Class 11 Mathematics
Exercise 4.1
NCERT Solutions
JEE Mains
NEET
Board Exam
Chapter 4 — COMPLEX NUMBERS AND QUADRATIC EQUATIONS
Step-by-step NCERT solutions with stress–strain analysis and exam-oriented hints for Boards, JEE & NEET.
📋 14 questions
⏱ Ideal time: 45-60 min
📍 Now at: Q1
Concept Used
\(i = \sqrt{-1}\)
\(i^2 = -1\)
Multiplication rule: \((ai)(bi) = ab \cdot i^2\)
Standard form: \(a + ib\), where \(a, b \in \mathbb{R}\)
Solution Roadmap
Step 1: Multiply constants separately
Step 2: Multiply \(i \cdot i = i^2\)
Step 3: Replace \(i^2 = -1\)
Step 4: Express result in \(a + ib\) form
Solution
We simplify the given expression step by step:
$$\begin{aligned}
(5i)\left(-\dfrac{3}{5}i\right)
&= 5i \cdot \dfrac{-3}{5}i \
&= -3 i^2 \
&= -3(-1) \
&= 3
\end{aligned}$$
Writing in standard form:
\[
3 = 3 + 0i
\]
Final Answer: \(3 + 0i\)
Geometric Interpretation (Argand Plane)
Re
Im
(3, 0)
The result lies on the real axis, showing it is a purely real number.
Why This Question Matters
Builds foundation of complex number algebra
Frequently tested in CBSE board exams (1–2 marks direct question)
Essential for simplifying expressions in JEE Main & NEET
Helps in understanding powers of \(i\), a core concept for higher problems
Concept Used
Powers of \(i\) follow a cyclic pattern of 4:
\(i^1 = i,\; i^2 = -1,\; i^3 = -i,\; i^4 = 1\)
General rule: \(i^n = i^{(n \bmod 4)}\)
Standard form: \(a + ib\)
Solution Roadmap
Step 1: Reduce powers using modulo 4
Step 2: Replace using cycle values
Step 3: Add results
Step 4: Express in \(a + ib\) form
Solution
Using cyclicity of powers of \(i\):
$$\begin{aligned}
i^9 &= i^{(9 \bmod 4)} = i^1 = i \\
i^{19} &= i^{(19 \bmod 4)} = i^3 = -i
\end{aligned}$$
$$\begin{aligned}
i^9 + i^{19} &= i + (-i) \\
&= 0
\end{aligned}$$
Writing in standard form:
\[
0 = 0 + 0i
\]
Final Answer: \(0 + 0i\)
Cyclic Pattern of Powers of \(i\)
1
i
-1
-i
Every increase of power rotates the point by \(90^\circ\) on the Argand plane.
Why This Question Matters
Very common pattern-based question in CBSE board exams
Direct application of modulo arithmetic (important for JEE)
Strengthens speed and accuracy in simplifying powers of \(i\)
Foundation for solving higher-level complex number identities
Concept Used
Powers of \(i\) are cyclic with period 4
\(i^4 = 1\), so \(i^n = i^{(n \bmod 4)}\)
Negative powers: \(i^{-n} = \dfrac{1}{i^n}\)
Useful identity: \(i^{-1} = -i\)
Solution Roadmap
Step 1: Convert negative exponent into positive form
Step 2: Reduce power using modulo 4
Step 3: Use cyclic values of \(i\)
Step 4: Express in \(a + ib\) form
Solution
First, rewrite the negative power:
$$\begin{aligned}
i^{-39} &= \frac{1}{i^{39}}
\end{aligned}$$
Now reduce the exponent using modulo 4:
$$\begin{aligned}
39 \bmod 4 = 3 \Rightarrow i^{39} = i^3 = -i
\end{aligned}$$
$$\begin{aligned}
i^{-39} &= \frac{1}{-i} \
&= i
\end{aligned}$$
Writing in standard form:
\[
i = 0 + 1i
\]
Final Answer: \(0 + i\)
Negative Power as Reverse Rotation
Re
Im
i
Reverse rotation
Negative powers correspond to clockwise rotation on the Argand plane.
