IIT JEE Main Maths -Unit 2- Algebra of complex numbers- Study Notes-New Syllabus
IIT JEE Main Maths -Unit 2- Algebra of complex numbers – Study Notes – New syllabus
IIT JEE Main Maths -Unit 2- Algebra of complex numbers – Study Notes -IIT JEE Main Maths – per latest Syllabus.
Key Concepts:
- Algebra of Complex Numbers
- Geometrical Representation of Addition and Multiplication of Complex Numbers
- Multiplication and Division in Polar/Exponential Form (De Moivre’s Theorem)
- Finding Powers and Roots of Complex Numbers (De Moivre’s Theorem) & Square Root of a Complex Number
Algebra of Complex Numbers
Let \( z_1 = a + ib \) and \( z_2 = c + id \) be two complex numbers. The following algebraic operations can be defined on complex numbers just like on real numbers.
1. Addition:
The sum of two complex numbers \( z_1 = a + ib \) and \( z_2 = c + id \) is given by:
\( z_1 + z_2 = (a + c) + i(b + d) \)
Rule: Add real parts together and imaginary parts together.
2. Subtraction:
\( z_1 – z_2 = (a – c) + i(b – d) \)
Rule: Subtract real parts and imaginary parts separately.
3. Multiplication:
\( z_1 \times z_2 = (a + ib)(c + id) = (ac – bd) + i(ad + bc) \)
Rule: Multiply like algebraic expressions using \( i^2 = -1 \).
4. Division:
To divide \( z_1 \) by \( z_2 \), multiply numerator and denominator by the conjugate of the denominator.
\( \dfrac{z_1}{z_2} = \dfrac{a + ib}{c + id} \times \dfrac{c – id}{c – id} = \dfrac{(ac + bd) + i(bc – ad)}{c^2 + d^2} \)
5. Conjugate of a Complex Number:
The conjugate of \( z = a + ib \) is \( \overline{z} = a – ib \).
- \( z \cdot \overline{z} = a^2 + b^2 = |z|^2 \)
- \( \overline{(z_1 + z_2)} = \overline{z_1} + \overline{z_2} \)
- \( \overline{(z_1 z_2)} = \overline{z_1} \, \overline{z_2} \)
6. Modulus Properties:
- \( |z_1 z_2| = |z_1| \cdot |z_2| \)
- \( \left|\dfrac{z_1}{z_2}\right| = \dfrac{|z_1|}{|z_2|} \)
- \( |z_1 + z_2| \le |z_1| + |z_2| \) (Triangle inequality)
- \( |z_1 – z_2| \ge \big| |z_1| – |z_2| \big| \)
Example
Find the sum and product of \( z_1 = 2 + 3i \) and \( z_2 = 1 – 2i \).
▶️ Answer / Explanation
Addition: \( z_1 + z_2 = (2 + 1) + i(3 – 2) = 3 + i \)
Multiplication: \( z_1 z_2 = (2 + 3i)(1 – 2i) = 2 – 4i + 3i – 6i^2 = 2 – i + 6 = 8 – i \)
Hence, \( z_1 + z_2 = 3 + i \), and \( z_1 z_2 = 8 – i \).
Example
If \( z_1 = 4 + 3i \) and \( z_2 = 2 – i \), find \( \dfrac{z_1}{z_2} \) in the form \( x + iy \).
▶️ Answer / Explanation
Step 1: Multiply numerator and denominator by the conjugate of \( z_2 \):
\( \dfrac{z_1}{z_2} = \dfrac{(4 + 3i)(2 + i)}{(2 – i)(2 + i)} \)
Step 2: Simplify numerator and denominator.
Denominator: \( (2 – i)(2 + i) = 4 + 1 = 5 \)
Numerator: \( (4 + 3i)(2 + i) = 8 + 4i + 6i + 3i^2 = 8 + 10i – 3 = 5 + 10i \)
\( \Rightarrow \dfrac{z_1}{z_2} = \dfrac{5 + 10i}{5} = 1 + 2i \)
Hence, \( \dfrac{z_1}{z_2} = 1 + 2i \).
Example
Let \( z_1 = 2 + i \) and \( z_2 = 3 – 2i \). Verify that \( |z_1 z_2| = |z_1| \cdot |z_2| \).
