (10+2a)i
Step 1 :Expand LHS: 2(a+b) + (a+b)i
Step 2 :Equate real: 2(a+b) = b+1
∴ 2a + 2b = b + 1 ∴ 2a + b = 1 ... (1)
Step 3 :Equate imaginary: a + b = 10 + 2a
∴ b − a = 10 ∴ b = a + 10 ... (2)
Step 4 :Substitute (2) in (1):
2a + (a+10) = 1 ∴ 3a = −9 ∴ a = −3
b = −3 + 10 = 7
∴ a = −3, b = 7
iv) abi = 3a − b + 12i
Step 1 :LHS = 0 + abi. Equate real: 0 = 3a − b ∴ b = 3a ... (1)
Step 2 :Equate imaginary: ab = 12 ... (2)
Step 3 :Substitute (1) in (2): a(3a) = 12 ∴ 3a² = 12 ∴ a² = 4 ∴ a = ±2
For a = 2 : b = 6 For a = −2 : b = −6
∴ a = 2, b = 6 OR a = −2, b = −6
v) 1/(a+ib) = 3−2i
Step 1 :a + ib = 1/(3−2i)
Step 2 :Rationalise: = (3+2i) / [(3−2i)(3+2i)] = (3+2i) / (9+4) = (3+2i) / 13
= 3/13 + (2/13)i
∴ a = 3/13, b = 2/13
vi) (a+ib)(1+i) = 2+i
Step 1 :a + ib = (2+i) / (1+i)
Step 2 :Rationalise by (1−i)/(1−i):
= (2+i)(1−i) / (1+1) = (2−2i+i−i²) / 2
= (2 − i + 1) / 2 = (3−i) / 2 = 3/2 − (1/2)i
∴ a = 3/2, b = −1/2
Q.4 – Express in a+ib form IMP
i) (1+2i)(−2i)
= −2i − 4i²
= −2i − 4(−1)
= 4 − 2i
∴ a = 4, b = −2
ii) (1+i)(1−i)⁻¹
Step 1 := (1+i) / (1−i)
Step 2 :Multiply by (1+i)/(1+i):
= (1+i)² / (1+1) = (1 + 2i + i²) / 2
= (1 + 2i − 1) / 2 = 2i / 2 = i
∴ a = 0, b = 1
iii) −i(4+3i) / (1−i)
Step 1 :Numerator: −i(4+3i) = −4i − 3i² = −4i + 3 = 3 − 4i
Step 2 := (3−4i) / (1−i). Multiply by (1+i)/(1+i):
= (3−4i)(1+i) / (1+1)
= (3 + 3i − 4i − 4i²) / 2
= (3 − i + 4) / 2 = (7−i) / 2
= 7/2 − (1/2)i
∴ a = 7/2, b = −1/2
iv) (2+i) / [(3−i)(1+2i)]
Step 1 :Denominator: (3−i)(1+2i) = 3+6i−i−2i² = 3+5i+2 = 5+5i
Step 2 := (2+i) / (5+5i). Multiply by (5−5i)/(5−5i):
= (2+i)(5−5i) / (25+25)
= (10−10i+5i−5i²) / 50
= (10 − 5i + 5) / 50 = (15−5i) / 50
= 3/10 − (1/10)i
∴ a = 3/10, b = −1/10
v) [(1+i)/(1−i)]²
Step 1 :(1+i)/(1−i) = (1+i)²/[(1−i)(1+i)] = (1+2i+i²)/2 = (1+2i−1)/2 = 2i/2 = i
Step 2 :∴ [(1+i)/(1−i)]² = i² = −1
∴ −1 + 0i (a = −1, b = 0)
vi) (3+2i)/(2−5i) + (−2i)/(2+5i)
Step 1 :Part 1: (3+2i)(2+5i) / [(2−5i)(2+5i)]
= (6+15i+4i+10i²) / (4+25) = (6+19i−10) / 29 = (−4+19i) / 29
Step 2 :Part 2: (−2i)(2−5i) / [(2+5i)(2−5i)]