Why This Question Matters
Tests understanding of negative exponents — frequently asked in exams
Important for JEE Main where powers of \(i\) appear in complex simplifications
Builds strong conceptual clarity for cyclic patterns
Helps avoid common mistakes in sign handling
Concept Used
Distributive property: \(a(b+c) = ab + ac\)
Multiplication with \(i\): \(i^2 = -1\)
Combine like terms (real and imaginary separately)
Standard form: \(a + ib\)
Solution Roadmap
Step 1: Expand both terms using distributive law
Step 2: Replace \(i^2 = -1\)
Step 3: Combine real and imaginary parts
Step 4: Write final answer in \(a + ib\) form
Solution
Expand both terms:
$$\begin{aligned}
3(7+7i) + i(7+7i)
&= 21 + 21i + 7i + 7i^2 \\
&= 21 + 28i + 7(-1) \\
&= 21 + 28i - 7 \\
&= 14 + 28i
\end{aligned}$$
Final Answer: \(14 + 28i\)
Geometric Interpretation (Vector Addition)
Re
Im
(14, 28)
The result represents a vector in the complex plane combining real and imaginary components.
Why This Question Matters
Tests algebraic manipulation of complex numbers (very common in boards)
Strengthens distributive expansion — key for JEE simplification problems
Builds understanding of real vs imaginary separation
Forms base for vector interpretation of complex numbers
Concept Used
Subtraction of complex numbers: distribute the minus sign
\(-(a + ib) = -a - ib\)
Combine real and imaginary parts separately
Standard form: \(a + ib\)
Solution Roadmap
Step 1: Remove brackets carefully (change signs)
Step 2: Group real and imaginary terms
Step 3: Simplify
Step 4: Express in \(a + ib\) form
Solution
Carefully remove the brackets:
$$\begin{aligned}
(1 - i) - (-1 + 6i)
&= 1 - i + 1 - 6i \\
&= (1 + 1) + (-i - 6i) \\
&= 2 - 7i
\end{aligned}$$
Final Answer: \(2 - 7i\)
Geometric Interpretation (Subtraction as Vector Difference)
Re
Im
(2, -7)
Subtraction shifts the point downward (negative imaginary direction).
Why This Question Matters
Sign mistakes are one of the most common errors in exams
Frequently asked in CBSE boards as direct simplification
Essential for JEE where expressions become multi-layered
Builds clarity in handling brackets and negative terms
Concept Used
Subtraction of complex numbers: distribute negative sign carefully
Combine real and imaginary parts separately
Use LCM to simplify fractional expressions
Standard form: \(a + ib\)
Solution Roadmap
Step 1: Remove brackets and apply sign change
Step 2: Separate real and imaginary parts
Step 3: Use LCM to simplify fractions
Step 4: Write in \(a + ib\) form
Solution
Remove brackets carefully:
$$\begin{aligned}
\left(\frac{1}{5} + i\frac{2}{5}\right) - \left(4 + i\frac{5}{2}\right)
&= \frac{1}{5} + i\frac{2}{5} - 4 - i\frac{5}{2}
\end{aligned}$$
Group real and imaginary parts:
$$\begin{aligned}
&= \left(\frac{1}{5} - 4\right) + i\left(\frac{2}{5} - \frac{5}{2}\right)
\end{aligned}$$
Simplify each part:
$$\begin{aligned}
\frac{1}{5} - 4 &= \frac{1 - 20}{5} = -\frac{19}{5} \\
\frac{2}{5} - \frac{5}{2} &= \frac{4 - 25}{10} = -\frac{21}{10}
\end{aligned}$$
Final result:
\[
-\frac{19}{5} - \frac{21}{10}i
\]
Final Answer: \(-\dfrac{19}{5} - \dfrac{21}{10}i\)
Geometric Interpretation (Quadrant III Position)
Re
Im
(-19/5, -21/10)
Both real and imaginary parts are negative, so the point lies in the third quadrant.
Why This Question Matters
Tests accuracy in handling fractions — a common source of mistakes
Frequently appears in CBSE board exams (direct simplification)
Important for JEE where fraction-heavy expressions are common
Improves algebraic precision and speed under exam conditions
Concept Used
Addition and subtraction of complex numbers
Careful handling of multiple brackets
Combine real and imaginary parts separately
Use LCM for fractional simplification
Standard form: \(a + ib\)
Solution Roadmap
Step 1: Simplify expression inside square brackets
Step 2: Distribute negative sign in last bracket
Step 3: Group real and imaginary parts
Step 4: Use LCM and simplify
Step 5: Express in \(a + ib\)
Solution
First simplify the expression inside the square brackets:
$$\begin{aligned}
\left(\frac{1}{3}+i\frac{7}{3}\right)+\left(4+i\frac{1}{3}\right)
&= \left(\frac{1}{3}+4\right) + i\left(\frac{7}{3}+\frac{1}{3}\right) \\
&= \frac{1}{3} + 4 + i\frac{8}{3}
\end{aligned}$$
Now subtract the last bracket:
$$\begin{aligned}
-\left(-\frac{4}{3}+i\right) = \frac{4}{3} - i
\end{aligned}$$
Combine all terms:
$$\begin{aligned}
&= \left(\frac{1}{3}+4+\frac{4}{3}\right) + i\left(\frac{8}{3}-1\right)
\end{aligned}$$
Simplify:
$$\begin{aligned}
\frac{1}{3} + \frac{4}{3} + 4 &= \frac{5}{3} + 4 = \frac{17}{3} \\
\frac{8}{3} - 1 &= \frac{8 - 3}{3} = \frac{5}{3}
\end{aligned}$$
Final result:
\[
\frac{17}{3} + \frac{5}{3}i
\]
Final Answer: \(\dfrac{17}{3} + \dfrac{5}{3}i\)
Geometric Interpretation (Resultant Vector)
Re
Im
(17/3, 5/3)
Both components are positive, so the point lies in the first quadrant.
