This MCQ module is based on: 4.1 Introduction
TOPIC 13 OF 45
4.1 Introduction
🎓 Class 11
Mathematics
CBSE
Theory
Ch 4 — Complex Numbers and Quadratic Equations
⏱ ~15 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: 4.1 Introduction
Targeting Class 11 level in Algebra, with Advanced difficulty.
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4.1 Introduction
In earlier classes you solved linear and quadratic equations in one variable, but only when real solutions existed. The equation \(x^2+1=0\) has no real solution because \(x^2\ge 0\) for every real \(x\), so \(x^2+1\) is at least 1 — never zero. To make every quadratic solvable, we extend the real number system to include a new "imaginary" quantity \(i\) satisfying \(i^2=-1\).
This idea, formalised by William Rowan Hamilton? in 1833, builds the field of complex numbers? — a single algebraic system in which every polynomial of degree \(n\) has exactly \(n\) roots (the Fundamental Theorem of Algebra).
William Rowan Hamilton
1805 – 1865
The Irish mathematician who, in 1833, gave the first rigorous algebraic definition of a complex number as an ordered pair \((a,b)\) of real numbers, with addition and multiplication rules that produce \(i^2=-1\) without any mystical "imaginary" leap. His later invention of quaternions generalised the idea to four-dimensional rotations — used today in computer graphics and aerospace.
4.2 Complex Numbers
Definition
A complex number is an expression of the form
\[z=a+ib\quad\text{with } a,b\in\mathbb R\text{ and }i^2=-1.\]
Here \(a\) is the real part \(\text{Re}(z)\), and \(b\) is the imaginary part \(\text{Im}(z)\). For example, \(z=2+(-1)i\) gives \(\text{Re}(z)=2\) and \(\text{Im}(z)=-1\). A purely real number is \(z=a+0i=a\); a purely imaginary number is \(z=0+ib=ib\).
Equality
Two complex numbers \(z_1=a+ib\) and \(z_2=c+id\) are equal iff both real parts agree and both imaginary parts agree:
\[a+ib=c+id\iff a=c\ \text{and}\ b=d.\]
4.3 Algebra of Complex Numbers
4.3.1 Addition
If \(z_1=a+ib\) and \(z_2=c+id\), then
\[z_1+z_2=(a+c)+i(b+d).\]Add real parts and imaginary parts separately. The set of complex numbers with this addition satisfies closure, commutativity, associativity, has additive identity \(0+i0\), and every \(z=a+ib\) has additive inverse \(-z=-a+i(-b)\).
4.3.2 Difference
\[z_1-z_2=z_1+(-z_2)=(a-c)+i(b-d).\]4.3.3 Multiplication
Just multiply out using \(i^2=-1\):
\[z_1\cdot z_2=(a+ib)(c+id)=ac+iad+ibc+i^2bd=(ac-bd)+i(ad+bc).\]The multiplicative identity is \(1=1+i0\). For any non-zero \(z=a+ib\), the multiplicative inverse is
Multiplicative inverse
\[z^{-1}=\dfrac{1}{a+ib}=\dfrac{a}{a^2+b^2}+i\dfrac{-b}{a^2+b^2},\quad (a,b)\ne (0,0).\]
Obtained by multiplying numerator and denominator of \(1/(a+ib)\) by the conjugate \(a-ib\).
4.3.4 Division
Use the inverse: \(z_1/z_2=z_1\cdot z_2^{-1}\). The standard recipe — multiply by the conjugate of the denominator:
\[\dfrac{a+ib}{c+id}=\dfrac{(a+ib)(c-id)}{(c+id)(c-id)}=\dfrac{(ac+bd)+i(bc-ad)}{c^2+d^2}.\]4.3.5 Powers of \(i\)
Since \(i^2=-1\),
\[i^3=i^2\cdot i=-i,\quad i^4=(i^2)^2=1,\quad i^5=i^4\cdot i=i,\quad i^6=-1,\ldots\]The pattern repeats every four exponents:
| n mod 4 | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| iⁿ | 1 | i | −1 | −i |
Negative powers: \(i^{-1}=\dfrac{1}{i}=\dfrac{1}{i}\cdot\dfrac{i}{i}=\dfrac{i}{-1}=-i\), and similarly \(i^{-2}=-1,\ i^{-3}=i,\ i^{-4}=1\).
4.3.6 Square roots of negative real numbers
For any positive real \(a\), define \(\sqrt{-a}=\sqrt a\cdot i\). For example \(\sqrt{-9}=3i\). The two square roots of \(-a\) are \(\pm\sqrt a\,i\).