▶️ Answer / Explanation
Step 1: Compute \( z_1 z_2 \): \( (2 + i)(3 – 2i) = 6 – 4i + 3i – 2i^2 = 6 – i + 2 = 8 – i \)
Step 2: Find \( |z_1 z_2| = \sqrt{8^2 + (-1)^2} = \sqrt{65} \)
Step 3: Find \( |z_1| \) and \( |z_2| \):
\( |z_1| = \sqrt{2^2 + 1^2} = \sqrt{5}, \quad |z_2| = \sqrt{3^2 + (-2)^2} = \sqrt{13} \)
Step 4: \( |z_1| \cdot |z_2| = \sqrt{5} \times \sqrt{13} = \sqrt{65} \)
Hence verified: \( |z_1 z_2| = |z_1| \cdot |z_2| \).
Geometrical Representation of Addition and Multiplication of Complex Numbers
Every complex number \( z = a + ib \) can be represented as a point \( P(a, b) \) or as a vector \( \overrightarrow{OP} \) in the Argand plane, where:
- The x-axis represents the real part \( a \)
- The y-axis represents the imaginary part \( b \)
1. Geometrical Representation of Addition
Let \( z_1 = a_1 + ib_1 \) and \( z_2 = a_2 + ib_2 \).
Their sum is:
\( z_1 + z_2 = (a_1 + a_2) + i(b_1 + b_2) \)
This means the point representing \( z_1 + z_2 \) is obtained by adding the corresponding coordinates of \( z_1 \) and \( z_2 \).
Geometrically: The addition of two complex numbers corresponds to the vector addition of their position vectors in the Argand plane.
- Represent \( z_1 \) and \( z_2 \) as vectors \( \overrightarrow{OA} \) and \( \overrightarrow{OB} \).
- The diagonal of the parallelogram formed by these vectors represents \( z_1 + z_2 \).
2. Geometrical Representation of Multiplication
Let \( z_1 = r_1 (\cos \theta_1 + i \sin \theta_1) \) and \( z_2 = r_2 (\cos \theta_2 + i \sin \theta_2) \).
Their product is:
\( z_1 z_2 = r_1 r_2 [\cos(\theta_1 + \theta_2) + i \sin(\theta_1 + \theta_2)] \)
Interpretation:
- The modulus of the product \( |z_1 z_2| = |z_1| \times |z_2| \).
- The argument (angle) of the product \( \arg(z_1 z_2) = \arg(z_1) + \arg(z_2) \).
Geometrically: Multiplying two complex numbers means:
- Multiplying their moduli (scales the vector’s length).
- Adding their arguments (rotates the vector by that angle).
Example
Let \( z_1 = 3 + 2i \) and \( z_2 = 1 + 4i \). Represent \( z_1 + z_2 \) geometrically.
▶️ Answer / Explanation
Step 1: Calculate the sum:
\( z_1 + z_2 = (3 + 1) + i(2 + 4) = 4 + 6i \)
Step 2: Representation:
Plot points \( A(3, 2) \) and \( B(1, 4) \) on the Argand plane. Draw vectors \( \overrightarrow{OA} \) and \( \overrightarrow{OB} \). The diagonal of the parallelogram formed represents the point \( (4, 6) \), i.e. \( z_1 + z_2 \).
Example
Let \( z_1 = 2(\cos 30^\circ + i \sin 30^\circ) \) and \( z_2 = 3(\cos 45^\circ + i \sin 45^\circ) \). Find \( z_1 z_2 \) and interpret geometrically.
▶️ Answer / Explanation
Step 1: Multiply moduli: \( r_1 r_2 = 2 \times 3 = 6 \)
Step 2: Add arguments: \( \theta_1 + \theta_2 = 30^\circ + 45^\circ = 75^\circ \)
Step 3: Write the result:
\( z_1 z_2 = 6[\cos 75^\circ + i \sin 75^\circ] \)
Geometrically: The product is a vector of modulus 6 obtained by rotating \( z_1 \) through an additional \( 45^\circ \) (the argument of \( z_2 \)).
Example
If \( z_1 = 4(\cos 60^\circ + i \sin 60^\circ) \) and \( z_2 = \dfrac{1}{2}(\cos 120^\circ + i \sin 120^\circ) \), find \( z_1 z_2 \) and describe its effect geometrically.
▶️ Answer / Explanation
Step 1: Multiply moduli: \( r_1 r_2 = 4 \times \dfrac{1}{2} = 2 \)
Step 2: Add arguments: \( \theta_1 + \theta_2 = 60^\circ + 120^\circ = 180^\circ \)
Step 3: Write result:
\( z_1 z_2 = 2[\cos 180^\circ + i \sin 180^\circ] = 2(-1 + i0) = -2 \)
Geometrical Meaning: The product represents a vector of length 2 rotated by 180°, that is, the result is a point on the negative real axis (a rotation of half-turn about the origin).