Why This Question Matters
Tests multi-layer bracket handling — very common in exams
Combines fractions + complex arithmetic
Important for CBSE boards (step-mark based questions)
Builds accuracy for JEE multi-step simplification problems
Enhances structured problem-solving approach
Concept Used
Use algebraic expansion or repeated squaring
\(i^2 = -1\)
Powers of complex numbers can be simplified efficiently using \((z^2)^2\)
Standard form: \(a + ib\)
Solution Roadmap
Step 1: Compute \((1-i)^2\)
Step 2: Square the result
Step 3: Use \(i^2 = -1\)
Step 4: Express in \(a + ib\) form
Solution
First compute the square:
$$\begin{aligned}
(1 - i)^2
&= 1 - 2i + i^2 \\
&= 1 - 2i - 1 \\
&= -2i
\end{aligned}$$
Now square again:
$$\begin{aligned}
(1 - i)^4
&= (-2i)^2 \\
&= 4i^2 \\
&= -4
\end{aligned}$$
Writing in standard form:
\[
-4 = -4 + 0i
\]
Final Answer: \(-4 + 0i\)
Geometric Interpretation (Rotation & Scaling)
Re
Im
(1,-1)
(-4,0)
Repeated multiplication causes rotation and scaling, resulting in a real negative number.
Why This Question Matters
Demonstrates efficient use of repeated squaring
Common in CBSE boards (power simplification)
Highly relevant for JEE where powers of complex numbers are frequent
Builds intuition for rotation and magnitude in complex plane
Concept Used
Binomial expansion: \((a+b)^3 = a^3 + b^3 + 3a^2b + 3ab^2\)
\(i^2 = -1,\; i^3 = -i\)
Separate real and imaginary parts carefully
Standard form: \(a + ib\)
Solution Roadmap
Step 1: Apply binomial expansion
Step 2: Simplify each term using powers of \(i\)
Step 3: Combine real and imaginary parts
Step 4: Write in \(a + ib\) form
Solution
Apply binomial expansion:
$$\begin{aligned}
\left(\frac{1}{3} + 3i\right)^3
&= \left(\frac{1}{3}\right)^3 + (3i)^3
+ 3\left(\frac{1}{3}\right)^2(3i)
+ 3\left(\frac{1}{3}\right)(3i)^2
\end{aligned}$$
Simplify each term:
$$\begin{aligned}
\left(\frac{1}{3}\right)^3 &= \frac{1}{27} \\
(3i)^3 &= 27i^3 = -27i \\
3\cdot\left(\frac{1}{3}\right)^2\cdot 3i &= i \\
3\cdot\frac{1}{3}\cdot(3i)^2 &= 9i^2 = -9
\end{aligned}$$
Combine terms:
$$\begin{aligned}
&= \frac{1}{27} - 9 + (-27i + i) \\
&= \frac{1 - 243}{27} - 26i \\
&= -\frac{242}{27} - 26i
\end{aligned}$$
Final Answer: \(-\dfrac{242}{27} - 26i\)
Geometric Interpretation (Scaling & Rotation)
Re
Im
(-242/27, -26)
Large imaginary component dominates, placing the result deep in the third quadrant.