Pitfall — \(\sqrt a\sqrt b=\sqrt{ab}\) FAILS for two negatives
The familiar identity \(\sqrt a\sqrt b=\sqrt{ab}\) holds only when at least one of \(a,b\) is non-negative. With both negative, you must factor out \(i\) first:
\[\sqrt{-1}\cdot\sqrt{-1}=i\cdot i=-1,\qquad \text{not}\quad\sqrt{(-1)(-1)}=\sqrt 1=1.\]
The correct manipulation: \(\sqrt{-2}\cdot\sqrt{-3}=(\sqrt 2\,i)(\sqrt 3\,i)=\sqrt 6\,i^2=-\sqrt 6\).
Interactive: Complex Multiplication on the Plane
Drag the sliders for two complex numbers \(z_1=a+ib\) and \(z_2=c+id\). The simulation shows the product \(z_1z_2=(ac-bd)+i(ad+bc)\) updating live, with both inputs and the output plotted on the Argand plane.
2
1
1
2
z₁ = 2 + 1i | z₂ = 1 + 2i | z₁·z₂ = (2−2) + (4+1)i = 0 + 5i
Worked Examples
Example 1. Express \((-5i)\!\left(\dfrac{1}{8}i\right)\) in standard form.
\((-5i)(\tfrac{1}{8}i)=-\dfrac{5}{8}\,i^2=-\dfrac{5}{8}(-1)=\dfrac{5}{8}+0i\).
Example 2. Express \(i^{-35}\) in standard form.
\(i^{-35}=\dfrac{1}{i^{35}}=\dfrac{1}{i^{32}\cdot i^3}=\dfrac{1}{1\cdot(-i)}=\dfrac{-1}{i}\cdot\dfrac{i}{i}=\dfrac{-i}{-1}=i\).
Example 3. Express the product \((3+2i)(2-3i)\) in standard form.
\((3+2i)(2-3i)=6-9i+4i-6i^2=6-5i+6=12-5i\).
Example 4. Express \(\dfrac{(5+\sqrt 3 i)}{(1-\sqrt 3 i)}\) in standard form.
Multiply by conjugate \(1+\sqrt 3 i\):
\[\dfrac{(5+\sqrt 3 i)(1+\sqrt 3 i)}{(1-\sqrt 3 i)(1+\sqrt 3 i)}=\dfrac{5+5\sqrt 3 i+\sqrt 3 i+3i^2}{1-3i^2}=\dfrac{2+6\sqrt 3 i}{4}=\dfrac{1}{2}+\dfrac{3\sqrt 3}{2}i.\]
Example 5. Express \(\sqrt{-3}\cdot\sqrt{-2}\) in standard form.
Factor out \(i\) first: \(\sqrt{-3}=\sqrt 3\,i\) and \(\sqrt{-2}=\sqrt 2\,i\). So
\[\sqrt{-3}\cdot\sqrt{-2}=(\sqrt 3\,i)(\sqrt 2\,i)=\sqrt 6\,i^2=-\sqrt 6.\]
Note: applying \(\sqrt a\sqrt b=\sqrt{ab}\) directly would give the WRONG answer \(\sqrt 6\).
Activity: Powers-of-\(i\) Clock
L3 ApplyMaterials: Paper, pen, optional calculator.
Predict: What is \(i^{2025}\)? Without computing \(i^{2024}\), can you find it just from \(2025 \bmod 4\)?
- Tabulate \(i^0, i^1, i^2, \ldots, i^{12}\) and observe the cycle of length 4.
- Now find \(i^{50},\ i^{99},\ i^{2025}\) and \(i^{-7}\) using only \(n\bmod 4\).
- Verify with a sum: compute \(i^1+i^2+i^3+\ldots+i^{100}\). What's the trick?
- Compute \(i^{592}\cdot i^{592}\). Does the answer equal \((i^{1184})\)? (Check both.)
2025 mod 4 = 1, so \(i^{2025}=i\). For the sum: every block of 4 consecutive powers (\(i+i^2+i^3+i^4=i-1-i+1=0\)). 100 powers = 25 such blocks, so the sum is 0. \(i^{592}\cdot i^{592}=i^{1184}\); 1184 mod 4 = 0, so the answer is 1. The cycle period 4 is the central organising principle of all powers of \(i\).
Competency-Based Questions
Scenario: An electrical engineer represents AC circuit impedance as a complex number \(Z=R+iX\), where \(R\) is resistance (real, positive) and \(X\) is reactance (positive for inductive, negative for capacitive). The total impedance of two components in series is \(Z_1+Z_2\); in parallel it is \(\dfrac{Z_1Z_2}{Z_1+Z_2}\).
