Exercise 4.1
Express i⁹ + i¹⁹ in the form a + ib.
We use the pattern: for any integer k, i^(4k) = 1, i^(4k+1) = i, i^(4k+2) = −1, i^(4k+3) = −i.
i⁹ = i^(4×2 + 1) = i
i¹⁹ = i^(4×4 + 3) = −i
Therefore i⁹ + i¹⁹ = i + (−i) = 0 = 0 + 0i.
Answer: 0 + 0i
Back to Mathematics
Class 11 Mathematics
Chapter 4 Solutions — Complex Numbers and Quadratic Equations
Open Solutions PDFReads in your browser→Also seeTextbook page→Get the full experience in the app
Free practice, offline PDFs & solutions.
Already installed? Open in the app
Solutions
Overview
Step-by-step NCERT solutions for Complex Numbers and Quadratic Equations (Chapter 4, NCERT Class 11 Mathematics) — every question and answer worked out in full, not just the final result. You can also read the Complex Numbers and Quadratic Equations textbook chapter.
Solved
What these solutions cover
All 28 questions in Complex Numbers and Quadratic Equations are solved in the PDF. Here's what's inside, exercise by exercise:
Exercise 4.1
- Express the complex number (5i)(−3/5 i) in the form a + ib.
- Express i⁹ + i¹⁹ in the form a + ib.
- Express i^(−39) in the form a + ib.
- Express 3(7 + i7) + i(7 + i7) in the form a + ib.
- Express (1 − i) − (−1 + i6) in the form a + ib.
- Express (1/5 + i 2/5)(−4 + i 5/2) in the form a + ib.
- Express {(1/3 + i√3) + (4 + i/3)} − (−4/3 + i) in the form a + ib.
- Express (1 − i)⁴ in the form a + ib.
- Express (1/3 + 3i)³ in the form a + ib.
- Express (−2 − i/3)³ in the form a + ib.
- Find the multiplicative inverse of 4 − 3i.
- Find the multiplicative inverse of √5 + 3i.
- Find the multiplicative inverse of −i.
- Express the following expression in the form a + ib: (3 + i√5)(3 − i√5) / [(√3 + √2 i) − (√3 − i√2)]
Miscellaneous Exercise
- Evaluate: [i¹⁸ + (1/i)²⁵]³.
- For any two complex numbers z_1 and z_2, prove that Re(z_1 z_2) = Re z_1 Re z_2 − Im z_1 Im z_2.
- Reduce [1/(1 − 4i) − 2/(1 + i)] × [(3 − 4i)/(5 + i)] to the standard form.
- If x − iy = √[(a − ib)/(c − id)], prove that (x² + y²)² = (a² + b²)/(c² + d²).
- If z_1 = 2 − i, z_2 = 1 + i, find |(z_1 + z_2 + 1)/(z_1 − z_2 + 1)|.
- If a + ib = (x + i)²/(2x² + 1), prove that a² + b² = (x² + 1)²/(2x² + 1)².
- Let z_1 = 2 − i, z_2 = −2 + i. Find
- (i) Re(z_1 z_2 / z̄_1),
- (ii) Im(1/(z_1 z̄_1)).
- Find the real numbers x and y if (x − iy)(3 + 5i) is the conjugate of −6 − 24i.
- Find the modulus of (1 + i)/(1 − i) − (1 − i)/(1 + i).
- If (x + iy)³ = u + iv, then show that u/x + v/y = 4(x² − y²).
- If α and β are different complex numbers with |β| = 1, then find |(β − α)/(1 − ᾱβ)|.
- Find the number of non-zero integral solutions of the equation |1 − i|^x = 2^x.
- If (a + ib)(c + id)(e + if)(g + ih) = A + iB, then show that (a² + b²)(c² + d²)(e² + f²)(g² + h²) = A² + B².
- If [(1 + i)/(1 − i)]^m = 1, then find the least positive integral value of m.
Sample
Worked examples
3 questions from Complex Numbers and Quadratic Equations, solved step by step.
Miscellaneous Exercise
Evaluate: [i¹⁸ + (1/i)²⁵]³.
First, simplify i¹⁸ and (1/i)²⁵.
i¹⁸ = i^(4×4 + 2) = i² = −1.
(1/i)²⁵ = i^(−25). Since −25 = 4×(−7) + 3, we get i^(−25) = i³ = −i.
Alternatively: (1/i)²⁵ = (1/i)²⁵. Note 1/i = −i, so (−i)²⁵ = (−1)²⁵ × i²⁵ = (−1)(i) = −i.
So i¹⁸ + (1/i)²⁵ = −1 + (−i) = −1 − i.
Now compute (−1 − i)³ using (a + b)³ = a³ + 3a²b + 3ab² + b³ with a = −1, b = −i:
a³ = −1
3a²b = 3(1)(−i) = −3i
3ab² = 3(−1)(i²) = 3(−1)(−1) = 3
b³ = (−i)³ = −(i³) = −(−i) = i
Sum: (−1 + 3) + (−3i + i) = 2 − 2i.
Therefore [i¹⁸ + (1/i)²⁵]³ = 2 − 2i.
Answer: 2 − 2i
Exercise 4.1
Express i^(−39) in the form a + ib.
We write −39 = 4×(−10) + 1, so −39 ≡ 1 (mod 4).
Therefore i^(−39) = i¹ = i = 0 + 1i.
Alternatively: i^(−39) = 1/i³⁹ = 1/i^(4×9+3) = 1/(−i) = 1/(−i) × (i/i) = i/(−i²) = i/1 = i.
Answer: 0 + i
All 28 questions in Complex Numbers and Quadratic Equations are solved step by step in the Solutions PDF above.
Keep solving
More solutions in Mathematics
Explore
More NCERT Solutions for Class 11
Read the Complex Numbers and Quadratic Equations textbook chapter / PDF, or browse all NCERT Class 11 Mathematics solutions.
Solve offline with notes, solutions & mock tests
CBSE Prepmaster — free on iOS & Android