= (−4i+10i²) / 29 = (−10−4i) / 29
Step 3 :Sum = (−4+19i−10−4i) / 29 = (−14+15i) / 29
∴ −14/29 + (15/29)i
vii) (1+i)⁻³
Step 1 :(1+i)² = 1+2i+i² = 1+2i−1 = 2i
Step 2 :(1+i)³ = (1+i)(2i) = 2i+2i² = 2i−2 = −2+2i
Step 3 :(1+i)⁻³ = 1/(−2+2i). Multiply by (−2−2i)/(−2−2i):
= (−2−2i) / (4+4) = (−2−2i) / 8 = −1/4 − (1/4)i
∴ a = −1/4, b = −1/4
x) (2+3i)(2−3i)
= 4 − 6i + 6i − 9i²
= 4 − 9(−1)
= 4 + 9 = 13
∴ 13 + 0i (a = 13, b = 0)
Q.7 – Evaluate Powers of i IMP
All parts — Shortcut: divide power by 4, use remainder
Rule :rem 0 → 1 | rem 1 → i | rem 2 → −1 | rem 3 → −i
i) i³⁵: 35 ÷ 4 = 8 rem 3 ∴ i³⁵ = −i
ii) i⁸⁸⁸: 888 ÷ 4 = 222 rem 0 ∴ i⁸⁸⁸ = 1
iii) i⁹³: 93 ÷ 4 = 23 rem 1 ∴ i⁹³ = i
iv) i¹¹⁶: 116 ÷ 4 = 29 rem 0 ∴ i¹¹⁶ = 1
v) i⁴⁰³: 403 ÷ 4 = 100 rem 3 ∴ i⁴⁰³ = −i
vi) 1/i⁵⁸: 58 ÷ 4 = 14 rem 2 ∴ i⁵⁸ = −1 ∴ 1/i⁵⁸ = 1/(−1) = −1
vii) i⁻⁸⁸⁸: 888 rem 0 ∴ i⁸⁸⁸ = 1 ∴ i⁻⁸⁸⁸ = 1/1 = 1
viii) i³⁰+i⁴⁰+i⁵⁰+i⁶⁰: rem 2,0,2,0 → −1+1+(−1)+1 = 0
Answers: −i, 1, i, 1, −i, −1, 1, 0
Q.9 – Find Values IMP
i) i⁴⁹ + i⁶⁸ + i⁸⁹ + i¹¹⁰
i⁴⁹ :49 ÷ 4 = 12 rem 1 ∴ i⁴⁹ = i
i⁶⁸ :68 ÷ 4 = 17 rem 0 ∴ i⁶⁸ = 1
i⁸⁹ :89 ÷ 4 = 22 rem 1 ∴ i⁸⁹ = i
i¹¹⁰ :110 ÷ 4 = 27 rem 2 ∴ i¹¹⁰ = −1
Sum := i + 1 + i + (−1) = 2i
Answer: 2i
ii) i + i² + i³ + i⁴
= i + (−1) + (−i) + 1
= (i − i) + (−1 + 1) = 0
Answer: 0
Q.14 – Evaluate (i³⁷ + 1/i⁶⁷) IMP
i³⁷ + 1/i⁶⁷
i³⁷ :37 ÷ 4 = 9 rem 1 ∴ i³⁷ = i
i⁶⁷ :67 ÷ 4 = 16 rem 3 ∴ i⁶⁷ = −i
1/i⁶⁷ := 1/(−i) = (−i)/(−i×−i) = −i/(i²) ... or multiply by (i/i): = i/(−i²) = i/1 = i
Sum := i + i = 2i
Answer: 2i
Q.24 – Find x and y, x,y ∈ R IMP
i) (x+2y) + (2x−y)i + 4i = 5
Step 1 :Group: (x+2y) + (2x−y+4)i = 5 + 0i
Step 2 :Equate real: x + 2y = 5 ... (1)
Step 3 :Equate imaginary: 2x − y + 4 = 0 ∴ 2x − y = −4 ... (2)
Step 4 :From (2): y = 2x+4. Substitute in (1):
x + 2(2x+4) = 5 ∴ 5x + 8 = 5 ∴ x = −3/5
y = 2(−3/5)+4 = −6/5 + 20/5 = 14/5
∴ x = −3/5, y = 14/5
v) If x + 2i + 15i⁶y = 7x + i³(y+4), find x+y
Step 1 :Simplify: i⁶ = −1, i³ = −i
x + 2i − 15y = 7x − (y+4)i
(x−15y) + 2i = 7x − (y+4)i
Step 2 :Equate real: x − 15y = 7x ∴ −6x = 15y ... (i)
