NCERT Solutions for Class 11 Maths Exercise 4.1 provides detailed solutions based on the fundamental concepts of complex numbers, imaginary numbers, algebraic operations, and powers of iota. These solutions help students understand the properties and representation of complex numbers in a simple and step-by-step manner. Exercise 4.1 is extremely important for building a strong foundation in algebra and is highly useful for CBSE board examinations as well as competitive exams like JEE Main and JEE Advanced.
NCERT Solutions of Exercise 4.1 : Express each of the complex number given in the Questions 1 to 10 in the form a + ib.
Question.1 : Write the complex number in the form a + ib : $(5i)\left(\dfrac{-3i}{5}\right)$
Solution :
$5i \times \left(\dfrac{-3i}{5}\right) = -3 \times i^2$
Since, $i^2 = -1$. Therefore,
$-3 \times i^2 = -3 \times (-1) = 3$
Hence,
$$(5i)\left(\frac{-3i}{5}\right) = 3 + i0$$
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Question 2: Write the complex number in the form a + ib : $i^9 + i^{19}$
Solution :
$i^9 + i^{19} = i(i^2)^4 + i(i^2)^9$
Since, $i^2 = -1$. Therefore,
$i(i^2)^4 + i(i^2)^9 = i(-1)^4 + i(-1)^9$
$= i – i = 0$
Hence,
$$i^9 + i^{19} = 0 + i0$$
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Question 3: Write the complex number in the form a + ib : $i^{-39}$
Solution :
$i^{-39} = \dfrac{1}{i^{39}} = \dfrac{1}{i^3 \cdot (i^4)^9}$
Since, $i^3 = -i$ and $i^4 = 1$. Therefore,
$$\frac{1}{i^3 \cdot (i^4)^9} = \frac{1}{(-i)(1)} = \frac{1}{-i}$$
Multiply and divide by $i$:
$$\frac{1}{-i} \times \frac{i}{i} = \frac{i}{-i^2}$$
Since, $i^2 = -1$ and $-i^2 = 1$ Therefore,
$$\frac{i}{-i^2} = i$$
Hence,
$$i^{-39} = 0 + i1$$
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Question 4: Write the complex number in the form a + ib : $3(7 + i7) + i(7 + i7)$
Solution :
$3(7 + i7) + i(7 + i7) = 21 + i21 + i7 + i^2 7$
Since, $i^2 = -1$. Therefore,
$$21 + i21 + i7 + i^2 7 = 21 + i28 + (-1)7 = 21 – 7 + i28$$
$$= 14 + i28$$
Hence,
$$3(7 + i7) + i(7 + i7) = 14 + i28$$
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Question 5: Write the complex number in the form a + ib : $(1 – i) – (-1 + i6)$
Solution :
$(1 – i) – (-1 + i6) = 1 – i + 1 – i6 = 2 – i7$
$$(1 – i) – (-1 + i6)= 2 – i7$$
Question 6: Write the complex number in the form a + ib : $\left(\dfrac{1}{5} + i\dfrac{2}{5}\right) – (4 + i\dfrac{5}{2})$
Solution :
$\left(\dfrac{1}{5} + i\dfrac{2}{5}\right) – (4 + i\dfrac{5}{2}) = \dfrac{1}{5} + i\dfrac{2}{5} – 4 – i\dfrac{5}{2}$
$$= \dfrac{1}{5} – 4 + i\dfrac{2}{5} – i\dfrac{5}{2}$$
$$= -\dfrac{19}{5} – i\dfrac{21}{10}$$
Hence,
$$\left(\dfrac{1}{5} + i\dfrac{2}{5}\right) – (4 + i\dfrac{5}{2}) = -\dfrac{19}{5} – i\dfrac{21}{10}$$
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Question 7: Write the complex number in the form a + ib : $$\left[\left(\frac{1}{3}+i\frac{7}{3}\right)+(4+i\frac{1}{3})\right]-\left(-\frac{4}{3}+i\right)$$
