Edexcel IAL - Further Pure Mathematics 1- 1.2 Operations on Complex Numbers- Study notes - New syllabus
Edexcel IAL – Further Pure Mathematics 1- 1.2 Operations on Complex Numbers -Study notes- New syllabus
Edexcel IAL – Further Pure Mathematics 1- 1.2 Operations on Complex Numbers -Study notes -Edexcel A level Maths- per latest Syllabus.
Key Concepts:
- 1.2 Operations on Complex Numbers
Sum of Complex Numbers
The sum of two complex numbers is found by adding their real parts and imaginary parts separately.
Let:
\( z_1 = a + ib \), \( z_2 = c + id \)
Then the sum is:
\( z_1 + z_2 = (a + c) + i(b + d) \)
Geometric Interpretation
On an Argand diagram, the sum of two complex numbers corresponds to vector addition.
Each complex number is represented by a point \( (a,b) \). The sum is the point obtained by adding the coordinates.
Important Properties
| Property | Statement |
| Commutative | \( z_1 + z_2 = z_2 + z_1 \) |
| Associative | \( (z_1 + z_2) + z_3 = z_1 + (z_2 + z_3) \) |
| Additive identity | \( z + 0 = z \), where \( 0 = 0 + 0i \) |
Example
Find \( (3 + 2i) + (1 – 5i) \).
▶️ Answer / Explanation
Add real parts and imaginary parts:
\( (3 + 1) + i(2 – 5) \)
\( = 4 – 3i \)
Example
Given \( z_1 = 4 – 3i \) and \( z_2 = -2 + 7i \), find \( z_1 + z_2 \).
▶️ Answer / Explanation
\( z_1 + z_2 = (4 – 2) + i(-3 + 7) \)
\( = 2 + 4i \)
Example
If \( z_1 = x + 2i \) and \( z_2 = 3 – yi \), and \( z_1 + z_2 = 5 + i \), find \( x \) and \( y \).
▶️ Answer / Explanation
Write the sum:
\( (x + 2i) + (3 – yi) = (x + 3) + i(2 – y) \)
Compare with \( 5 + i \):
\( x + 3 = 5 \Rightarrow x = 2 \)
\( 2 – y = 1 \Rightarrow y = 1 \)
Product of Complex Numbers
The product of two complex numbers is found by multiplying them algebraically using the distributive law.
Let:
\( z_1 = a + ib,\quad z_2 = c + id \)
Then:
\( z_1 z_2 = (a + ib)(c + id) \)
Expand:
\( = ac + aid + ibc + i^2 bd \)
Since \( i^2 = -1 \):
\( z_1 z_2 = (ac – bd) + i(ad + bc) \)
Important Result for Modulus
The modulus of a product satisfies:
\( |z_1 z_2| = |z_1||z_2| \)
This means that multiplying two complex numbers multiplies their distances from the origin.
Example
Find \( (2 + 3i)(1 – i) \).
▶️ Answer / Explanation
\( (2 + 3i)(1 – i) = 2 – 2i + 3i – 3i^2 \)
Since \( i^2 = -1 \):
\( = 2 + i + 3 \)
\( = 5 + i \)
Example
Find \( (4 – 2i)(3 + i) \).
▶️ Answer / Explanation
\( (4 – 2i)(3 + i) = 12 + 4i – 6i – 2i^2 \)
\( = 12 – 2i + 2 \)
\( = 14 – 2i \)
Example
Given \( z_1 = 1 + 2i \) and \( z_2 = 3 – 4i \), verify that \( |z_1 z_2| = |z_1||z_2| \).
▶️ Answer / Explanation
First find \( z_1 z_2 \):
\( (1 + 2i)(3 – 4i) = 3 – 4i + 6i – 8i^2 \)
\( = 3 + 2i + 8 = 11 + 2i \)
\( |z_1 z_2| = \sqrt{11^2 + 2^2} = \sqrt{121 + 4} = \sqrt{125} \)
Now find individual moduli:
\( |z_1| = \sqrt{1^2 + 2^2} = \sqrt{5} \)
\( |z_2| = \sqrt{3^2 + 4^2} = 5 \)
\( |z_1||z_2| = 5\sqrt{5} = \sqrt{125} \)
So:
\( |z_1 z_2| = |z_1||z_2| \)
Quotient of Complex Numbers
The quotient of two complex numbers is found by removing the imaginary number from the denominator. This is done by multiplying the numerator and denominator by the conjugate of the denominator.
Let:
\( z_1 = a + ib,\quad z_2 = c + id \), where \( z_2 \neq 0 \)
Then:
\( \dfrac{z_1}{z_2} = \dfrac{a + ib}{c + id} \)
Multiply top and bottom by the conjugate \( c – id \):
\( \dfrac{a + ib}{c + id} = \dfrac{(a + ib)(c – id)}{(c + id)(c – id)} \)
The denominator becomes:
\( (c + id)(c – id) = c^2 + d^2 \)
which is a real number.
Final Formula
\( \dfrac{a + ib}{c + id} = \dfrac{(ac + bd) + i(bc – ad)}{c^2 + d^2} \)
Example
Find \( \dfrac{3 + i}{1 – i} \).
▶️ Answer / Explanation
Multiply top and bottom by the conjugate \( 1 + i \):
\( \dfrac{3 + i}{1 – i} = \dfrac{(3 + i)(1 + i)}{(1 – i)(1 + i)} \)
Numerator:
\( (3 + i)(1 + i) = 3 + 3i + i + i^2 = 2 + 4i \)
Denominator:
\( (1 – i)(1 + i) = 1 + 1 = 2 \)
\( \dfrac{2 + 4i}{2} = 1 + 2i \)
Example
Find \( \dfrac{4 – 2i}{3 + i} \).
▶️ Answer / Explanation
Multiply by the conjugate \( 3 – i \):
\( \dfrac{4 – 2i}{3 + i} = \dfrac{(4 – 2i)(3 – i)}{(3 + i)(3 – i)} \)
Numerator:
\( 12 – 4i – 6i + 2i^2 = 10 – 10i \)
Denominator:
\( 9 + 1 = 10 \)
\( \dfrac{10 – 10i}{10} = 1 – i \)
Example
Find \( \dfrac{1 + 2i}{2 – 3i} \) in the form \( a + ib \).
▶️ Answer / Explanation
Multiply by the conjugate \( 2 + 3i \):
\( \dfrac{1 + 2i}{2 – 3i} = \dfrac{(1 + 2i)(2 + 3i)}{(2 – 3i)(2 + 3i)} \)
Numerator:
\( 2 + 3i + 4i + 6i^2 = -4 + 7i \)
Denominator:
\( 4 + 9 = 13 \)
\( \dfrac{-4 + 7i}{13} = -\dfrac{4}{13} + \dfrac{7}{13}i \)