Why This Question Matters
Tests binomial expansion with complex numbers — very common in boards
Important for JEE where higher powers are frequently used
Builds accuracy in handling multiple terms and fractions
Strengthens control over powers of \(i\)
Foundation for De Moivre’s theorem (advanced topics)
Concept Used
Binomial expansion: \((a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\)
\(i^2 = -1,\; i^3 = -i\)
Careful handling of negative signs and fractions
Standard form: \(a + ib\)
Solution Roadmap
Step 1: Identify \(a = -2,\; b = -\frac{1}{3}i\)
Step 2: Apply binomial expansion
Step 3: Simplify each term using powers of \(i\)
Step 4: Combine real and imaginary parts
Step 5: Express in \(a + ib\) form
Solution
Using binomial expansion:
$$\begin{aligned}
\left(-2 - \frac{1}{3}i\right)^3
&= (-2)^3
+ 3(-2)^2\left(-\frac{1}{3}i\right)
+ 3(-2)\left(-\frac{1}{3}i\right)^2
+ \left(-\frac{1}{3}i\right)^3
\end{aligned}$$
Simplify each term:
$$\begin{aligned}
(-2)^3 &= -8 \\
3(-2)^2\left(-\frac{1}{3}i\right) &= 12\cdot\left(-\frac{1}{3}i\right) = -4i \\
3(-2)\left(-\frac{1}{3}i\right)^2 &= -6 \cdot \frac{1}{9}i^2 = -\frac{6}{9}(-1) = \frac{2}{3} \\
\left(-\frac{1}{3}i\right)^3 &= -\frac{1}{27}i^3 = \frac{1}{27}i
\end{aligned}$$
Combine real and imaginary parts:
$$\begin{aligned}
\text{Real part} &= -8 + \frac{2}{3} = -\frac{24}{3} + \frac{2}{3} = -\frac{22}{3} \\
\text{Imaginary part} &= -4i + \frac{1}{27}i = \left(-\frac{108}{27} + \frac{1}{27}\right)i = -\frac{107}{27}i
\end{aligned}$$
Final result:
\[
-\frac{22}{3} - \frac{107}{27}i
\]
Final Answer: \(-\dfrac{22}{3} - \dfrac{107}{27}i\)
Geometric Interpretation (Third Quadrant)
Re
Im
(-22/3, -107/27)
Both components are negative, so the complex number lies in the third quadrant.
Why This Question Matters
Classic binomial expansion with fractions — common in CBSE exams
High error probability due to signs and fractions (JEE trap type)
Strengthens multi-term algebraic control
Builds base for higher power expansions and complex identities
Concept Used
Multiplicative inverse: \(z^{-1} = \dfrac{1}{z}\)
For \(z = a + ib\): \(z^{-1} = \dfrac{a - ib}{a^2 + b^2}\)
Complex conjugate: \(\overline{z} = a - ib\)
Modulus squared: \(|z|^2 = a^2 + b^2\)
Solution Roadmap
Step 1: Identify \(a\) and \(b\)
Step 2: Find conjugate \(\overline{z}\)
Step 3: Compute \(|z|^2\)
Step 4: Apply formula \(z^{-1} = \dfrac{\overline{z}}{|z|^2}\)
Step 5: Express in \(a + ib\) form
Solution
Given:
\[
z = 4 - 3i
\]
Conjugate:
\[
\overline{z} = 4 + 3i
\]
Modulus squared:
$$\begin{aligned}
|z|^2 &= 4^2 + (-3)^2 \\
&= 16 + 9 = 25
\end{aligned}$$
Multiplicative inverse:
$$\begin{aligned}
z^{-1} &= \frac{\overline{z}}{|z|^2} \\
&= \frac{4 + 3i}{25} \\
&= \frac{4}{25} + \frac{3}{25}i
\end{aligned}$$
Final Answer: \(\dfrac{4}{25} + \dfrac{3}{25}i\)
Geometric Interpretation (Inverse Reflection & Scaling)
Re
Im
(4, -3)
(4/25, 3/25)
The inverse reflects across the real axis (conjugate) and scales by \(\frac{1}{|z|^2}\).