Step 3 :Equate imaginary: 2 = −(y+4) ∴ y = −6
Step 4 :−6x = 15(−6) = −90 ∴ x = 15
∴ x = 15, y = −6 → x + y = 9
✏️ Exercise 1.2 – Square Roots & Quadratic Equations
Q.1 – Find Square Roots IMP
i) √(−8−6i)
Let :√(−8−6i) = a+ib
Step 1 :Squaring: a²−b² = −8 ... (1) and 2ab = −6 ∴ ab = −3 ... (2)
Step 2 :(a²+b²)² = (a²−b²)² + (2ab)² = 64 + 36 = 100
∴ a² + b² = 10 ... (3)
Step 3 :Adding (1) and (3): 2a² = 2 ∴ a = ±1
Step 4 :For a = 1 : b = −3/1 = −3
For a = −1 : b = −3/(−1) = 3
∴ √(−8−6i) = 1−3i or −1+3i = ±(1−3i)
ii) √(7+24i)
Let :√(7+24i) = a+ib
Step 1 :a²−b² = 7 ... (1) and 2ab = 24 ∴ ab = 12 ... (2)
Step 2 :(a²+b²)² = 49 + 576 = 625 ∴ a²+b² = 25 ... (3)
Step 3 :Adding (1) and (3): 2a² = 32 ∴ a² = 16 ∴ a = ±4
Step 4 :For a = 4 : b = 12/4 = 3
For a = −4 : b = 12/(−4) = −3
∴ √(7+24i) = 4+3i or −4−3i = ±(4+3i)
iii) √(1+4√3·i)
Let :√(1+4√3·i) = a+ib
Step 1 :a²−b² = 1 and 2ab = 4√3 ∴ ab = 2√3
Step 2 :(a²+b²)² = 1 + 48 = 49 ∴ a²+b² = 7 ... (3)
Step 3 :2a² = 8 ∴ a = ±2
Step 4 :For a = 2 : b = 2√3/2 = √3
For a = −2 : b = −√3
∴ ±(2+√3·i)
iv) √(3+2√10·i)
Step 1 :a²−b² = 3, ab = √10
Step 2 :(a²+b²)² = 9+40 = 49 ∴ a²+b² = 7
Step 3 :2a² = 10 ∴ a = ±√5
Step 4 :For a=√5 : b = √10/√5 = √2 | For a=−√5 : b = −√2
∴ ±(√5 + √2·i)
v) √[2(1−√3·i)]
Step 1 := √(2−2√3·i)
Step 2 :a²−b² = 2, ab = −√3
Step 3 :(a²+b²)² = 4+12 = 16 ∴ a²+b² = 4
Step 4 :2a² = 6 ∴ a = ±√3
Step 5 :For a=√3 : b=−1 | For a=−√3 : b=1
∴ ±(√3 − i)
Q.2 – Solve Quadratic (Real Coefficients)
i) 8x² + 2x + 1 = 0
a=8, b=2, c=1
Step 1 :D = b²−4ac = 4 − 32 = −28 < 0
Step 2 :x = (−2 ± √(−28)) / 16 = (−2 ± 2√7·i) / 16 = (−1 ± √7·i) / 8
∴ x = (−1+√7·i)/8 and (−1−√7·i)/8
ii) 2x² − √3·x + 1 = 0
a=2, b=−√3, c=1
Step 1 :D = 3 − 8 = −5
Step 2 :x = (√3 ± √(−5)) / 4 = (√3 ± √5·i) / 4
∴ x = (√3+√5·i)/4 and (√3−√5·i)/4
iii) 3x² − 7x + 5 = 0
Step 1 :D = 49 − 60 = −11
Step 2 :x = (7 ± √(−11)) / 6 = (7 ± √11·i) / 6
∴ x = (7+√11·i)/6 and (7−√11·i)/6
iv) x² − 4x + 13 = 0
Step 1 :D = 16 − 52 = −36
Step 2 :x = (4 ± √(−36)) / 2 = (4 ± 6i) / 2 = 2 ± 3i