Solution :
$$\left[\frac{1}{3}+i\frac{7}{3}+4+i\frac{1}{3}\right]+ \frac{4}{3}-i$$
$$= \frac{13}{3} + i\frac{8}{3} + \frac{4}{3} – i$$
$$= \frac{17}{3} + i\frac{5}{3}$$
Hence,
$$\left[\left(\frac{1}{3}+i\frac{7}{3}\right)+(4+i\frac{1}{3})\right]-\left(-\frac{4}{3}+i\right)=\frac{17}{3}+i\frac{5}{3}$$
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Question 8: Write the complex number in the form a + ib : $(1-i)^4$
Solution :
$$(1-i)^4=(1-i)^2(1-i)^2$$
$$(1-i)^2 = 1 + i^2 – 2i$$
Since $i^2=-1$,
$$(1-i)^2 = 1 -1 -2i = -2i$$
So
$$(1-i)^4 = (-2i)^2 = 4i^2 = 4(-1) = -4$$
Hence,
$$(1-i)^4 = -4 + i0$$
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Question 9: Write the complex number in the form a + ib : $\left(\frac{1}{3}+3i\right)^3$
Solution:
Using expansion,
$$\left(\frac{1}{3}+3i\right)^3 = \left(\frac{1}{3}\right)^3 + (3i)^3 + 3\left(\frac{1}{3}\right)^2(3i) + 3(3i)^2\left(\frac{1}{3}\right)$$
$$= \frac{1}{27} + 27i^3 + 3\cdot\frac{1}{9}\cdot 3i + 3\cdot 9 i^2 \cdot \frac{1}{3}$$
$$= \frac{1}{27} + 27i^3 + i + 9i^2$$
Since $i^2=-1$ and $i^3=-i$,
$$= \frac{1}{27} -27i + 9(-1) + i$$
$$= \frac{1}{27} -9 -26i$$
$$= -\frac{242}{27} – 26i$$
Hence,
$$\left(\frac{1}{3}+3i\right)^3 = -\frac{242}{27} – i26$$
Question 10: Write the complex number in the form a + ib : $\left(-2 – i\frac{1}{3}\right)^3$
Solution :
Using expansion,
$$\left(-2 – i\frac{1}{3}\right)^3 = (-2)^3 + \left(-i\frac{1}{3}\right)^3 + 3(-2)\left(-i\frac{1}{3}\right)^2 + 3\left(-i\frac{1}{3}\right)(-2)^2$$
$$= -8 – i^3\frac{1}{27} – 6\left(\frac{i^2}{9}\right) – i\cdot 4$$
$$= -8 – i^3\frac{1}{27} – \frac{2 i^2}{3} – 4i$$
Since $i^2=-1$ and $i^3=-i$,
$$= -8 + i\frac{1}{27} + \frac{2}{3} – 4i$$
$$= -\frac{22}{3} + i\left(\frac{1}{27} – 4\right)$$
$$= -\frac{22}{3} – i\frac{107}{27}$$
Hence,
$$\left(-2 – i\frac{1}{3}\right)^3 = -\frac{22}{3} – i\frac{107}{27}$$
NCERT Solutions of Exercise 4.1 : Find the multiplicative inverse of each of the complex numbers given in the Questions 11 to 13.
Question.11 : Find the multiplicative inverse of $4 – 3i$
Solution
Let $4 – 3i = z$ (so $\overline{z} = 4 + 3i$).
$$
|z|^2 = 4^2 + (-3)^2 = 16 + 9 = 25
$$
Multiplicative inverse of $z$ is $z^{-1}$.
$$
z^{-1} = \frac{\overline{z}}{|z|^2} = \frac{4 + 3i}{25} = \frac{4}{25} + i\frac{3}{25}
$$
Question.12 : Find the multiplicative inverse of $\sqrt{5} + 3i$
Solution
Let $\sqrt{5} + 3i = z$ (so $\overline{z} = \sqrt{5} – 3i$).
$$
|z|^2 = (\sqrt{5})^2 + 3^2 = 5 + 9 = 14
$$
Multiplicative inverse of $z$ is $z^{-1}$.
$$
z^{-1} = \frac{\overline{z}}{|z|^2} = \frac{\sqrt{5} – 3i}{14} = \frac{\sqrt{5}}{14} – i\frac{3}{14}
$$
Question.13 : Find the multiplicative inverse of $-i$
Solution
Let $-i = z$ (so $\overline{z} = i$).
$$
|z|^2 = (-1)^2 = 1
$$
Multiplicative inverse of $z$ is $z^{-1}$.