Q1. \(i^{-39}=\)
L3 ApplyAnswer: (a) i. \(i^{-39}=i^{-39+40}=i^1=i\) (since \(i^{40}=1\)).
Q2. The standard form of \(\dfrac{3+2i}{2-i}\) is:
L3 ApplyAnswer: Multiply by conjugate \(2+i\): \(\dfrac{(3+2i)(2+i)}{4+1}=\dfrac{6+3i+4i+2i^2}{5}=\dfrac{4+7i}{5}=\dfrac{4}{5}+\dfrac{7}{5}i\).
Q3. (True/False) "\(\sqrt{-4}\cdot\sqrt{-9}=\sqrt{36}=6\)." Justify.
L5 EvaluateFalse. The identity \(\sqrt a\sqrt b=\sqrt{ab}\) requires at least one of \(a,b\) non-negative. Correct: \(\sqrt{-4}\cdot\sqrt{-9}=(2i)(3i)=6i^2=-6\), not +6.
Q4. Two AC components have impedances \(Z_1=4+3i\,\Omega\) and \(Z_2=2-i\,\Omega\). Find \(Z_1+Z_2\) and \(Z_1\cdot Z_2\).
L3 ApplySolution: \(Z_1+Z_2=(4+2)+(3-1)i=6+2i\,\Omega\). \(Z_1Z_2=(4)(2)+(4)(-i)+(3i)(2)+(3i)(-i)=8-4i+6i-3i^2=11+2i\,\Omega\).
Q5. Design: derive a closed-form expression for \(z\cdot\bar z\) where \(z=a+ib\) and \(\bar z=a-ib\). Why is this expression always a non-negative real number?
L6 CreateSolution: \(z\bar z=(a+ib)(a-ib)=a^2-(ib)^2=a^2-i^2 b^2=a^2+b^2\). The right side is a sum of squares of real numbers, hence \(\ge 0\), and equals 0 only when \(a=b=0\). This is the key calculation that defines the modulus \(|z|=\sqrt{a^2+b^2}\) in Part 2.
Assertion–Reason Questions
Assertion (A): \(i^4=1\).
Reason (R): \(i^2=-1\), so \(i^4=(i^2)^2=(-1)^2=1\).
Reason (R): \(i^2=-1\), so \(i^4=(i^2)^2=(-1)^2=1\).
Answer: (a). R is exactly the calculation that yields A.
Assertion (A): \(\sqrt{-2}\cdot\sqrt{-3}=\sqrt 6\).
Reason (R): \(\sqrt a\cdot\sqrt b=\sqrt{ab}\) for any real numbers \(a,b\).
Reason (R): \(\sqrt a\cdot\sqrt b=\sqrt{ab}\) for any real numbers \(a,b\).
Answer: (d). A is false: actually \(\sqrt{-2}\sqrt{-3}=-\sqrt 6\). R is also false in general — it requires at least one of \(a,b\) non-negative. Sometimes phrased as "both false"; closest single option (d).
Assertion (A): Every non-zero complex number has a multiplicative inverse.
Reason (R): For \(z=a+ib\ne 0\), the inverse is \(\dfrac{a-ib}{a^2+b^2}\).
Reason (R): For \(z=a+ib\ne 0\), the inverse is \(\dfrac{a-ib}{a^2+b^2}\).
Answer: (a). Verify: \((a+ib)\cdot\dfrac{a-ib}{a^2+b^2}=\dfrac{a^2+b^2}{a^2+b^2}=1\). R provides the explicit construction making A true.
Frequently Asked Questions
What is a complex number?
A complex number is an expression of the form a + ib where a and b are real numbers and i is defined by i² = −1. Here a is the real part Re(z) and b is the imaginary part Im(z).
Why was the imaginary unit i needed?
Equations like x² + 1 = 0 have no real solutions because the square of any real number is non-negative. Defining i with i² = −1 extends the real numbers and lets every quadratic (and indeed every polynomial) have a root.
How do you multiply two complex numbers?
(a + ib)(c + id) = (ac − bd) + i(ad + bc). Just expand using i² = −1.
What are the powers of i?
i⁰ = 1, i¹ = i, i² = −1, i³ = −i, then the cycle repeats: i⁴ = 1, i⁵ = i, etc. So i^n only depends on n mod 4.
What is √(−a) for positive real a?
√(−a) = √a · i. So √(−9) = 3i and √(−2) = √2 · i.
Is √a · √b = √(ab) for negative numbers?
NO. The identity √a · √b = √(ab) holds only when at least one of a, b is non-negative. For example, √(−1) · √(−1) = i · i = −1, but √((−1)(−1)) = √1 = 1. Always factor out i first.
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