Multiplication and Division in Polar/Exponential Form (De Moivre’s Theorem)
When complex numbers are written in polar form \( z = r(\cos \theta + i \sin \theta) \) or exponential form \( z = r e^{i\theta} \), multiplication and division become simpler.
1. Multiplication of Two Complex Numbers
If \( z_1 = r_1(\cos \theta_1 + i \sin \theta_1) \) and \( z_2 = r_2(\cos \theta_2 + i \sin \theta_2) \), then
\( z_1 z_2 = r_1 r_2 [\cos(\theta_1 + \theta_2) + i \sin(\theta_1 + \theta_2)] \)
In exponential form:
\( z_1 z_2 = (r_1 e^{i\theta_1})(r_2 e^{i\theta_2}) = r_1 r_2 e^{i(\theta_1 + \theta_2)} \)
Interpretation: Multiply the moduli and add the arguments.
2. Division of Two Complex Numbers
If \( z_1 = r_1(\cos \theta_1 + i \sin \theta_1) \) and \( z_2 = r_2(\cos \theta_2 + i \sin \theta_2) \), then
\( \dfrac{z_1}{z_2} = \dfrac{r_1}{r_2} [\cos(\theta_1 – \theta_2) + i \sin(\theta_1 – \theta_2)] \)
In exponential form:
\( \dfrac{z_1}{z_2} = \dfrac{r_1}{r_2} e^{i(\theta_1 – \theta_2)} \)
Interpretation: Divide the moduli and subtract the arguments.
3. De Moivre’s Theorem
For any real number \( \theta \) and integer \( n \):
\( (\cos \theta + i \sin \theta)^n = \cos(n\theta) + i \sin(n\theta) \)
Exponential form:
\( (e^{i\theta})^n = e^{in\theta} \)
Uses: To find powers and roots of complex numbers easily in trigonometric or exponential form.
Example
Find the product of \( z_1 = 2(\cos 30^\circ + i \sin 30^\circ) \) and \( z_2 = 3(\cos 45^\circ + i \sin 45^\circ) \).
▶️ Answer / Explanation
Step 1: Multiply the moduli: \( r = 2 \times 3 = 6 \)
Step 2: Add the arguments: \( \theta = 30^\circ + 45^\circ = 75^\circ \)
Step 3: Write result in polar form: \( z = 6(\cos 75^\circ + i \sin 75^\circ) \)
Example
Divide \( z_1 = 8(\cos 60^\circ + i \sin 60^\circ) \) by \( z_2 = 4(\cos 30^\circ + i \sin 30^\circ) \).
▶️ Answer / Explanation
Step 1: Divide moduli: \( \dfrac{8}{4} = 2 \)
Step 2: Subtract arguments: \( 60^\circ – 30^\circ = 30^\circ \)
Step 3: Result: \( \dfrac{z_1}{z_2} = 2(\cos 30^\circ + i \sin 30^\circ) \)
Example
Using De Moivre’s theorem, find \( (1 + i)^6 \).
▶️ Answer / Explanation
Step 1: Express \( 1 + i \) in polar form:
\( r = \sqrt{1^2 + 1^2} = \sqrt{2}, \quad \theta = 45^\circ = \dfrac{\pi}{4} \)
\( 1 + i = \sqrt{2}(\cos \dfrac{\pi}{4} + i \sin \dfrac{\pi}{4}) \)
Step 2: Apply De Moivre’s theorem:
\( (1 + i)^6 = (\sqrt{2})^6 [\cos(6 \times \dfrac{\pi}{4}) + i \sin(6 \times \dfrac{\pi}{4})] \)
\( = 8 [\cos(\dfrac{3\pi}{2}) + i \sin(\dfrac{3\pi}{2})] \)
Step 3: Simplify:
\( \cos(\dfrac{3\pi}{2}) = 0, \, \sin(\dfrac{3\pi}{2}) = -1 \)
\( \Rightarrow (1 + i)^6 = 8(0 – i) = -8i \)
Finding Powers and Roots of Complex Numbers (De Moivre’s Theorem) & Square Root of a Complex Number
1. Powers of a Complex Number
If \( z = r(\cos \theta + i \sin \theta) \), then by De Moivre’s theorem:
\( z^n = [r(\cos \theta + i \sin \theta)]^n = r^n [\cos(n\theta) + i \sin(n\theta)] \)
Steps:
- Express the complex number in polar form \( r(\cos \theta + i \sin \theta) \).