∴ x = 2+3i and 2−3i
Q.3 – Solve Quadratic (Complex Coefficients)
i) x² + 3ix + 10 = 0
Step 1 :D = (3i)² − 4(10) = −9 − 40 = −49
Step 2 :√(−49) = 7i
Step 3 :x = (−3i ± 7i) / 2
x = (−3i+7i)/2 = 4i/2 = 2i or x = (−3i−7i)/2 = −10i/2 = −5i
∴ x = 2i and −5i
ii) 2x² + 3ix + 2 = 0
Step 1 :D = (3i)² − 4(2)(2) = −9 − 16 = −25
Step 2 :√(−25) = 5i
Step 3 :x = (−3i ± 5i) / 4
x = 2i/4 = i/2 or x = −8i/4 = −2i
∴ x = i/2 and −2i
iii) x² + 4ix − 4 = 0
Step 1 :D = (4i)² − 4(1)(−4) = −16 + 16 = 0
Step 2 :x = −4i / 2 = −2i (repeated root)
∴ x = −2i (repeated)
iv) ix² − 4x − 4i = 0
Step 1 :Multiply throughout by i: i²x² − 4ix − 4i² = 0
→ −x² − 4ix + 4 = 0 → x² + 4ix − 4 = 0
Step 2 :D = (4i)² + 16 = −16+16 = 0 ∴ x = −4i/2 = −2i
∴ x = −2i (repeated)
Q.4 – Solve (Mixed Complex Coefficients)
i) x² − (2+i)x − (1−7i) = 0
Step 1 :D = (2+i)² + 4(1−7i)
= 4+4i+i² + 4−28i = 4+4i−1+4−28i = 7−24i
Step 2 :Find √(7−24i) = a+ib:
a²−b² = 7, ab = −12
(a²+b²)² = 49+576 = 625 ∴ a²+b² = 25
2a² = 32 ∴ a = ±4
For a=4: b = −12/4 = −3 ∴ √(7−24i) = 4−3i
Step 3 :x = [(2+i) ± (4−3i)] / 2
x = (6−2i)/2 = 3−i or x = (−2+4i)/2 = −1+2i
∴ x = 3−i and −1+2i
iii) x² − (5−i)x + (18+i) = 0
Step 1 :D = (5−i)² − 4(18+i)
= 25−10i+i² − 72−4i = 25−10i−1−72−4i = −48−14i
Step 2 :Find √(−48−14i):
a²−b² = −48, ab = −7
(a²+b²)² = 2304+196 = 2500 ∴ a²+b² = 50
2a² = 2 ∴ a = ±1. For a=1: b = −7 ∴ √(−48−14i) = 1−7i
Step 3 :x = [(5−i) ± (1−7i)] / 2
x = (6−8i)/2 = 3−4i or x = (4+6i)/2 = 2+3i
∴ x = 3−4i and 2+3i
✏️ Exercise 1.3 – Modulus, Argument, Polar Form
Q.1 – Modulus and Amplitude IMP
i) z = 7−5i
a = 7, b = −5 (Q IV)
|z| := √(7²+(−5)²) = √(49+25) = √74
arg(z) :Q IV → θ = tan⁻¹(−5/7) + 2π
|z| = √74 | arg(z) = tan⁻¹(5/7) measured from +X clockwise (Q IV)
iii) z = −8+15i
a = −8, b = 15 (Q II)
|z| := √(64+225) = √289 = 17
arg(z) :Q II → θ = tan⁻¹(15/−8) + π
|z| = 17 | arg(z) = π − tan⁻¹(15/8)
v) z = −4−4i
a = −4, b = −4 (Q III)
|z| := √(16+16) = √32 = 4√2
arg(z) :Q III → θ = tan⁻¹(−4/−4) + π = tan⁻¹(1) + π = π/4 + π = 5π/4
|z| = 4√2 | arg(z) = 5π/4
vi) z = √3 − i