Why This Question Matters
Direct formula-based question in CBSE board exams
Very common in JEE (especially in division of complex numbers)
Builds foundation for rationalization techniques
Helps understand geometric meaning of inverse in complex plane
Frequently used in solving equations involving complex numbers
Concept Used
Multiplicative inverse: \(z^{-1} = \dfrac{1}{z}\)
For \(z = a + ib\): \(z^{-1} = \dfrac{a - ib}{a^2 + b^2}\)
Conjugate: \(\overline{z} = a - ib\)
Modulus squared: \(|z|^2 = a^2 + b^2\)
Solution Roadmap
Step 1: Identify real and imaginary parts
Step 2: Find conjugate
Step 3: Compute modulus squared
Step 4: Apply inverse formula
Step 5: Express in \(a + ib\) form
Solution
Given:
\[
z = \sqrt{5} + 3i
\]
Conjugate:
\[
\overline{z} = \sqrt{5} - 3i
\]
Modulus squared:
$$\begin{aligned}
|z|^2 &= (\sqrt{5})^2 + (3)^2 \\
&= 5 + 9 = 14
\end{aligned}$$
Multiplicative inverse:
$$\begin{aligned}
z^{-1} &= \frac{\overline{z}}{|z|^2} \\
&= \frac{\sqrt{5} - 3i}{14} \\
&= \frac{\sqrt{5}}{14} - \frac{3}{14}i
\end{aligned}$$
Final Answer: \(\dfrac{\sqrt{5}}{14} - \dfrac{3}{14}i\)
Geometric Interpretation (Reflection & Scaling)
Re
Im
(√5, 3)
(√5/14, -3/14)
The inverse reflects across the real axis and shrinks by factor \(\frac{1}{|z|^2}\).
Why This Question Matters
Combines surds with complex numbers — common in CBSE exams
Important for JEE where irrational components appear frequently
Strengthens conceptual clarity of conjugate method
Essential for division and rationalization problems
Improves algebraic accuracy with mixed number types
Concept Used
Multiplicative inverse: \(z^{-1} = \dfrac{1}{z}\)
\(i^2 = -1 \Rightarrow \dfrac{1}{i} = -i\)
Special identity: \(\dfrac{1}{-i} = i\)
Standard form: \(a + ib\)
Solution Roadmap
Step 1: Write inverse as \(\dfrac{1}{-i}\)
Step 2: Use identity \(\dfrac{1}{i} = -i\)
Step 3: Simplify directly
Step 4: Express in \(a + ib\) form
Solution
Given:
\[
z = -i
\]
Multiplicative inverse:
$$\begin{aligned}
z^{-1}
&= \frac{1}{-i} \\
&= -\frac{1}{i} \\
&= -(-i) \\
&= i
\end{aligned}$$
Writing in standard form:
\[
i = 0 + 1i
\]
Final Answer: \(0 + i\)
Geometric Interpretation (Rotation by 180°)
Re
Im
-i
i
The inverse of \(-i\) is \(i\), which is a reflection across the origin.
Why This Question Matters
Tests shortcut understanding — no need for full formula
Very common conceptual MCQ in JEE
Builds fluency with powers and reciprocals of \(i\)
Helps in rapid simplification during exams
Reduces calculation time significantly
Concept Used
Identity: \((a+ib)(a-ib) = a^2 + b^2\)
Simplification of complex expressions
Rationalization using \(i^{-1} = -i\)
Standard form: \(a + ib\)
Solution Roadmap
Step 1: Simplify numerator using identity
Step 2: Simplify denominator carefully
Step 3: Reduce the fraction
Step 4: Eliminate \(i\) from denominator
Step 5: Express in \(a + ib\)
Solution
Simplify the numerator using identity:
$$\begin{aligned}
(3 + i\sqrt{5})(3 - i\sqrt{5})
&= 3^2 - (i\sqrt{5})^2 \\
&= 9 - (-5) \\
&= 14
\end{aligned}$$
Simplify the denominator:
$$\begin{aligned}
(\sqrt{3} + \sqrt{2}i) - (\sqrt{3} - \sqrt{2}i)
&= \sqrt{3} + \sqrt{2}i - \sqrt{3} + \sqrt{2}i \\
&= 2\sqrt{2}i
\end{aligned}$$
Now divide:
$$\begin{aligned}
\frac{14}{2\sqrt{2}i}
&= \frac{7}{\sqrt{2}i}
\end{aligned}$$
Rationalize:
$$\begin{aligned}
\frac{7}{\sqrt{2}i}
&= \frac{7}{\sqrt{2}} \cdot \frac{1}{i} \\
&= \frac{7}{\sqrt{2}}(-i) \\
&= -\frac{7\sqrt{2}}{2}i
\end{aligned}$$
Writing in standard form:
\[
0 - \frac{7\sqrt{2}}{2}i
\]
Final Answer: \(-\dfrac{7\sqrt{2}}{2}i\)
Geometric Interpretation (Pure Imaginary Result)
Re
Im
(0, -7√2/2)
The result lies purely on the imaginary axis (no real component).
Why This Question Matters
Combines identities + rationalization — very common exam pattern
Tests ability to simplify complex fractions efficiently
Important for CBSE (multi-step structured solution)
Highly relevant for JEE where mixed concepts appear together
Builds speed using identities instead of expansion
🎓
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