$$
z^{-1} = \frac{\overline{z}}{|z|^2} = \frac{i}{1} = i
$$
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Question.14 : Express the following expression in the form $a+ib$:
$$
\frac{(3 + i\sqrt{5})(3 – i\sqrt{5})}{(\sqrt{3} + \sqrt{2}\,i) – (\sqrt{3} – \sqrt{2}\,i)}
$$
Solution
$$
\frac{(3 + i\sqrt{5})(3 – i\sqrt{5})}{(\sqrt{3} + \sqrt{2}i) – (\sqrt{3} – \sqrt{2}i)}
= \frac{9 – (i\sqrt{5})^2}{\sqrt{3} + \sqrt{2}i – \sqrt{3} + \sqrt{2}i}
= \frac{9 – (i\sqrt{5})^2}{2\sqrt{2}\,i}
$$
Since $i^2 = -1$, we have $(i\sqrt{5})^2 = i^2\cdot 5 = -5$, so
$$
9 – (i\sqrt{5})^2 = 9 – (-5) = 14
$$
Thus
$$
\frac{9 – (i\sqrt{5})^2}{2\sqrt{2}\,i} = \frac{14}{2\sqrt{2}\,i} = \frac{7}{\sqrt{2}\,i}
$$
Multiply numerator and denominator by $\sqrt{2}\,i$:
$$
\frac{7}{\sqrt{2}\,i}\cdot\frac{\sqrt{2}\,i}{\sqrt{2}\,i}
= \frac{7\sqrt{2}\,i}{(\sqrt{2}\,i)^2}
= \frac{7\sqrt{2}\,i}{2i^2}
$$
Since $i^2 = -1$,
$$
\frac{7\sqrt{2}\,i}{2i^2} = \frac{7\sqrt{2}\,i}{2(-1)} = -\frac{7\sqrt{2}}{2}\,i
$$
So the expression in the form $a+ib$ is
$$
0 – i\frac{7\sqrt{2}}{2}\quad\text{or}\quad -\frac{7\sqrt{2}}{2}\,i.
$$
FAQs Based on NCERT Solutions of Complex Numbers and Quadratic Equations
What is Chapter 4 Complex Numbers and Quadratic Equations mainly about?
This chapter introduces imaginary numbers, complex numbers, algebra of complex numbers, and quadratic equations. Students learn how to perform operations on complex numbers and solve quadratic equations using different methods. The chapter also explains concepts like discriminant, nature of roots, modulus, conjugate, and Argand plane representation.
What is the value of iota (i) in complex numbers?
The imaginary unit (i) is defined by the relation (i^2=-1). It is used to represent imaginary numbers and helps solve equations that do not have real solutions. Powers of (i) follow a repeating cyclic pattern after every four powers.
Why are complex numbers important in Class 11 Maths?
Complex numbers form the foundation for advanced mathematics, engineering, and physics. They are important for solving algebraic equations, understanding higher-level mathematical concepts, and preparing for competitive exams like JEE, NDA, IMUCET, and Merchant Navy entrance examinations.
What is a quadratic equation?
A quadratic equation is an equation of degree 2 generally written in the form (ax^2+bx+c=0), where (a\neq0). The solutions of quadratic equations are called roots, and they can be real or complex depending on the discriminant.
What methods are used to solve quadratic equations?
Quadratic equations can be solved using factorization, completing the square, and quadratic formula methods. Among these, the quadratic formula is the most commonly used method for solving all types of quadratic equations.
What is the discriminant of a quadratic equation?
The discriminant is the expression (b^2-4ac) in a quadratic equation. It helps determine the nature of roots. If the discriminant is positive, roots are real and distinct; if zero, roots are real and equal; and if negative, roots are complex.
How many exercises are included in the latest NCERT pattern?
According to the latest NCERT rationalised syllabus, the chapter generally includes Exercise 4.1, Exercise 4.2, Exercise 4.3, and Miscellaneous Exercise. Some older NCERT editions may also contain additional exercises.
Are NCERT solutions enough for CBSE Class 11 preparation?
NCERT solutions are very important for CBSE school examinations because they cover all fundamental concepts and textbook problems. They also help students build a strong base for competitive exams when combined with additional practice questions and PYQs.
Is this chapter important for JEE Main and Advanced?
Yes, Complex Numbers and Quadratic Equations is one of the most important algebra chapters for JEE Main and Advanced. Questions based on complex numbers, roots of equations, discriminant, and properties of quadratic equations are frequently asked.
What is the best way to prepare this chapter?
Students should first understand all formulas and concepts from NCERT, then practice solved examples, NCERT exercises, exemplar problems, and previous year questions. Regular practice improves speed, accuracy, and conceptual clarity.
Mathematics explanations by Maths Anand Classes and Written by Er. Neeraj Anand.