a = √3, b = −1 (Q IV)
|z| := √(3+1) = √4 = 2
arg(z) :Q IV → θ = tan⁻¹(−1/√3) + 2π = −π/6 + 2π = 11π/6
|z| = 2 | arg(z) = 11π/6 (or −π/6)
ix) z = 1 + i√3
a = 1, b = √3 (Q I)
|z| := √(1+3) = √4 = 2
arg(z) :Q I → θ = tan⁻¹(√3/1) = tan⁻¹(√3) = π/3
|z| = 2 | arg(z) = π/3
Q.4 – Polar and Exponential Form IMP
i) z = −1 + √3·i
a = −1, b = √3 (Q II)
r := √(1+3) = √4 = 2
θ :Q II → tan⁻¹(√3/−1) + π = −π/3 + π = 2π/3
Polar :z = 2(cos 2π/3 + i sin 2π/3)
Exponential :z = 2·e^(2πi/3)
Polar: 2(cos 2π/3 + i sin 2π/3) | Exp: 2e^(i·2π/3)
ii) z = −i
a = 0, b = −1 (−Y axis)
r := √(0+1) = 1
θ :On negative imaginary axis → θ = 3π/2
Polar :z = 1(cos 3π/2 + i sin 3π/2)
Exponential :z = e^(3πi/2)
Polar: cos 3π/2 + i sin 3π/2 | Exp: e^(i·3π/2)
iii) z = −1
a = −1, b = 0 (−X axis)
r := √(1+0) = 1
θ :On negative real axis → θ = π
Polar :z = 1(cos π + i sin π)
Exponential :z = e^(iπ)
Polar: cos π + i sin π | Exp: e^(iπ)
iv) z = 1/(1+i)
Step 1 :Rationalise: (1−i)/[(1+i)(1−i)] = (1−i)/2 = 1/2 − (1/2)i
Now: a = 1/2, b = −1/2 (Q IV)
r := √(1/4+1/4) = √(1/2) = 1/√2
θ :Q IV → tan⁻¹(−1/2 ÷ 1/2) + 2π = tan⁻¹(−1) + 2π = −π/4 + 2π = 7π/4
Polar :z = (1/√2)(cos 7π/4 + i sin 7π/4)
Exponential :z = (1/√2)·e^(7πi/4)
Polar: (1/√2)(cos 7π/4 + i sin 7π/4) | Exp: (1/√2)e^(i·7π/4)
Q.6 – Modulus and Argument of (1+2i)/(1−3i) IMP
Find modulus and argument of z = (1+2i)/(1−3i)
Step 1 :Rationalise: multiply by (1+3i)/(1+3i)
= (1+2i)(1+3i) / [(1−3i)(1+3i)]
= (1+3i+2i+6i²) / (1+9)
= (1+5i−6) / 10 = (−5+5i) / 10 = −1/2 + (1/2)i
Step 2 :a = −1/2, b = 1/2 (Q II)
|z| := √(1/4+1/4) = √(1/2) = 1/√2
arg(z) :Q II → tan⁻¹(1/2 ÷ −1/2) + π = tan⁻¹(−1) + π = −π/4 + π = 3π/4
|z| = 1/√2 | arg(z) = 3π/4
Q.7 – Convert to Polar Form: z = (i−1) / (cos π/3 + i sin π/3)
z = (i−1) / (cos π/3 + i sin π/3)
Step 1 :Denominator = e^(iπ/3)
Step 2 :Numerator: i−1 = −1+i. r=√2, θ=3π/4 → = √2·e^(i·3π/4)
Step 3 :z = √2·e^(i·3π/4) / e^(iπ/3) = √2·e^(i(3π/4 − π/3))
= √2·e^(i(9π/12 − 4π/12)) = √2·e^(i·5π/12)
Polar :z = √2(cos 5π/12 + i sin 5π/12)
z = √2(cos 5π/12 + i sin 5π/12)