- Raise the modulus to the nth power.
- Multiply the argument by \( n \).
2. nth Roots of a Complex Number
If \( z = r(\cos \theta + i \sin \theta) \), then the nth roots are given by:
\( z_k = \sqrt[n]{r} \left[\cos\left(\dfrac{\theta + 2k\pi}{n}\right) + i \sin\left(\dfrac{\theta + 2k\pi}{n}\right)\right], \quad k = 0, 1, 2, …, n-1 \)
Meaning: There are \( n \) distinct roots of a complex number, equally spaced on a circle of radius \( \sqrt[n]{r} \).
3. Square Root of a Complex Number
The square root of a complex number is a special case where \( n = 2 \):
\( z = r(\cos \theta + i \sin \theta) \Rightarrow \sqrt{z} = \sqrt{r}\left[\cos\left(\dfrac{\theta + 2k\pi}{2}\right) + i \sin\left(\dfrac{\theta + 2k\pi}{2}\right)\right], \; k = 0, 1 \)
Thus, every complex number has two square roots.
Example
Find \( (1 + i)^5 \).
▶️ Answer / Explanation
Step 1: Convert \( 1 + i \) to polar form:
\( r = \sqrt{1^2 + 1^2} = \sqrt{2}, \quad \theta = 45^\circ = \dfrac{\pi}{4} \)
Step 2: Apply De Moivre’s theorem:
\( (1 + i)^5 = (\sqrt{2})^5 [\cos(5 \times \dfrac{\pi}{4}) + i \sin(5 \times \dfrac{\pi}{4})] \)
\( = 4\sqrt{2}[\cos(\dfrac{5\pi}{4}) + i \sin(\dfrac{5\pi}{4})] \)
Step 3: Simplify:
\( \cos(\dfrac{5\pi}{4}) = -\dfrac{1}{\sqrt{2}}, \; \sin(\dfrac{5\pi}{4}) = -\dfrac{1}{\sqrt{2}} \)
\( \Rightarrow (1 + i)^5 = 4\sqrt{2}(-\dfrac{1}{\sqrt{2}} – i\dfrac{1}{\sqrt{2}}) = -4 – 4i \)
Example
Find the cube roots of unity, i.e., roots of \( z^3 = 1 \).
▶️ Answer / Explanation
Step 1: Write \( 1 \) in polar form:
\( 1 = 1(\cos 0 + i \sin 0) \)
Step 2: Use formula for cube roots:
\( z_k = \sqrt[3]{1}[\cos(\dfrac{0 + 2k\pi}{3}) + i \sin(\dfrac{0 + 2k\pi}{3})], \; k = 0, 1, 2 \)
\( z_0 = 1, \quad z_1 = \cos\dfrac{2\pi}{3} + i \sin\dfrac{2\pi}{3}, \quad z_2 = \cos\dfrac{4\pi}{3} + i \sin\dfrac{4\pi}{3} \)
Hence, the cube roots of unity are:
\( 1, \; -\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2}, \; -\dfrac{1}{2} – i\dfrac{\sqrt{3}}{2} \)
Example
Find the square roots of \( -3 + 4i \).
▶️ Answer / Explanation
Step 1: Find \( r \) and \( \theta \):
\( r = \sqrt{(-3)^2 + 4^2} = 5 \)
\( \tan \theta = \dfrac{4}{-3} \Rightarrow \theta = 180^\circ – \tan^{-1}\left(\dfrac{4}{3}\right) = 180^\circ – 53.13^\circ = 126.87^\circ \)
Step 2: Use formula for square roots:
\( \sqrt{z} = \sqrt{5}\left[\cos\left(\dfrac{126.87^\circ + 2k\pi}{2}\right) + i \sin\left(\dfrac{126.87^\circ + 2k\pi}{2}\right)\right], \; k = 0, 1 \)
For k = 0: \( \theta_1 = 63.43^\circ \)
\( z_1 = \sqrt{5}(\cos 63.43^\circ + i \sin 63.43^\circ) = 2 + i \)
For k = 1: \( \theta_2 = 63.43^\circ + 180^\circ = 243.43^\circ \)
\( z_2 = \sqrt{5}(\cos 243.43^\circ + i \sin 243.43^\circ) = -2 – i \)
Hence, the square roots of \( -3 + 4i \) are \( 2 + i \) and \( -2 – i \).